ࡱ> e`abcdn,mأUTPqv[-PNG  IHDRX n+PLTE333---@@3f-be3f )B\ H9kR{ZL+`9* HMylsZ#[DLDsR}be|RmLxM#Aql8j']SAcDV {fE2| N#[`Y~A~iV)|S~0,rrtcN[Hvk{>E_=暋ѐQXҥUTМ5uGus'o:enK[7Awx*Q3l#n?o=IK,j-zcpvlooo\ KvlO󗶁(s3iK{ql\dt+!l]@} 8PKqig?~D՟LBdT~)5AZ%r:fXdt:Lټy|LԚ+_ 7=YEH9˵ y [ϝ"Ƽ]p]2)v3ךΦhY7Z]5 p.|3nN^ڢ\gSMso6l-k ;}LU3%]P|.- _%̈́"ɏb+5ֹYlk&Pe.?5lrs`K=#;Ptq V1Lb`=6[6w퀟35e)y`:D.Ϩ#JY`"|hv6g]|c;1F{Bs?L=#dZJ 2%km>~?2 ܀[StkBwr PVkܼwUTF%_-WK]iK]_疫j`)Η?OXkd:Ը/-jc՟nt.`]=՛㾺G[f>Kf{o`csgb_zЍ4Zk+iS `5徦]~6k`5T2ڲW{fyk[&YD߻:R]l"ױ7+U{ I+m4xamaXi k,tֿʋ9t5{3B"0#ؒ֯bʹh.4pvlX]Op lQseQt?qgW<P8Rb`/ vc#rH7ޗ{0velFQ 2uoVcۆ깛atۗʨXÎA&aްGjM^SD )3Fn[C|b [i -W5ZXc H/ P;nU`h 7Ŗ,{LF=[e\aOu8V mkڰB>ؽctlKú9Vtxzwǭ {lap,|0, dqLaK"Ꮠ@l®>,mLa*/, {rcE)lv+^7Ua>8{/!PրZ=l e;|wWL&`j؉G&",?,8=v4ςܰ^Ѭ V&2%í_`L. ,uow;S>jܰ(; `#øTubǑn#M4wZX͝} CfUu,KÂnjml.ܗbs `a[ }3[A'&{Xخڏa]q7vtDV-&ÂcW*L0uI"Kr8Z48؊I-YaaǵZV>,2 v|w>7k~`_ -,ƭܰթ#X![懽>Fφgtͯ懭x7 F6x:ZuL^<,2 0ޢ [|\)ÎR-!Sê ҅1*ޟ33J2,8~$ ct~}aa2aPz1F\v?İaj`ǹcOq¾rX ,J:gP1®J=,="Ybt(j}0wz`%)ѿAgJهBz`'S:p*eFX~XZEmh/^+evɳ t@#''7_eI'h~.'xwaؑva.+W'xu::QenuVH}{ޜTz;[~\dyX vS6}rhUK6_ l"Uco&_ JȰ`avHvn(kiu{İ XM 5 46"l/ru 4NEEd@@H3lOw΋i ^7MaVl#P kDN+5-,ل*v+>4VwZa;6$C%ky { ɡ ;1ksaQc+Z 46ԺqM:/ha aF-\sBawsgaCؕ+f~'ΘF`ݵlQuZ3椑6F4nMifF'9cNi^쥻PY1'NmXJ:cNi{+C+n"D\eDh͉af]jk`l4٫ [V v!bٮ@{[9s ~?n6ݮ H,m*)Ö g>|R ;-^\zR ;Z}A!)FFȗ-Xaݞ6n¥f<(CRBE6[,ua\)zːa{&K"l{whg `Sv3'Xڔ`X䰍x &(vX:rK ǽ{K 6]6l[{hMM 'Xdg,mɰE,mɰ :Ewdk%,mɰ7KXڒ`Kjw,mIc蝵K[#7b $ I&,mR؊m,mR!t@z. M {2dM ˓~Ia/W:`i=t,&d慤KѷOI`o<'eKgGɓ61-_ɓ61KNSaoYvaMnd6wKWgXD])&?7dM * " /_p1- %`?>0`iKނ>0`iK҉b7+]`iQi20d-콊Xⰷת4`ig6*^Y `i~xx-{c>K[ 3+_Z `i^ `CWEXڢK'K[]K]o/U %hť6v rE_}!U66@O!uoۨX8!t@ˆK;OjKj.mOm2F`!,mk&~hyK6 `i#甪NXڲ|\ue-x T e-ƵZu2ҖO~K[eow\Ojom'T,}Ywl }  }莿]N1R}'apR'{NS0Rͱ(u-GvQep-IENDB`nT _ewPNG  IHDR;~n^sBIT&C`PLTEUf""U"f"DDff"ww3fffDDUUww""33U"fDDUUffwẅ̈ݙ*F tRNS\\bKGDHIDATxv0U/`P1wKKN8 BcE3R ;sOp7O.XA9?e8c[ mUB'j`"K,5 |-U3ByG={81YYWbZEhCu,gg(Ќ#cus$ c _5hi @ qc<o荏l.,(>])o*R РKh+UWV luNˑrpP bYBägz7zǀy!1o0w6̓%96SE]cC~iy:LY~H*cauK2LӒ̷X'7W؟˴LZ7N>Zs>9U{UKue'zHWUe^W\Pfil8.v8քjL,ӡd;WIJzN:kиJ_]+,,mPb`Yw*YR>tnZ~Q@m6pO*X+vDtgIu8o %;t/ukʑֽoGXuDƿ #RE]30Eg) [86}Y̝q*GM68}se!D:1Xׅ#3*GLMe-R`3wݭWœO]+CunǞ횕tvgu~Ou' >m4"JFIFKKMSO Palette C   ")$+*($''-2@7-0=0''8L9=CEHIH+6OUNFT@GHEC !!E.'.EEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEE"!!1AQqa!1 ?F3]'IƓe5,J$3?O 5LXyYL4tO%fxM\DXL`: a2HS=rtd+f\UtV5m7L?cf7 `Zx5\knw/$q i ## 1PNG  IHDRTPLTE333---̛ ee22eee 22ee!! lll{{ZZLLLll999++ճ!!**KKllBBBYYY::AAJJzzzzzZZ99ŽAA!!!SSSSStttSStttt''I!!c~##\F((s ++: Qbb))> Ӣ,,292,,|-`^""f%%R $~ kiKKK##)kk[[CC^^ooBBB44]]//WWW{{TTTkkk꽽{{{```yy'000NKqquuu]]{nn<]llrrrrrnnnބéLM©B7UKc%@>S=L{+tRNS@f pHYsod]IDATxgCÐDzGRTl;*콭s{{LH 5$adaٙw)$ >w#Oߍ<}7w#Oߍ<}7w#Oߍ<}7w#Oߍ<}7w#Oߍ<}7w#.|LR"` k]l!G-/>i]\!?.7+ZWH3ˀ+$qp!#3DSZSЦօIr^׺x;պx> "5u)=#bѺX]SԺXBK(j] ,!A\MZG~NSo߁F}$ F bӺ a3E "vQ!V.#G  [ư/c&A TLTK#N-A!%/k ټԘ 9<¿OwNc0c%'n WN(:$ݎ̪D`Zw>|F?SQ%vXG2*1g*Bt|F?驮?K=PyG Z3(ݖz 5QC4`)tFK 祉CՁuXUꢛB876*-lJyrN@n^--7؇Q9תG_'!-@q.M r\U0A&rΜZ'~FoPYV[[Uݨ;~PBU+FWtS>;j}1Se"&Mn57*9GU%1]iHg {ܽcf 7#ǽOqvШwanxnnS2Z(r{0{^]T.ɓ9⻱n,`[AcbAn/<%{miYLM#6 tpCvUMt@S;q9:ݼrC^RUWt`7 7ZZinN@7z[] F2t&Ėl}'[cUqQݸ%A[Zq:nPufkWz֌XibmKK# *Z&\nµ["ZI'0(/m1oFJE 7Eo/`M' 7I"W1 7BcLjF>n2ˎ'/@NWKm܈}آ2[G]V@n iY.u@n̓$EG蠉A0691\ Ұ! >`4PLF_oO_DWqxO\wa5N+~߿8X^2 MsnfCv]`b*f܄eW8&s`]9Pd'4Pf]&$Ɠ2xA7۶/dDIxOMN \Ս+EM?m&U;Ac*ne 8|c"qjːúmatEϺ¤ 0g^MB&{HaaI9}mÀ\S51|^22!3/ sn DQs3[XtR2Fj g-{ō}*΍XQlQ" FbRYa Io\2U7Θhf}ͯNQr͐x8׆`! u[M&:57!'ꦀhԠ}y͆R `nuY@kU憘mO9Lƌ<7Fn@rX\9?NFsCxT[㻹+ ȵ},YNvq=>nΜHW(MHGg6u{WDlO+b]IjXO&3m!t-JءqBM8qḉG RunjцH_bh.r.~亾e=l.!FݔX[*0 ^b7?ϯnl@_}J?铂g}<0dG& o #Oَ$d JK&MG^Zvܔ@9?CN7A²M6o)@IWMf6JW٩~*X ˬ2MIA9Ytѩ,ovb;  K$(%S@!8\MdSԳ Gz*٤7jnND39;N034ܔ}og{s8N z0V嚛@r$b[S̪O 7R{'@f`ĪTݜ =>HTAWMYn;QL,S+$h^U7R!U=mk,@GC7ZȄ ؎ZŶڤQMpnv f'(m :sn^_*bcWryf[Uu=ݨ܁qnH>lnj񹣽d44NBTƺ9v{.6N"(m S Aus-9<S1בHѱ.$#259f~0tX,76mGn{Nl-Uz0XסЊزXR !se킲6 HBg)iydԁ8^*nZ UL|̍Oݑ%#)$ ))9%|a*`g ;bnE7 o+$i).Y7Y&l1%Qo;SءfwzMMttwϦ 79ه9b%ӱX:a0nj7ܡ˚&q3!4#^N+23nN޴{7Pa4iٽ3{*ȍn~~.R mmp}<~6An}~J1tCQG#>}On NT>/iv> :ۑxÖ t8fn /Y+t9bh!gz~tC,/5} ݴƠ< Y,r(5bh;1G0V}٘߃!(uncim1(Ksy1LHi{Ayio*{, r13h77WY**11b TQURˈ7E&Qdt~6uK*b{n:0by0;QQKK{lHZB1jLܫPcffx'Uj::4p(1WW4Ate| )[rddwv.9kh~Vm8;6nIZEnդAAak|^v!-mAP hOlKf߅Ң//j'!!#),<Kakr_\~ax6pٷ18,֣ʬ||p67ر^dV^M!3"Nm2UfѓKAn5!dV^Ma^esw9w {}P^_8@&%׿K~Y<s}\?n&΢AI_t+gVUV7vU52+2F#3.=`$_d Ӧߪ(d%3m"3^]_봓g3d 35̐df2V>f\5 ^̝=^ݫ8+Һn1R+xfΞ0vGޚfuwrs:{=ySQȬrycZӺ_DfR>z2'wQf=$2Ӕ,]NNj{^y'Y'c9M ԨK˪'SӢgQ{zev\w3MiABp7+uxvǿ=4fv{Qyl)K߷|w*Df1QB.!}Jvz}6'@f11z,2sd 3].yqp'qfٺ2szRd!3XOxd6UOxd6 h}̊x ]GqcHΉsd%3Nm 2zm 2 sDfb 3UOUM \B`7d|JK8eF9j9"3G1;Gd(fdh 9Y9c/?gaB?\q41xjTkv8+QnQ5fɑȌf5V*dy5/Vo l^='c11## Y\zb-ʞ_^Ec=Ԅohm3ϐH?qSH;FvEX<< Ȍe taېX |$pzOO!3[j[~5y'چlac3[7l6dVNfYuo湫_|1xΑ$2fȬR>6*Y,42s3)>*lk=xs}g*`72܏̜l26d#K:xdFy?d%ǒlSuv>>NhG0θ9 9re*%d fG5ql@\~ȌSfD~ 9'ק;co\=b0qf&Y'[ 2svi{!3;ic&=Y5v3ocFYuɞ35&mK!J1ڔcn$qqU.iA/wCdf?Ln+4v Sfh>^xZ-0SRO8Kw㬢}#;VkBfcvwMVJ.2ǡgF̪2YKdD"OCf01noGf62+^bٺ>ދ^b,A}lZ~Ȍsf;}>%̒9d(fd2Xl\@f62+=ק?Wr}3*ïk|0*7WNzEW3!2;rˉ@ pXW3!2xG pfgBfev9Áe]͂τ*Mu5 >2[,3Kr~@r}0g8oj_̼RO*l!3G1#35jU+Gf60&ח+b_v>Þ18o {Dxks:̒됙nn,N2#놓>f- Ǭ "w 3ۘ5fE./ĊfFDf1[-ˌ2 5y9HC"lHp V)c 2+Y\[.fUrlNz^-⫆z^ؖ3uXqf}H=4eugjnHtDk1̖o3 3ks SX,fJ.2Y\B-<a6]p)64ԘcZ>fދQrPcƏi ދQ3E4֘cV8…Ә a2s q>ۮ~s}Gͻ㌫5Ȍ/f 5Ȍ/fd묱8gAf|15֘cqԞƌc jf1qLMj1ָ/ŌLxd(fIdVM9C?Fdf,EfcFvsȬʕ볗`\8oz#7]%3?̬e+YKʑ)}df)3i-tFc)N3KfFoAfdeQv6Jfv̓,duGZ: Jf>c1Ǹmc͌O>RZnz2R u]Sv,x}j:0֘cXwztXygsO'@8#ƙ<rQ>P3vXcƏQǬ+(ZjEa XFf幦|ڒ<Fe5lRIC?ms!3Owx~Hgu4fs)dVf:4fd͆dfZS{1iⱿ3rߋɘ-}3F(b6'ۧJ6dV``FT2+}K;)gɑi3gdzx9MC^2Y?>fs[,n)3H%~,]yVe*MlzzP!VBmw߰ďks !~`2f8`g0H"uŭOmN3pH2+4A7eYi !F6KecFJfP1KfY i A={fɬ`~$| cFf"ŬwڒU7;k^LvďBc rY-~?.U :wVWfߒ,bf:fe-c\_6;2՟gNi~9YcԌ9qtq~grdnQ~ly"AHm|^&9u\,]A{\+uN멵4?zj#M^=u-#n v`ZK3%`;Y-SP2DSII4yk{2:{goxf6_Ng32*ѧ lc24Fլ#31#S2s3r^~ty?csW5_x\!/(s`9o{3}& 35S flVNd(fdLTJCfcF琙3>s}yo+1wN/͍b9odߞ1LŘu[Vj3r*5mAfj,_=36q+x s=WDgp2smCfx㶹YT`4fAY-NͅjggUB0#zfr ГhPv{!Ե B^{4o_5;v_=u$ ] M?te{6Sh Y~(0,,ڼBTYOK"fu4%ÒIdV67v@(3BBfzq&3/iub葏Y ! /0#[Ƶ>f;qd0fɱܴ/ȬpˌLԥȌWfopP}jf\Phs}~+ό<˪/ 4̸mvj<+GfFUwLlaFC3~LkxaVh2G>`fd">FfY9gO䵎Œ\L)vhRv;!GFuw]=u-Vf [> /j&Δ_<2y6A]I9qROmض& 5Ŭp0@(mm*f}Cu3ϨPcƏf ցgŞMbFdha 1la[b:ƌ, &Ak;4HڜƌBf{1k7ɣJ=KC bI/]3CM܋Y=ibMّ,!锿ƈ d~dY!CCfNb&#3g1ІYD|9+u??hWUɦ'ƌ- %jTG'>1GKJԨHY!30w.2y 3;282pŌ9YuiS[1`Ie4FYu3#Iqƌ=)2+VcØ⤡ƌm)2+Vc-7Ҙ)EfjSVYUU+`o[`j ?|ę;yF&*Efjkd~1#dRYe̖7BHN()q=vA1#$+TYL냮Lo7?:ڜGVᣞ%b8sH=9͛Fę뼚5 dgN6f+Yamn I MΊS,~OZ%'K3oN5 %!s>EX8]>ytV'32p>5W 3K] O~>dGGe\{?9ֱT9qlekxoc(8+)>^IJw!!Q8!hRJBJMIJC hK_PJ*T_R_jUTHT3]|ٵ׻k{[w㷓W</3扬1ff5YGF 3,7hev/3fdޕ=={d]_y̺bݕu;/u]=3;bU涆`hRc)z?"!ćjh 3ȩ<|sWfTcvIQfse Af&5晱uAf&50#x{ߵY4?'tA_ߤƒzF gF̪ 3˘EK"bFV;?"&j,dFEշLj,eF&kן5O_SGIG'm]ϐٹ3[kjIʲ7hPƃ9]1#C{Ǵ56avs *BGcfiqd&f[ٯO^l73#eKب-)a9[$uyIS7}_i^GƬ9kTc=Ӽ}Ãǯպ63&NVX&MlW3'A:r:Z._hKf;K=ŭW>=:: Df5v`FȲa:=vk!jY]&f|j.<Tذ85#̐FYSj.AfbF][u41+oT&z}82)▱0;#+I#+Ȭ^>"}Q}M2[ɬ )/}HWvFfŵ5)a˃]芃~>xj_ǘS 9ZK~,b6Kz6Y5/Z#3 us5_c3X=Htn;ʐ5U/GĽ;o]XX*;(_}Ili'7l?ˈ[^23i|[GEq޿Cf4M`(GIlrlpdfHf/^_!5;̐ ȉÛ^Yv77:L 3d 3df%; IDAT^G_g ^\HV^d lplVLCiiL: lL2ִ][9 wO|? 6R0Z:6VJN2 5VYϴAٟYEY0$/!OBfu4mfV-Lc;R8q\_Kcqp"Q gyأmT"3d 2SK2Vd̬t3\wչ3gSϐ2k% YAfj 23o2Cf 3!3k5L-Afb>t_=CfȬe61IUvxCfZ0;rݍSܟZKcfJ?@G!3{2StvQ.~cx8u zBVjԥM̓6[Td[S?QNfWʌWbe>eFdZA.iuBv)9+gH$fAsI)f.G{f`H^گ $iwQ [TEȷyv3 S'9j/I>n'0 sOREh[S~Aя}j~?oQ;QdwgX]Y\DN;)}>BR2ǥ5d({>K5vt$pTX_3i%d. ~R*W \ &iXƦy$y',)!!03|&87suϧvcMlt }q77軇O;xW:o;rJp)/? 7ߖz&;yIe` @ (ílg'4lM۸GYZY.ped>7ͱyZ_̬Tbo213kS }3 )#3 %! aNjcB̧|"ܕaLB/g- F[uf7yVjRvytEXtOFIENDB`n}/ l]yO,3O'tPNG  IHDRT'.PLTEZf$yI̖KKK888ZZZ+++eee AAA!!!kkkyyy222﫫k甯 SSSttt挌22!!ۥכ徾**ͯ77KKԓ΄̴AA 33ssffPP\\||ᴴZZllެ@Pr}aa``HYьTT<>*2%?Ka}\t88g(*66,70;)3ȋvvot6=vp,v HHw1Scpq魭K[ +7ttR b11[ce}gk22켼``${Zo{0BՇIISSݗt㚚nD]2㊊sTj2>XXqqeeܚ~zd~!-XXյ 11))V CCHggM__}pDDAOnnġ,,֦ii{ztRNS@fbKGDH cmPPJCmp0712Hs+IDATx^}Ǚf71v40 7fɉ3"cR9$EzL=,K',[^˲di+NuZ=}9˷s{绫t7 =U~3CF(]*.@hRFOYN{ؘ80vد˼g6L09= oFCo֣!Հ٘~]=Kf9aVglط|\d@0`*kz*Z~q S",SI @%?{Xf=6/y;e޳dk&SmuƆ}[̷u[wn 1U} )HkA@]zPTUF _5-m9KCm 7 uTdS!BLeIAuyA!KAdܟ^E{YˈS549@ O{/ :t*kml4GmKpAwЩxrxlSAN :t*ff$1Uc*H  )4Ux5 e PG@:uԙ9=pbM鲲3NX(Y F?`}yؘ80vد˼g6L09= o붦Q!qm<6pЩSAN :t*v̀bRT,d) K`U#@RI$H  )@RhQ,1U=I$H 6OM xl )@R x`*tC'ybFZ?OTP?J>uGv1"#PG?{SQ.Krb_)xBaً:Hј"i`*JzPTՁ+;f›+LHvP~")0m,u&`6Щ\o̷K1"S5Ȥ1VX׾*kc'O R u٤S/(z~Z}(@@قA5 JV@PT-~60 L9sRʾv@5 L9BL/ ST럺?l`CY2b*x1U=|u#PGs!nΏVXQ hmj^V@R xl )@R8l xl )@R1IaǵYLPa'}PT$8M4`*0 LL :VЩS!R_#PG@ull`3H$ PT=?ݮtEL}J`XaVX a9 xxlP???>pR@A,:u+D ZDE|*c,f[IsCR _RX>]-yM^1"8F!lSdR'TT`*\)d~u0<6@mmӀxXLUg*H <<6RCBR@L/ ct1Y R@L/ NiHғ3_0,ac*55i4ITB9ꥬ d>jUFIҋ&fnV}ƾ7$XHWm<<6XuxL%-P%ℤ|>P{0qSO*&( T}b6]9HrA@J+cc=|m'ḥ/.2`6_yϒm؍36?|\s*G狓:& K.g#wX˵l'=E"tNȗ2EqBLP J7 dfJINL*Ӧw&ؙs% M `'EPUBa[*A1Uh>P7U )xÇlTܸZz;=xX5$1o F i,RW_d{`*0U[LEM*@P *uW>l(cF\ڧT`*0c*SS4S`*U}M; bi5 [*) O?DD9wQs e>XLuhv1U3YcFn9mz *4W@g^Cn R环TU_PyH@PWC 񳍐!)S} Y zɀ,nr LS9{)TշkPr!)8ASգ GPTA>WՓCQ#`ڥj@%t~ VT0Q͐{ @阾 h~S!EW4؛ͷ@s~$ڃ~c5ļJUn2ݎ))`[5ԭ눩 &+EN~k S`˃Tu1ʕdDƟsM^HH|1LNTfnxl ~OFB懯^ ?jkY秫_66yw}o}tM7swZ0Z{uRoj?ivUgxbTnɁ9s{wT+;Du/@ `T%~ؑ{nxĠ@?y]w:S*yJ r9v\@iS-[p=y\m0ݟ~fg噀=T*h]mv{tSF& z|"G"a7|:9r4^)T8'H-9H C*ɨRu T$x:U7ݖMfQOmw}a0I *9x[uX{#_ {7J=< $mǏ#xJ; *v PS.kZUX/p꡹.?@TB6 *׃%q-)nON5:Uei͚mR?R# `&E֞q|,b#r?>-v*% o(헨sAO&mtyЩ&p5h \LzLz| e~9m3R<Ȗ)x/ONfݟ2UʕsQ@l}(߁bzq0.@PTUZP5fuV>%^0qe((;~$o?ACR`7d2<:eI[$: LeL*U eO(TjWS_:*" T4G_TwLq*LpW^똚gY5IqO{چ=g$PM.cHOp׃tz)\gSg)t/YeX,RCJL%ʧSF>d)k4fÄYEVE),=gd75 ^Y}&Tw,0 d9\.zd{z~M1JI2LLS$PgTP}2/;> Ǖؓ?TIAFB݉%1>j[YO*J4P}Æ[\_[>W})z>PBo60sշJjylU p"~:*uv_ #P=? RuWrsIP! Y嫴l/B$~y)ʌ j秭FP d4 uZ-6JSBI uL zlB6 ]l4ge87'XqTmm(wdgdǦS!S T#k` %SII}iՠo=rfX]lI/c*@3aPL>S_*uqg} eY?*]Aʎŵe0U2t!GώJ]9^,`ׇхx6_ 0ɓ~Y|i{ˬ')Z]+T=i6'jh4ċ(#Ҩ9SzH = ڃD癊*4DVpܟGzZ;PVJUoÑT=h(' &H3@Ugk9`8W=q](Y EL%),Pd?ՓzM7 T@5-GtGR0d_{ےp2g΀#"ZMwif{gǾ^_`Vdݙxz[eloj G N޲[ $S)Tb*l^^4-I̓%= x铩 F{*^ ]>2v䞻[+ Rɀ`JHSuTf{WS*I3J3&#j#ߪƒ43"^? 9B?H j*TSp8O;+뚤@=۬`jyޙx0$ZȎ،_mW)L3%21xjW\{ W | * @*&ȗ [JUZ"Ps6(*JFO&Z?ꝏݮ#( NT*Ɖ}*}DV"$PE&2(!+m4FS\8׀Tdb_2Vue2M+G1!)\N~U7&`x`j2DqV'AMyn &Rqel8E zOhW^&Y^7]LPi(*im!{v˝@%IRHtP͋]Le7֫{/x$P-,ƳB]_\lT@ pJgS_H.dܟz߃/~^BJNkI ys~qŠV`˲@]"dTҏjŬnVO0C+?Io4O3 eS,~١@% TLZ q Ib=1IZ$;VT*8lRs; 9$PfBt*NCIuۯ'h! AG2ΖuApzx|nu&I Se*E2EzWslP h JU/TׯY!9Ps8VRk@[׵4'GT),ez"9`"} 6?/Ύxr@Hv ?#x)ҏ}lțe3 A <]y|l[0tFv,y0%)CVGj`2U$o,2p1~}Oȋ0Pa`Z=&U;>LjoNi}rV?EIF*d1wm]EczxlbK;’*lبu#`8P g*TؽW ׁKEf-RR ,]~-+ѣ>AMB2u1||g#zg&*pJ=W?єUHN/4#U3 H̾}ZlP`*^QfJH,m /S`:zI\ho2OӜ`%zhj1(ҏ|`%U⃓6'vB?0tfi8eD_4WտC##;<ڞ 5q45 O:mCJ.*C!ȒdMd*=P_ܫictyQyl0PiL)oj0I-S Fyq _ծ|GiG RV H ir-L)+82cRqOgSiM%4-TҔPJ8/i{z NE ňڝ +S]]>=}Q_ t@PMԝ^װHOkU*|T°?U=|/kkԐ2xr<'8?)%lѠo|Ʃ|w5-q\sm7m|w[ɇp8ѬSy@lX f,|`jImv0$ UTMގV'$jI힁˳T}yvg5g["6jF90U}jfک?VV@PyOm>*!G8Ēq3yM{Wl!&+Ku~ÏPN^L@5`?Lݩ#BLŕNoMn Mg u`ʵݪ{b+@eYxnΒq;X/tb0]xcU[?m%4X[9`*T|1U,m҉-V@ejL%TuV'6:[ PWU0%M{C_}KC*Ť=hՓ{V **UӮ<rTqH u9_Пz߃/~>]ڳ 8=d聤++r%ljky0u?slV嘀9 '_˞(wTAռy~]ӎ>h nPZ|̘;J:*JQxaGo8J(;ES'TW?iI@ P j~$%k5s9VJfHgL>ݴncӺ.=4ms1lt}A}R"PPLes^m߼Dӛwq Og*j"je5 ׮vLiAE oӵn`be/m% U9 T$v.H\Ƿ77/yqeյ3gNo^>_{Bf LU|2;Gk`')ԩ.rkgί-zwvid>?LUJ.c|$%9~ƚ aNH)+c\&CCRX,1{Y+*eR(1I_cQ-<) gؿ=N,O98=NK{ 2@%g,Mu 7w %JRP%fLv@h"a:F~%b&( ?"kJX%>*Q9IT(3Ke")|}a'g%_䌥 5TqT 'DƠ;j4O8 ! %e&f&hYTrRh-c%YS9d/NG|MJzqR{j LGs1FV!1ƍTȼq͏G/Y!>_dMвm 9*)cUHS%KP'c$;N!AA5{2JJ~% %'&itEɝ'ƪ$bXiX!-=O>'281" 83$Og)Y 㴂dP'"EfK1V*(sl:WG̖ #q 'ac0f3 `0f3͘W.BUiʌ*M7!zj>LSBr.PL^ D1Vֺ-g(ih0d*J(!sUB&clZЫlHZ.2ũbDxYߊ…π*|&Τ3i[0(d4g^SYjDmÉx&KR)Ƴxfi ayrFx\)s0*$R M\*IZ #b&ZmyzeizgDB'WKCBaZ ")p!\zRI0;nʒyX)Dꠚ"OS…π TtgJg,\2j* gZ*E^NXRشd67R-b-/ ᙁ2B)7HɅT"Br&4W$8& ĖQod>DٴIENDB`n* ,XߟûR'PNG  IHDRYjPLTEZf$IyࢢdddzzzՓ222lll +++ZZZ::: KKKsssBBBRRRH̔Lכjjͭ22̓77ᴴ徾Ԕffۤզ""LL**̳AAZZޭ}}++ TTЌ aassfϹQQ3@@@''<< (2IZCS|jj쿞haUUii88Rfj!!ss0;cz")``s}^^uk _|w]zz9H|u!ptv qmۀUjs--vn|~jxxzd~OOɖ ??2=2= "%,,j c{0022qk=LVVg!!^stt2=__KK7տllyl44㕕== ViMMšaňiMiww\\EWTT_\~%366$%%nn{㯯ZeSpt./55ۆN`!)EEwYwtRNS@fbKGDH IDATxiuǫ3g{Yx%w=C䒔HnEIuYR"Ye$ǖ[v|Wb8p N} @>AIA9jWWX87կ ].CذukE<ּN}M;n67F\AsЁL:dLPq]'k F NeBl28YHM\' ɲdKOGot{cn67F\ , NeBl28YHM\' ɲd!Y6qq¬¬]\' ɲd!Y6q,$&.dep, Nָdada..dep, NeBl28YHM\' ɲdK泀 M28YHM\' ɲd!Y6qql\'k b$epƍYZUsz)AƼ7 cچƈ24YcEpep, NeBl28YHM\' ɲdKu24Y..dep, NeBl28YHM\'k\0eh0`Bl28YHM\' ɲd!Y6q,$&.5.Yu24Yud!Y6q,$&.dep, Ne,:q,:ep, Ne 5.ɂ28Y<O28Y\AYNK/% cךr,YoQ&iXٳ:\'w5d-YdB"1 q~uJQJ\RF3,(.Շ,YuHLR#Y\' V8ÖVlG N,ru]fdAs,X(3y"'S0M Nds,YA5'3l'Ёjup ]4vq,$&.dep, NeBl28Y㒅Y .CY NeBl28YHM\' ɲd!Y6qq¬¬]\' ɲdKU`#YBcp,qȺ 5K/=%Ys=%Y͏ @u"kKsw߶!){YĬo?< ={$K{ÿYo\HɂC?mLHޘnp7cd!+7Kd|ё4 euqҏ!#Y`"$ +$ )ƈ?VjXuJ+'!#Yɒ Q0RX)M95@5 $Y}I&Z hĴ1VM+wbݧn:S8 ,V7Nj#Y r*r%wbV5m FɂHtpR!)v5LV*8z8HD|*]ck;~ GGV*Lq(O|Ьo7YK0z8HxTEBO]CiV8FCV}Y/ܭ=@n.( zvgg){YbVVñG$K>D,lC{k0p,ɚ: CFŐOAŬYU 9Lc*f5ZV@ñG$nڦ%4 28YƑ.ñG%ñGD%kd]JɂC AVVKj:/;KʽȚ#Y`Ȓ^6Yu$}1oνS[ ;u"nAL dM!kp,;E-I@ɚY&]'kbdSWŝ&]'/ThXS#Y d%W߲@2,,>)ӳ(ς"Dλ\rhؘEC7p1fw}Up,hD&]'kdCf#YgPiep4RN=^qt,ƇW+ԅ,NցuzX[x]&KEN#hEW} xp1fb_&޸xap,r'#'>tWcdN<#-3nSo,={$tRz|߫ek9ap,ɚ:_^k=% Oo~N}u1Eup'=#~%=Ր;:=,cV$9F!z8HL2`T_aŢ0ɢ, $tLIK샟d>=?1KHvAv,[-w2Ř%@b菬7gYAV7,R]L,d]$䪵vBBtYNo6Yg[k"YJHPz)m֢ blJٛOֱܣ-L,dqbV Y)V2+dr9ӄd@|+қ܍^>|Ϳ Y))IV^4=:T73"KnN%&YVŬ`8"k.WHh %|?`EOs2Ea񥉿$%mтLk_ɲ,ȕ&ΟQOoӌdYEV:я dQBb|? ˪.mzc/P"Iwh7k_o8c1ne1Ȣz]}uN*d54LɢtM壕c9^dAx:w C:Y\L{#YwQʊL]P}L YG%KN6w+S:˟{?]cHWCw=f F֡{??YC[YZ_Ϩg0fA{;z˛]"e,G8.x>kטxCM^tbQXPɢt{()؋NLqDE,, 8YSl5}()؋N?Q?YƼʩyP`v:t1s=]B;9Ǭ|@t ێ'kCM^tz7q>)tZݿdYFVHIJ$^AN:o!Zc"E $;J?MKDi`e!%ϮcbEY`kzkV.ۅ@BKZHj[c*l*? ,:uPg+oG3c 5$,_{wʑ)cz?IXðWI}Vo_{s|$CHjWXFwȻuJY-n]aſ;A<1!y5|ɶZ4VG9"74&xp1딎;Mzk eҮtT2dXh/|@f1qYΨ|< c@E|A%6MI[j7 z+69&$YwԻƀ&KP d3`P]j,J翯:\,9#Iش,;CdpOh1aɢW~t|\T"~ސ3m]n@.JV$H59TCMP. F9#X5Ɂt9TebCЕdMUGr3"Y$&ţɎ|ɢ#rϡ "RCdK~Q,eـ).z݆$SXbH1iГZjod9l5H]5dɏ>I.ut|ex,:]._Xnh֡cq"~o<,_;mZ2HHJ/>`׬&UןΝ? 3@B[/qڏ]l1KO2Ș*_'WWIrfc)Ym̪ edlǟ_Jt|D#K;]dѻ(ݷ_$Gǫq .w%w$xwpBHP1IQQ1ADHP_b ( . >/VU1w폪wU=u;]-Yq!=:lrS{;m/꾖o9]? TZyȶJ2!Y;O}[.gn-z`q!kܿj HlϬhHS5fn_GlgG?u׻qh{BV?c)=jLƟ kw]J\RdIS:IN?KHQ]$Yku\Xv{3˘T{,g5Tg̏_aT;b,ػm/x\CWn[5 39$D_^lqiuG,bUu8VD){2'9$k=A:ߡo_W*{&5ck˛_ÖסS2a,!YI~*zub8um{V;yU}HVgNzݽbyeK|p:~:WǾ/><=ڧoR^H>l[:u%_գt}KZ,o1 :@]xW)`ݻO-./?n%CkC[<{+ݳ:xlOf :tlebG+3lw/./v}_`×Ժ3V/Gd!Y'w/\mC+ˏ\XWd!YR~ /.-^Y0vvϬ[ܱ2.^͸V-͆N?HeЕƮ rH Jފ@"+G(3/'Y]]og5'`jt,k0p-ecֶT77OޠآoWf{LjMM?ŵBF]ՊEضyoSg]&=$D:Lz؛Tg]6 jjN`0(ƸDӸ Kc:TXM`BSaŮLZh0q,Պ]I{cPk,!K&U&U"=M*}{'E⻔&QPc6 O]eCVH=m3-a2\95jsN~{uQþò%:PcocUtEXtOFIENDB`(K    = 8  h  i y      / 0DTimes New RomanttD/ 0DArial BlackmanttD/ 0" DArialBlackmanttD/ 00DTahomalackmanttD/ 0"@DCourier NewmanttD/ 01PDSymbol NewmanttD/ 0@ .  @n?" dd@  @@``  | F+   ^55    J@aa^a^IKIGJGK>hH  ]1F0)2(((()('( '(&A ) # +$F<'F, wTn2RNO v(|RYYIB,    " # + +E*`8888$/%0&1(3'/)2$ !  &B1B % / -4#9387605ST#[ MLRONQPW 0 / [ - $ # " ) ( ' ] V U X Y Z \o$$b$,mأUTPqv[-$$$b$bpe# zt OR$FE-Qڛb>";$b$w/$q i ## 1$$$$$$$$b$̳!&z4,1$b$ l]yO,3O't//^b$Dȱ,=Hk#\* z0e0e    A A 8c8c     ?A)BCD|E|| f333f@d g4`d`d; 02pwpO <4BdBd@w 0t /g4:d:dw 0p% pY<4!d!d@w 0t /ʚ;ʚ;<4dddd@ x 0@r0___PPT10 pp2___PPT9/ 0? %O =TO <O_9`XWYZ:;[8\ijkmlnda ./01 234  567 '&()*+, !^-<=>"? P4   0` ̙33` ` ff3333f` 333MMM` f` f` 3>?" dd@,|?" dd@   " @ ` n?" dd@   @@``PR    @ ` ` p>>L0 VN (    6 P  T Click to edit Master title style! !  0ܳ   RClick to edit Master text styles Second level Third level Fourth level Fifth level!     S  0L ``  X*  0俍 `   Z*  0|ō `   Z*z  bA޽h @ ?Parchment ̙33 Default Design 0 zr@ ( )   0> <7   P*    0(n v 7  R*  d  c $ ?Em    0s  :xy  RClick to edit Master text styles Second level Third level Fourth level Fifth level!     S  6ty <K   P*    6T~ v K  R*  H  0g彖 ? ̙3380___PPT10.@Z 0X(   $+ X X 0 <7   X*  X 0 v 7  Z*  X 6 <K   X*  X 6 v K  Z* H X 0g彖 ? ̙3380___PPT10.D\ 0L0 P* ((    0 .THE RATE EQUATION A guide for A level studentsR/(26f<ff ( 0eT OKNOCKHARDY PUBLISHING(2 fJ ) C "A KTRE: . * 0@$"`>B u2008 SPECIFICATIONS2(2$   B  s *޽h ? ̙33y___PPT10Y+D=' = @B +{  0L0   `  (  I  0op INTRODUCTION This Powerpoint show is one of several produced to help students understand selected topics at AS and A2 level Chemistry. It is based on the requirements of the AQA and OCR specifications but is suitable for other examination boards. Individual students may use the material at home for revision purposes or it may be used for classroom teaching if an interactive white board is available. Accompanying notes on this, and the full range of AS and A2 topics, are available from the KNOCKHARDY SCIENCE WEBSITE at... www.knockhardy.org.uk/sci.htm Navigation is achieved by... either clicking on the grey arrows at the foot of each page or using the left and right arrow keys on the keyboard (2 2(2 2  75,   0 MTHE RATE EQUATION (2$fjB  BD)@0A  6Pp@ 0 0jB @ BD)@X@  6X0@ 0 0dB  <D?    0`  OKNOCKHARDY PUBLISHING(2B  s *޽h ? ̙33  0L0   p(0 J (  jB   BD)@0A   6Pp@ 0 0jB  @ BD)@X@   6X0@ 0 0  08 MTHE RATE EQUATION (2$f  6Lh  ? CONTENTS Collision theory Methods for increasing rate Rate changes during a reaction The rate equation Worked examples Graphical methods for determining rate Half-life Rate determining stepH0}2 (33  <?@ @ 0 0  B? @y 0 0  B?r f@ 0 0  B?   @ 0 0  B?h t @ 0 0  B?pO @i 0 0 $ B3? @h  0 0 ' <?. @ 0 0 ( <?z F@ 0 0B  s *޽h ? ̙33y___PPT10Y+D=' = @B + 0L0  pY(  pD p 0,( Before you start it would be helpful to& know how the energy changes during a chemical reaction know the basic ideas of Kinetic Theory know the importance of catalysts in industrial chemistry<*(2 2*3dB p <D)@0A p 0Pp@ 0 0dB p@ <D)@X@ p 0X0@ 0 0  p 0 MTHE RATE EQUATION (2$fB p s *޽h ? ̙33y___PPT10Y+D=' = @B + 0L0  y(    0Xۋ JCOLLISION THEORY(2fdB  <D)@0A  0Pp@ 0 0dB @ <D)@X@  0X0@ 0 0   s *ً?N C+Collision theory states that... particles must COLLIDE before a reaction can take place not all collisions lead to a reaction reactants must possess at least a minimum amount of energy - ACTIVATION ENERGY plus particles must approach each other in a certain relative way - the STERIC EFFECT !@@HS@@!33F3 ,    H<̙?= JREVISION& (2B  s *޽h ? ̙33y___PPT10Y+D=' = @B +  0L0  d(  d d 0x JCOLLISION THEORY(2fdB d <D)@0A d 0Pp@ 0 0dB d@ <D)@X@ d 0X0@ 0 0L d s *>?N 4Collision theory states that... particles must COLLIDE before a reaction can take place not all collisions lead to a reaction reactants must possess at least a minimum amount of energy - ACTIVATION ENERGY plus particles must approach each other in a certain relative way - the STERIC EFFECT According to collision theory, to increase the rate of reaction you therefore need... more frequent collisions increase particle speed or have more particles present more successful collisions give particles more energy or lower the activation energy!@@HS@@WHW@`#Z@!33F3V33?33>4 dB d <D??  d H$>̙?= JREVISION& (2B d s *޽h ? ̙33y___PPT10Y+D=' = @B + 0L0  *(    0p%> MINCREASING THE RATE(2fdB  <D)@0A  0Pp@ 0 0dB @ <D)@X@  0X0@ 0 0d   B<,>̙o?    INCREASE THE SURFACE AREA OF SOLIDS INCREASE TEMPERATURE SHINE LIGHT ADD A CATALYST INCREASE THE PRESSURE OF ANY GASES INCREASE THE CONCENTRATION OF REACTANTS,p   02>?d MThe following methods may be used to increase the rate of a chemical reactionN(2N   HX6>̙?= JREVISION& (2B  s *޽h ? ̙33 0L0 &(  dB  <D)@0A  0Pp@ 0 0dB @ <D)@X@  0X0@ 0 0  s *=>?p 0\Reactions are fastest at the start and get slower as the reactants concentration drops. In a reaction such as A + 2B   > C the concentrations might change as shownZH@TH[3(    0`K> WRATE CHANGE DURING A REACTION(2f  s *Q>?R  'CReactants (A and B) Concentration decreases with time Product (C) Concentration increases with time the steeper the curve the faster the rate of the reaction reactions start off quickly because of the greater likelihood of collisions reactions slow down with time as there are fewer reactants to collide j@&@@(@)@"@(@ 333333333%3333"=NJC "8  p    R  C "A  abcp p_   s *8k>?h y  HTIME(@    0p>?   Q CONCENTRATION(@     s *u>?   IB,@3f   s *s>?` p  IA,@3   s *X~>?`   EC(@   H>̙?= JREVISION& (2B  s *޽h ? ̙33U 0L0  ( w dB  <D)@0A  0Pp@ 0 0dB @ <D)@X@  0X0@ 0 0o   s *\>?  ?Experimental Investigation the variation in concentration of a reactant or product is followed with time the method depends on the reaction type and the properties of reactants/products e.g. Extracting a sample from the reaction mixture and analysing it by titration. - this is often used if an acid is one of the reactants or products Using a colorimeter or UV / visible spectrophotometer. Measuring the volume of gas evolved. Measuring the change in conductivity. More details of these and other methods can be found in suitable text-books.H@@4@MHKff?    0Р> LMEASURING THE RATE(2fB  s *޽h ? ̙33E  0L0   @[ ( Ow@ dB  <D)@0A  0Pp@ 0 0dB @ <D)@X@  0X0@ 0 0   s *$>?f : PRATE How much concentration changes with time. It is the equivalent of velocity.8Q@3LP    0> LMEASURING THE RATE(2fR   S *A yovxp4G   0ܱ>?b 7y@  6h>?r2)  C CONCENTRATION@8  h    + 7   0>? h   Z gradient = y x(@xB  HDo?: X:   0>? Xy  7x@  0>? ) h  :TIME@  s *,>?!  a the rate of change of concentration is found from the slope (gradient) of the curve the slope at the start of the reaction will give the INITIAL RATE the slope gets less (showing the rate is slowing down) as the reaction proceeds>` 3R   s *$>?# UTHE SLOPE OF THE GRADIENT OF THE CURVE GETS LESS AS THE REACTION SLOWS DOWN WITH TIMEV@V B  s *޽h ? ̙33  0L0 / '   ( w   0xϋ KTHE RATE EQUATION(2fdB  <D)@0A  0Pp@ 0 0dB @ <D)@X@  0X0@ 0 0F   s *x?N) Format links the rate of reaction to the concentration of reactants it can only be found by doing actual experiments it cannot be found by just looking at the equation the equation... A + B   > C + D might have a rate equation like this r = k [A] [B]2 r rate of reaction units of conc. / time usually mol dm-3 s-1 k rate constant units depend on the rate equation [ ] concentration units of mol dm-3 Interpretation The above rate equation tells you that the rate of reaction is... proportional to the concentration of reactant A doubling [A] doubles rate proportional to the square of the concentration of B doubling [B] quadruples (22) rate@HB`#@<e3 Et   dB   <D?- - B  s *޽h ? ̙33C 0L0 RJ (    0D x LORDER OF REACTION (2fdB  <D)@0A  0Pp@ 0 0dB @ <D)@X@  0X0@ 0 0   s *x?  wIndividual order The power to which a concentration is raised in the rate equation Overall order The sum of all the individual orders in the rate equation.@B:33    s *x x?(P IOrder tells you how much the concentration of a reactant affects the rate JHJ3I B  s *޽h ? ̙33y___PPT10Y+D=' = @B + 0L0 `h( w h h 0|&x LORDER OF REACTION (2fdB h <D)@0A h 0Pp@ 0 0dB h@ <D)@X@ h 0X0@ 0 0 h s *p+x? p  &Individual order The power to which a concentration is raised in the rate equation Overall order The sum of all the individual orders in the rate equation. e.g. in the rate equation r = k [A] [B]2 the order with respect to A is 1 1st Order the order with respect to B is 2 2nd Order and the overall order is 3 3rd Order Value(s) need not be whole numbers can be zero if the rate is unaffected by how much of a substance is present @B;-f3f3f33k33  h s *x=x?(P IOrder tells you how much the concentration of a reactant affects the rate JHJ3I B h s *޽h ? ̙33y___PPT10Y+D=' = @B +T  0L0 c [ Pl ( w l l 0Ax LORDER OF REACTION (2fdB l <D)@0A l 0Pp@ 0 0dB l@ <D)@X@ l 0X0@ 0 0 l s *TIx?  Individual order The power to which a concentration is raised in the rate equation Overall order The sum of all the individual orders in the rate equation. e.g. in the rate equation r = k [A] [B]2 the order with respect to A is 1 1st Order the order with respect to B is 2 2nd Order and the overall order is 3 3rd Order Value(s) need not be whole numbers can be zero if the rate is unaffected by how much of a substance is present NOTES The rate equation is derived from experimental evidence not by looking at an equation. Species appearing in the stoichiometric equation sometimes aren t in the rate equation. Substances not in the stoichiometric equation can appear in the rate equation - CATALYSTS@@B;-f3f3f33l33333HV  G  5  l s *bx?(P IOrder tells you how much the concentration of a reactant affects the rate JHJ3I B l s *޽h ? ̙33y___PPT10Y+D=' = @B +X 0L0 g_  ( w   0  KTHE RATE EQUATION(2fdB  <D)@0A  0Pp@ 0 0dB @ <D)@X@  0X0@ 0 0  <̙? p @8  0 06   s *?No  Experimental determination of order Method 1 Plot a concentration / time graph and calculate the rate (gradient) at points on the curve Plot another graph of the rate (y axis) versus the concentration at that point (x axis) If it gives a straight line, the rate is directly proportional to concentration - 1st ORDER. If the plot is a curve then it must have another order. Try plotting rate v. (conc.)2. A straight line would mean 2nd ORDER. This method is based on trial and error. %H@$33 3X  3/  B  s *޽h ? ̙33y___PPT10Y+D=' = @B +  0L0 `d(  ` ` 0Hix KTHE RATE EQUATION(2fdB ` <D)@0A ` 0Pp@ 0 0dB `@ <D)@X@ ` 0X0@ 0 0Q ` s *px?N+ Experimental determination of order Method 1 Plot a concentration / time graph and calculate the rate (gradient) at points on the curve Plot another graph of the rate (y axis) versus the concentration at that point (x axis) If it gives a straight line, the rate is directly proportional to concentration - 1st ORDER. If the plot is a curve then it must have another order. Try plotting rate v. (conc.)2. A straight line would mean 2nd ORDER. This method is based on trial and error. Method 2 - The initial rates method. Do a series of experiments (at the same temperature) at different concentrations of a reactant but keeping all others constant. Plot a series of concentration / time graphs and calculate the initial rate (slope of curve at start) for each reaction. From the results calculate the relationship between concentration and rate and hence deduce the rate equation. To find order directly, logarithmic plots are required. %H@$3gT333 B ` s *޽h ? ̙33y___PPT10Y+D=' = @B +Z  0L0   0  ( Ow@   0x OTHE RATE CONSTANT (k)(2fdB  <D)@0A  0Pp@ 0 0dB @ <D)@X@  0X0@ 0 0q   s *x?v  =Units The units of k depend on the overall order of reaction. e.g. if the rate equation is... rate = k [A]2 the units of k will be dm3 mol-1 sec-1 Divide the rate by as many concentrations as appear in the rate equation. Overall Order 0 1 2 3 units of k mol dm-3 sec-1 sec-1 dm3 mol-1 sec-1 dm6 mol-2 sec-1 example in the rate equation r = k [A] k will have units of sec-1 in the rate equation r = k [A] [B]2 k will have units of dm6 mol-2 sec-1>@37 +    NO  3333333333 > B  s *޽h ? ̙33 0L0  @.<(  < < 0|x ^"RATE EQUATION - SAMPLE CALCULATION #(2#fdB < <D)@0A < 0Pp@ 0 0dB <@ <D)@X@ < 0X0@ 0 0  < s *x?tO   #In an experiment between A and B the initial rate of reaction was found for various starting concentrations of A and B. Calculate... the individual orders for A and B the overall order of reaction the rate equation the value of the rate constant (k) the units of the rate constant:@@33 $   < <x?vP ;0.5(2  < 6x?vP 91(2  < <$x?vVP 92(2  < <8x?P* ;1.5(2 < 6Lx?P* 91(2 < <x?PV* 96(2 < 6x?* ;0.5(2 < 6@x?* 92(2 < 6 x?*V 98(2 < 6 x?v*P 91(2 < 6x?P** 92(2 < 6x?** 93(2 < 6Dz?v ;[A](2 < 6z?v ;[B](2 < 6z?Vv HInitial rate (r)(2  .< 6z? w ,Pr initial rate of reaction mol dm-3 s-1 [ ] concentration mol dm-3Q@$3333333 Q B < s *޽h ? ̙33y___PPT10Y+D=' = @B +f 0L0 umP/@ ( C `H@6 @dB @ <D)@0A @ 0Pp@ 0 0dB @@ <D)@X@ @ 0X0@ 0 0*  @ s *z?  Compare Experiments 1 & 2 [B] same [A] 3 x bigger rate 3 x bigger \ rate [A] FIRST ORDER with respect to (wrt) AH9@$H+3333%&p     @ <#z?vP ;0.5(2  @ N'z??vP 91(2  @ <X+z?vVP 92(2  @ <H/z?P* ;1.5(2 @ N3z??P* 91(2 @ <T6z?PV* 96(2 @ 69z?* ;0.5(2 @ 6=z?* 92(2 @ 6Az?*V 98(2 @ N?z??v*P 91(2 @ N8Hz??P** 92(2 @ 6Kz?** 93(2 @ 6Oz?v ;[A](2 @ 6Sz?v ;[B](2 @ 6Vz?Vv HInitial rate (r)(2 Y .@ s *Zz?tO Y  -CALCULATING ORDER wrt A Choose any two experiments where... [A] is changed and, importantly, [B] is KEPT THE SAME See how the change in [A] affects the rate As you can see, tripling [A] has exactly the same effect on the rate so... THE ORDER WITH RESPECT TO A = 1 (it is FIRST ORDER) HP}@'&    /@ 0gz ^"RATE EQUATION - SAMPLE CALCULATION #(2#fB @ s *޽h ? ̙33y___PPT10Y+D=' = @B + 0L0 `0D( w DdB D <D)@0A D 0Pp@ 0 0dB D@ <D)@X@ D 0X0@ 0 0 D N,pz??zT ;0.5(2 D Nptz??zyT 91(2 D N@xz??zzLT 92(2 D N4|z? ?T. ;1.5(2 D 6z?Ty. 91(2 D N؃z? ?TzL. 96(2 D Nz??. ;0.5(2  D Nz??.y 92(2 !D <@z?.zL 98(2 "D N$z??z T 91(2 #D 6`z?T . 92(2 $D Nz??.  93(2 %D 6,z?z ;[A](2 &D 6z?yz ;[B](2 'D 68z?zLz HInitial rate (r)(2 a .D s *0z?t  CALCULATING ORDER wrt B Choose any two experiments where... [B] is changed and, importantly, [A] is KEPT THE SAME See how a change in [B] affects the rate As you can see, doubling [B] quadruples the rate so... THE ORDER WITH RESPECT TO B = 2 It is SECOND ORDERHP}m@'< 3<&    /D 0Hz ^"RATE EQUATION - SAMPLE CALCULATION #(2#f> 0D s *z?O   Compare Experiments 1 & 3 [A] same [B] 2 x bigger rate 4 x bigger \ rate [B]2 SECOND ORDER wrt BH8@H3(33&_   B D s *޽h ? ̙33y___PPT10Y+D=' = @B +J 0L0 YQp'.L(  LdB L <D)@0A L 0Pp@ 0 0dB L@ <D)@X@ L 0X0@ 0 0$  L s *z?  Compare Experiments 1 & 2 [B] same [A] 3 x bigger rate 3 x bigger \ rate [A] FIRST ORDER with respect to (wrt) AH;@$HF#&r     L <z?vP ?0.5 (2  L 6z?vP =1 (2  L N,sz? ?vVP =2 (2  L N0z? ?P* ?1.5 (2 L 64z?P* =1 (2 L NTz? ?PV* =6 (2 L 6lz?* ?0.5 (2 L 6hz?* =2 (2 L 6Xz?*V =8 (2 L 6,z?v*P =1 (2 L 6dz?P** =2 (2 L 6${?** =3 (2 L 6X{?v ?[A] (2 L 6p{?v ?[B] (2 L 6 {?Vv LInitial rate (r) (2  L N{? ?zT ?0.5 (2 L 6<{?zyT =1 (2 L N4{? ?zzLT =2 (2 L N{? ?T. ?1.5 (2 L 6{?Ty. =1 (2 L NT!{? ?TzL. =6 (2 L N{? ?. ?0.5 (2  L Nx({? ?.y =2 (2 !L N,{? ?.zL =8 (2 "L 60{?z T =1 (2 #L 6|4{?T . =2 (2 $L 6P8{?.  =3 (2 %L 6;{?z ?[A] (2 &L 6?{?yz ?[B] (2 'L 6B{?zLz LInitial rate (r) (2 " )L s *TG{?O   Compare Experiments 1 & 3 [A] same [B] 2 x bigger rate 4 x bigger \ rate [B]2 SECOND ORDER wrt BH8@HB&_   A *L s *R{?"`$ f7 IOVERALL ORDER = THE SUM OF THE INDIVIDUAL ORDERS = 1 + 2 = 3`J@ 9 J  .L 0Z{ ^"RATE EQUATION - SAMPLE CALCULATION #(2#fB L s *޽h ? ̙33y___PPT10Y+D=' = @B +  0L0 )2HR( w HdB H <D)@0A H 0Pp@ 0 0dB H@ <D)@X@ H 0X0@ 0 0  H <`{?vP ?0.5 (2  H 60e{?vP =1 (2  H Ni{? ?vVP =2 (2  H Nl{? ?P* ?1.5 (2 H 6p{?P* =1 (2 H Nt{? ?PV* =6 (2 H 6lx{?* ?0.5 (2 H 6{{?* =2 (2 H 6s{?*V =8 (2 H 6{?v*P =1 (2 H 6 {?P** =2 (2 H 6@{?** =3 (2 H 6{?v ?[A] (2 H 6{?v ?[B] (2 H 6D{?Vv LInitial rate (r) (2  H N<{? ?zT ?0.5 (2 H 6{?zyT =1 (2 H NH{? ?zzLT =2 (2 H N{? ?T. ?1.5 (2 H 6{?Ty. =1 (2 H NĪ{? ?TzL. =6 (2 H N{? ?. ?0.5 (2  H N{? ?.y =2 (2 !H N<{? ?.zL =8 (2 "H 6{?z T =1 (2 #H 6{?T . =2 (2 $H 6@{?.  =3 (2 %H 6{?z ?[A] (2 &H 6{?yz ?[B] (2 'H 6D{?zLz LInitial rate (r) (2 dB (H <D? L  +H s *{?M   .\ rate = k [A] [B]2 V@H     .H 0p{ ^"RATE EQUATION - SAMPLE CALCULATION #(2#f" /H s *,{?"`$ <   ^By combining the two proportionality relationships you can construct the overall rate equation,_@^3  _ " 1H s *4{?O   Compare Experiments 1 & 3 [A] same [B] 2 x bigger rate 4 x bigger \ rate [B]2 SECOND ORDER wrt BH8@HB&_   $ 2H s *{?  Compare Experiments 1 & 2 [B] same [A] 3 x bigger rate 3 x bigger \ rate [A] FIRST ORDER with respect to (wrt) AH;@$HF#&r   B H s *޽h ? ̙33y___PPT10Y+D=' = @B +" 0L0 "!+1P!(  PdB P <D)@0A P 0Pp@ 0 0dB P@ <D)@X@ P 0X0@ 0 0  P <|?vP ?0.5 (2  P 6D|?vP =1 (2  P N8|? ?vVP =2 (2  P N |? ?P* ?1.5 (2 P 6|?P* =1 (2 P NP|? ?PV* =6 (2 P 6|?* ?0.5 (2 P 6|?* =2 (2 P 6 |?*V =8 (2 P 6!|?v*P =1 (2 P 6%|?P** =2 (2 P 6)|?** =3 (2 P 6-|?v ?[A] (2 P 60|?v ?[B] (2 P 64|?Vv LInitial rate (r) (2  P N8|? ?zT ;0.5(2 P 6<|?zyT 91(2 P N?|? ?zzLT 92(2 P NC|? ?T. ;1.5(2 P 6G|?Ty. 91(2 P NK|? ?TzL. 96(2 P N( N,|??zyT 91(2 ?( N|??zzLT 92(2 @( N |? ?T. ;1.5(2 A( 6|?Ty. 91(2 B( N |? ?TzL. 96(2 C( <|?. ;0.5(2 D( N|??.y 92(2 E( <~?.zL 98(2 I( 6D~?z T 91(2 J( 68~?T . 92(2 K( 6 ~?.  93(2 M( 6~?z ;[A](2 N( 6P~?yz ;[B](2 O( 6~?zLz HInitial rate (r)(2  W( H~̙? ISUMMARY&(2 X( 0~ ^"RATE EQUATION - SAMPLE CALCULATION #(2#f> Y( s *"~?O   Compare Experiments 1 & 3 [A] same [B] 2 x bigger rate 4 x bigger \ rate [B]2 SECOND ORDER wrt BH8@H3(33&_   . Z( s *P1~?  Compare Experiments 1 & 2 [B] same [A] 3 x bigger rate 3 x bigger \ rate [A] FIRST ORDER with respect to (wrt) AH;@$H3,#33&r   dB [( <D? L  \( s *;~?% P Chose one experiment (e.g. Expt. 3) and substitute its values into the rate equation k = 8 = 4 dm6 mol-2 sec-1 (0.5) (2)2@Y &  w  ]( s *H~?%  X\ rate = k [A] [B]2 re-arranging k = rate [A] [B]2P@  *  P pB ^( HDo?A(AjB _( BDo? LB ( s *޽h ? ̙33y___PPT10Y+D=' = @B +  0L0  D(    0T~ cRATE EQUATION QUESTIONS0(2ffdB  <D)@0A  0Pp@ 0 0dB @ <D)@X@  0X0@ 0 0 > s *\~?.b  -CALCULATE THE ORDER WITH RESPECT TO A THE ORDER WITH RESPECT TO B THE OVERALL ORDER OF REACTION THE FORMAT OF THE RATE EQUATION THE VALUE AND UNITS OF THE RATE CONSTANT D@ 3   B s *Xc~?P kANSWER ON NEXT PAGE H 33 @ C 0Lg~?I+  [A] / mol dm-3 [B] / mol dm-3 Rate / mol dm-3 s-1 Expt 1 0.25 0.25 4 Expt 2 0.25 0.50 8 Expt 3 0.50 0.25 8o 24@       ObF        D Hw~?_0% FNo 1&(2B  s *޽h ? ̙33y___PPT10Y+D=' = @B +s  0L0   p  (    0h}~ SRATE EQUATION QUESTIONS (2fdB  <D)@0A  0Pp@ 0 0dB @ <D)@X@  0X0@ 0 0z   s *~? TExpts 1&2 [A] is constant [B] is doubled Rate is doubled Therefore rate [B] 1st order wrt B Explanation: What was done to [B] had exactly the same effect on the rate t@J3 @4 U  N v   0|~?Q#  [A] / mol dm-3 [B] / mol dm-3 Rate / mol dm-3 s-1 Expt 1 0.25 0.25 4 Expt 2 0.25 0.50 8 Expt 3 0.50 0.25 8o 24@ 4bF          H~?_0% FNo 1&(2   Hl~̙?= HANSWER&(2B  s *޽h ? ̙33m  0L0    D (  D D 0~ SRATE EQUATION QUESTIONS (2fdB D <D)@0A D 0Pp@ 0 0dB D@ <D)@X@ D 0X0@ 0 0f D s *T~?  Expts 1&2 [A] is constant [B] is doubled Rate is doubled Therefore rate [B] 1st order wrt B Explanation: What was done to [B] had exactly the same effect on the rate Expts 1&3 [B] is constant [A] is doubled Rate is doubled Therefore rate [A] 1st order wrt A Explanation: What was done to [A] had exactly the same effect on the rate Z@J3 3 @h U  N  Y  N  D 0~?Q# $ [A] / mol dm-3 [B] / mol dm-3 Rate / mol dm-3 s-1 Expt 1 0.25 0.25 4 Expt 2 0.25 0.50 8 Expt 3 0.50 0.25 8o 24@ bF         D H~?_0% FNo 1&(2  D HD~̙?= HANSWER&(2B D s *޽h ? ̙33  0L0 M E  @ (  @ @ 0x~ SRATE EQUATION QUESTIONS (2fdB @ <D)@0A @ 0Pp@ 0 0dB @@ <D)@X@ @ 0X0@ 0 0 @ s *8~?  Expts 1&2 [A] is constant [B] is doubled Rate is doubled Therefore rate [B] 1st order wrt B Explanation: What was done to [B] had exactly the same effect on the rate Expts 1&3 [B] is constant [A] is doubled Rate is doubled Therefore rate [A] 1st order wrt A Explanation: What was done to [A] had exactly the same effect on the rate Rate equation is r = k[A][B] x@J3 3 @ 3h U  N  Y  l V @ 0?I+  [A] / mol dm-3 [B] / mol dm-3 Rate / mol dm-3 s-1 Expt 1 0.25 0.25 4 Expt 2 0.25 0.50 8 Expt 3 0.50 0.25 8o 24@ ObF         @ H?_0% FNo 1&(2  @ H̙?= HANSWER&(2B @ s *޽h ? ̙33  0L0 _ W  < (  < < 0$ SRATE EQUATION QUESTIONS (2fdB < <D)@0A < 0Pp@ 0 0dB <@ <D)@X@ < 0X0@ 0 0 < s *~? Expts 1&2 [A] is constant [B] is doubled Rate is doubled Therefore rate [B] 1st order wrt B Explanation: What was done to [B] had exactly the same effect on the rate Expts 1&3 [B] is constant [A] is doubled Rate is doubled Therefore rate [A] 1st order wrt A Explanation: What was done to [A] had exactly the same effect on the rate Rate equation is r = k[A][B] Value of k Substitute numbers from Exp 1 to get value of k k = rate / [A][B] = 4 / 0.25 x 0.25 = 64 D@J3 3 @ 3 ]3h U  N  Y   V < 0A?I+  [A] / mol dm-3 [B] / mol dm-3 Rate / mol dm-3 s-1 Expt 1 0.25 0.25 4 Expt 2 0.25 0.50 8 Expt 3 0.50 0.25 8o 24@ ObF         < HM?_0% FNo 1&(2  < HR̙?= HANSWER&(2B < s *޽h ? ̙33 0L0  8)(  8 8 0X SRATE EQUATION QUESTIONS (2fdB 8 <D)@0A 8 0Pp@ 0 0dB 8@ <D)@X@ 8 0X0@ 0 0" 8 s *_?  Expts 1&2 [A] is constant [B] is doubled Rate is doubled Therefore rate [B] 1st order wrt B Explanation: What was done to [B] had exactly the same effect on the rate Expts 1&3 [B] is constant [A] is doubled Rate is doubled Therefore rate [A] 1st order wrt A Explanation: What was done to [A] had exactly the same effect on the rate Rate equation is r = k[A][B] Value of k Substitute numbers from Exp 1 to get value of k k = rate / [A][B] = 4 / 0.25 x 0.25 = 64 Units of k rate / conc x conc = dm3 mol-1 s-1@J3 3 @ 3 ]3 333333 U  N  Y       V 8 0?I+  [A] / mol dm-3 [B] / mol dm-3 Rate / mol dm-3 s-1 Expt 1 0.25 0.25 4 Expt 2 0.25 0.50 8 Expt 3 0.50 0.25 8o 24@ ObF         8 H?_0% FNo 1&(2  8 HX̙?= HANSWER&(2B 8 s *޽h ? ̙33P  0L0  A~( w   0 SRATE EQUATION QUESTIONS (2fdB  <D)@0A  0Pp@ 0 0dB @ <D)@X@  0X0@ 0 0 < s *?.WD  9CALCULATE THE ORDER WITH RESPECT TO C THE ORDER WITH RESPECT TO D THE OVERALL ORDER OF REACTION THE FORMAT OF THE RATE EQUATION THE VALUE AND UNITS OF THE RATE CONSTANT P@ 3   ? s *?P kANSWER ON NEXT PAGE H 33 5 @ 0 ?H*  [C] / mol dm-3 [D] / mol dm-3 Rate / mol dm-3 s-1 Expt 1 0.40 0.40 0.16 Expt 2 0.20 0.40 0.04 Expt 3 0.40 1.20 1.44p 26@     RbF        A H?_0% FNo 2&(2B  s *޽h ? ̙33  0L0 j b  P (  P P 0 € SRATE EQUATION QUESTIONS (2fdB P <D)@0A P 0Pp@ 0 0dB P@ <D)@X@ P 0X0@ 0 0 P 0ǀ?P" ' [C] / mol dm-3 [D] / mol dm-3 Rate / mol dm-3 s-1 Expt 1 0.40 0.40 0.16 Expt 2 0.20 0.40 0.04 Expt 3 0.40 1.20 1.44p 26@ bF        P Hـ?_0% FNo 2&(2  P s *ހ? nfExpts 1&3 [C] is constant [D] is tripled Rate is 9 x bigger Therefore rate [D]2 2nd order wrt D Explanation: Squaring what was done to D affected the rate (32 = 9) @S 3 3 4 c  I   P H̙?= HANSWER&(2B P s *޽h ? ̙33  0L0 D < 0 L (  L L 0 SRATE EQUATION QUESTIONS (2fdB L <D)@0A L 0Pp@ 0 0dB L@ <D)@X@ L 0X0@ 0 0y L 0?P"  [C] / mol dm-3 [D] / mol dm-3 Rate / mol dm-3 s-1 Expt 1 0.40 0.40 0.16 Expt 2 0.20 0.40 0.04 Expt 3 0.40 1.20 1.44p 26@ 6bF        L H ?_0% FNo 2&(2  L s *?"  VExpts 1&3 [C] is constant [D] is tripled Rate is 9 x bigger Therefore rate [D]2 2nd order wrt D Explanation: Squaring what was done to D affected the rate (32 = 9) Expts 1&2 [D] is constant [A] is halved Rate is quartered Therefore rate [C] 2 2nd order wrt C Explanation: One half squared = one quarter (R@S 3 3 [ 3 #h c  I  e  1   L H%̙?= HANSWER&(2B L s *޽h ? ̙33N  0L0   @ H| (  H H 0, SRATE EQUATION QUESTIONS (2fdB H <D)@0A H 0Pp@ 0 0dB H@ <D)@X@ H 0X0@ 0 0g H 0|0?H*  [C] / mol dm-3 [D] / mol dm-3 Rate / mol dm-3 s-1 Expt 1 0.40 0.40 0.16 Expt 2 0.20 0.40 0.04 Expt 3 0.40 1.20 1.44p 26@ QbF        H H@?_0% FNo 2&(2d  H s *G?  Expts 1&3 [C] is constant [D] is tripled Rate is 9 x bigger Therefore rate [D]2 2nd order wrt D Explanation: Squaring what was done to D affected the rate (32 = 9) Expts 1&2 [D] is constant [A] is halved Rate is quartered Therefore rate [C] 2 2nd order wrt C Explanation: One half squared = one quarter Rate equation is r = k[C]2[D]2 q@S 3 3 [ 3 "3333h c  I  e  P   H HLa̙?= HANSWER&(2B H s *޽h ? ̙334 0L0 P b(    0th SRATE EQUATION QUESTIONS (2fdB  <D)@0A  0Pp@ 0 0dB @ <D)@X@  0X0@ 0 0g   0Hd?H*  [C] / mol dm-3 [D] / mol dm-3 Rate / mol dm-3 s-1 Expt 1 0.40 0.40 0.16 Expt 2 0.20 0.40 0.04 Expt 3 0.40 1.20 1.44p 26@ QbF          Hy?_0% FNo 2&(2J   s *?t 2Expts 1&3 [C] is constant [D] is tripled Rate is 9 x bigger Therefore rate [D]2 2nd order wrt D Explanation: Squaring what was done to D affected the rate (32 = 9) Expts 1&2 [D] is constant [A] is halved Rate is quartered Therefore rate [C] 2 2nd order wrt C Explanation: One half squared = one quarter Rate equation is r = k[C]2[D]2 Value of k Substitute numbers from Exp 2 to get value of k k = rate / [C]2[D]2 = 0.04 / 0.22 x 0.42 = 6.25 Units of k rate / conc2 x conc2 = dm9 mol-3 s-1*@S 3 3 [ 3 "3333 A    3     333333p c  I  e     H䲁̙?= HANSWER&(2B  s *޽h ? ̙330  0L0 ` A^(    0 SRATE EQUATION QUESTIONS (2fdB  <D)@0A  0Pp@ 0 0dB @ <D)@X@  0X0@ 0 0 < s *8?.D  -CALCULATE THE ORDER WITH RESPECT TO E THE ORDER WITH RESPECT TO F THE OVERALL ORDER OF REACTION THE FORMAT OF THE RATE EQUATION THE VALUE AND UNITS OF THE RATE CONSTANT D@ 3  > Hǁ?_0% FNo 3&(2 @ s *d́?P kANSWER ON NEXT PAGE H 33 ! A 0 Ё?H*  [E] / mol dm-3 [F] / mol dm-3 Rate / mol dm-3 s-1 Expt 1 0.40 0.40 0.16 Expt 2 0.80 0.80 0.32 Expt 3 0.80 1.20 0.32\ 26@     Rb=       B  s *޽h ? ̙33  0L0 ) ! p \(  \ \ 0 SRATE EQUATION QUESTIONS (2fdB \ <D)@0A \ 0Pp@ 0 0dB \@ <D)@X@ \ 0X0@ 0 0e \ 0?P"  [E] / mol dm-3 [F] / mol dm-3 Rate / mol dm-3 s-1 Expt 1 0.40 0.40 0.16 Expt 2 0.80 0.80 0.32 Expt 3 0.80 1.20 0.32\ 26@ 6b=        \ Hd?_0% FNo 3&(2  \ s *? OExpts 2&3 [E] is constant [F] is x 1.5 Rate unchanged Rate is UNAFFECTED ZERO order wrt F Explanation: Concentration of [F] has no effect on the rate V@T3 24 Z  @   \ H ̙?= HANSWER&(2B \ s *޽h ? ̙33  0L0 P H  X (  X X 0 SRATE EQUATION QUESTIONS (2fdB X <D)@0A X 0Pp@ 0 0dB X@ <D)@X@ X 0X0@ 0 0e X 0?P"  [E] / mol dm-3 [F] / mol dm-3 Rate / mol dm-3 s-1 Expt 1 0.40 0.40 0.16 Expt 2 0.80 0.80 0.32 Expt 3 0.80 1.20 0.32\ 26@ 6b=        X Hd$?_0% FNo 3&(2  X s *)?  vExpts 2&3 [E] is constant [F] is x 1.5 Rate unchanged Rate is UNAFFECTED ZERO order wrt F Explanation: Concentration of [F] has no effect on the rate Expts 1&2 [E] is doubled [F] is doubled Rate is doubled Therefore rate [E] 2 2nd order wrt E Explanation: Although both concentrations have been doubled, we know [F] has no effect. The change must be all due to [E] @T3  3 rh Z  @  e     X H ;̙?= HANSWER&(2B X s *޽h ? ̙33  0L0    T8 (  T T 0E SRATE EQUATION QUESTIONS (2fdB T <D)@0A T 0Pp@ 0 0dB T@ <D)@X@ T 0X0@ 0 0E T 0K?H*  [E] / mol dm-3 [F] / mol dm-3 Rate / mol dm-3 s-1 Expt 1 0.40 0.40 0.16 Expt 2 0.80 0.80 0.32 Expt 3 0.80 1.20 0.32\ 26@ Rb=        T H|Z?_0% FNo 3&(2B  T s *_?V  RExpts 2&3 [E] is constant [F] is x 1.5 Rate unchanged Rate is UNAFFECTED ZERO order wrt F Explanation: Concentration of [F] has no effect on the rate Expts 1&2 [E] is doubled [F] is doubled Rate is doubled Therefore rate [E] 2 2nd order wrt E Explanation: Although both concentrations have been doubled, we know [F] has no effect. The change must be all due to [E] Rate equation is r = k[E]  @T3  3 r3h Z  @  e     T Hs̙?= HANSWER&(2B T s *޽h ? ̙33 0L0 VN  (    0y SRATE EQUATION QUESTIONS (2fdB  <D)@0A  0Pp@ 0 0dB @ <D)@X@  0X0@ 0 0E   0`?H*  [E] / mol dm-3 [F] / mol dm-3 Rate / mol dm-3 s-1 Expt 1 0.40 0.40 0.16 Expt 2 0.80 0.80 0.32 Expt 3 0.80 1.20 0.32\ 26@ Rb=          H莂?_0% FNo 3&(2   s *? XExpts 2&3 [E] is constant [F] is x 1.5 Rate unchanged Rate is UNAFFECTED ZERO order wrt F Explanation: Concentration of [F] has no effect on the rate Expts 1&2 [E] is doubled [F] is doubled Rate is doubled Therefore rate [E] 2 2nd order wrt E Explanation: Although both concentrations have been doubled, we know [F] has no effect. The change must be all due to [E] Rate equation is r = k[E] Value of k Substitute numbers from Exp 1 to get value of k k = rate / [E] = 0.16 / 0.4 = 0.40 Units of k rate / conc = s-1-@T3  3 r3 U3 33 Z  @  e       H̙?= HANSWER&(2B  s *޽h ? ̙33  0L0   @ c (    0\ \ GRAPHICAL DETERMINATION OF RATE  !(2!fN  C &Agr111g!pj   B?!~   B?  2RATE CALCULATION The rate of reaction at any moment can be found from the gradient of the tangent at that point. The steeper the gradient, the faster the rate of reaction Place a rule on the outside of the curve and draw a line as shown on the graph. y x gradient = y / x (2 2(2 29  3  3 3  3 pb   H8c?4 oAx x  B?"`   B,͂? ,*In the reaction& A(aq) + B(aq)   > C(aq) + D(aq) the concentration of B was measured every 200 minutes. The reaction is obviously very slow!j 2'6&3 3t         _ u  s * ؂? The variation in rate can be investigated by measuring the change in concentration of one of the reactants or products, plotting a graph and then finding the gradients of the curve at different concentrations.@  H  0޽h ? ̙33y___PPT10Y+D=' = @B +  0L0   P b (     0, \ GRAPHICAL DETERMINATION OF RATE  !(2!fN   C &Agr111g!ppB   HDjJ?6i j  B?!~  BT?  2RATE CALCULATION The rate of reaction at any moment can be found from the gradient of the tangent at that point. The steeper the gradient, the faster the rate of reaction Place a rule on the outside of the curve and draw a line as shown on the graph. y x gradient = y / x (2 2(2 29  3  3 3  3 pb   H8c?4 oAx x  B?"`Cf   B?  7concentration = 1.20 mol dm-3 gradient = - 1.60 mol dm-3 1520 min rate = - 1.05 x 10-3 mol dm-3 The rate is negative because the reaction is slowing down 6 2( & %    (  3:3 3  pB  HDo?HO 5 Hy  s *( ? The variation in rate can be investigated by measuring the change in concentration of one reactants or product, plotting a graph and then finding the gradients of tangents to the curve at different concentrations.@  H  0޽h ? ̙33y___PPT10Y+D=' = @B +  0L0     (    0| \ GRAPHICAL DETERMINATION OF RATE  !(2!fN  C &Agr111g!pj  B?!~  B?  2RATE CALCULATION The rate of reaction at any moment can be found from the gradient of the tangent at that point. The steeper the gradient, the faster the rate of reaction Place a rule on the outside of the curve and draw a line as shown on the graph. y x gradient = y / x (2 2(2 29  3  3 3  3 pb  H8c?4 oAx x  B?"`]   Bp*?L+w The gradients of tangents at several other concentrations are calculated. Notice how the gradient gets less as the reaction proceeds, showing that the reaction is slowing down. The tangent at the start of the reaction is used to calculate the initial rate of the reaction.L 2 33  u   s *1? The variation in rate can be investigated by measuring the change in concentration of one of the reactants or products, plotting a graph and then finding the gradients of the curve at different concentrations.@  jb   BjJ?F jb  BjJ? 0 jb  BjJ? h jb  BjJ? @ 8 H  0޽h ? ̙33y___PPT10Y+D=' = @B + 0L0 Q( w X  c 0A g gr110nd~  0: _#FIRST ORDER REACTIONS AND HALF LIFE $(2$fx  B?"` V  B??&m 'One characteristic of a FIRST ORDER REACTION is that it is similar to radioactive decay. It has a half-life that is independent of the concentration. It should take the same time to drop to one half of the original concentration as it does to drop from one half to one quarter of the original. ,( 2'3 ( j  B? #    HhG?N C   ?The concentration of a reactant falls as the reaction proceeds @(2@ @ H  0޽h ? ̙33y___PPT10Y+D=' = @B + 0L0 (  ( w X  c 0A g gr110nd~>F  D   b fbN  D    D ~B  NDjJ? D ~B  NDjJ?0 0 ~B   NDjJ?0 0 j  B?`  HO?7P ?The concentration of reactant A falls as the reaction proceeds @(2@ @   BU?d\  2The concentration drops from 4 to 2 in 17 minutes >3 233 3   0X[ _#FIRST ORDER REACTIONS AND HALF LIFE $(2$f|B  TD33Ԕ?  H  0޽h ? ̙33y___PPT10Y+D=' = @B +  0L0 U(  X  c 0A g gr110nd~>F  D   b fbN  D    D ~B  NDjJ? D ~B  NDjJ?0 0 ~B   NDjJ?0 0 ZF  x    X 8 `~B   NDjJ? x ~B   NDjJ?x x M  Bb?d\  QThe concentration drops from 4 to 2 in 17 minutes 2 to 1 in a further 17 minutes ZR 23 R   0k _#FIRST ORDER REACTIONS AND HALF LIFE $(2$f|B  TDԔ? @ j  B?`  Ho?7P ?The concentration of reactant A falls as the reaction proceeds @(2@ @ H  0޽h ? ̙33y___PPT10Y+D=' = @B +Y  0L0 h `  (  x  c 0A g gr110C"?nd~>F  D   b fbN  D    D ~B  NDjJ? D ~B  NDjJ?0 0 ~B   NDjJ?0 0 ZF  x    X 8 `~B   NDjJ? x ~B   NDjJ?x x ZF     T  h~B  NDjJ? ~B  NDjJ? n  By?d\  rThe concentration drops from 4 to 2 in 17 minutes 2 to 1 in a further 17 minutes 1 to 0.5 in a further 17 minutesZs 23!3 s   0P _#FIRST ORDER REACTIONS AND HALF LIFE $(2$f|B  TDԔ? 0 j  B?`  H ?7P ?The concentration of reactant A falls as the reaction proceeds @(2@ @ H  0޽h ? ̙33y___PPT10Y+D=' = @B +  0L0   " ( w X  c 0A g gr110nd~>F  D   b fbN  D    D ~B  NDjJ? D ~B  NDjJ?0 0 ~B   NDjJ?0 0 ZF  x    X 8 `~B   NDjJ? x ~B   NDjJ?x x ZF     T  h~B  NDjJ? ~B  NDjJ? <  Bh?d\  rThe concentration drops from 4 to 2 in 17 minutes 2 to 1 in a further 17 minutes 1 to 0.5 in a further 17 minutes(s 2r s   0얄 _#FIRST ORDER REACTIONS AND HALF LIFE $(2$fj  B?`  H?7P ?The concentration of reactant A falls as the reaction proceeds @(2@ @ H  0޽h ? ̙33y___PPT10Y+D=' = @B +  0L0   m (  X  c 0A g gr110nd~>F  D   b fbN  D    D ~B  NDjJ? D ~B  NDjJ?0 0 ~B  NDjJ?0 0 ZF  x   X 8 `~B   NDjJ? x ~B   NDjJ?x x ZF     T  h~B   NDjJ? ~B   NDjJ? 8 3K 3Kr  B?3K  HT?7  GUA useful relationship k t = loge 2 = 0.693 where t = the half life8F22#x3     U   0| _#FIRST ORDER REACTIONS AND HALF LIFE $(2$fs8 "-   "-' +  B?"-   Half life = 17 minutes k t = 0.693 k = 0.693 t k = 0.693 = 0.041 min-1 17j 2022,02 8F2!  / ' , xB  HDjJ? xB  HDjJ?   H  0޽h ? ̙33y___PPT10Y+D=' = @B + 0L0 -%( `h   0τ XORDER OF REACTION  GRAPHICAL DETERMINATION  -(2-f   s * Ԅ? /The order of reaction can be found by measuring the rate at different times during the reaction and plotting the rate against either concentration or time. The shape of the curve provides an indication of the order. &@  H  0޽h ? ̙33y___PPT10Y+D=' = @B +  0L0    ( ( w ( ( 0߄ XORDER OF REACTION  GRAPHICAL DETERMINATION  -(2-fF "  ( bS xB ( HDo?"xB ( HDo?"H   (  0e0e    B CDE(F3 o 8c8c     ?A)BCD|E|| t:K0 H) 0"   @   "5 VN *  ( " xB ( HDjJ?** xB  (B HDjJ?*    ( s *?a sThe order of reaction can be found by measuring the rate at different times during the reaction and plotting the rate against either concentration or time. The shape of the curve provides an indication of the order. PLOTTING RATE AGAINST CONCENTRATIONF@$H$3    ( 0Ä?&  RATE OF REACTION / mol dm-3 s-1H @          ( s *?6c  rCONCENTRATION / mol dm-3*@     H ( 0޽h ? ̙33y___PPT10Y+D=' = @B +u  0L0  |    ( w     0D XORDER OF REACTION  GRAPHICAL DETERMINATION  -(2-fF "    bS xB   HDo?"xB   HDo?"H      0e0e    B CDE(F3 o 8c8c     ?A)BCD|E|| t:K0 H) 0"   @   "5 VN *    " xB   HDjJ?** xB  B HDjJ?*     s *L?a sThe order of reaction can be found by measuring the rate at different times during the reaction and plotting the rate against either concentration or time. The shape of the curve provides an indication of the order. PLOTTING RATE AGAINST CONCENTRATIONF@$H$3     0p?&  RATE OF REACTION / mol dm-3 s-1H @           s *?6c  rCONCENTRATION / mol dm-3*@        s *?   &ZERO ORDER  the rate does not depend on the concentration. The line is parallel to the x axis.8`@ U  ` H   0޽h ? ̙33y___PPT10Y+D=' = @B + 0L0   0H (  0 0 0L" XORDER OF REACTION  GRAPHICAL DETERMINATION  -(2-fF "  0 bS xB 0 HDo?"xB 0 HDo?"H   0  0e0e    B CDE(F3 o 8c8c     ?A)BCD|E|| t:K0 H) 0"   @   "5 VN *  0 " xB 0 HDjJ?** xB  0B HDjJ?*    0 s *D(?a sThe order of reaction can be found by measuring the rate at different times during the reaction and plotting the rate against either concentration or time. The shape of the curve provides an indication of the order. PLOTTING RATE AGAINST CONCENTRATIONF@$H$3    0 00?&  RATE OF REACTION / mol dm-3 s-1H @          0 s *5?6c  rCONCENTRATION / mol dm-3*@       0 s *p;?   2ZERO ORDER  the rate does not depend on the concentration. The line is parallel to the x axis.D`@ U `   0 s *@? p ZFIRST ORDER  the rate is proportional to the concentration so you get a straight line of fixed gradient. The gradient of the line equals the rate constant for the reaction.<@    H 0 0޽h ? ̙33y___PPT10Y+D=' = @B +a 0L0 ph,(  , , 0TM XORDER OF REACTION  GRAPHICAL DETERMINATION  -(2-fF "  , bS xB , HDo?"xB , HDo?"H   ,  0e0e    B CDE(F3 o 8c8c     ?A)BCD|E|| t:K0 H) 0"   @   "5 VN *  , " xB , HDjJ?** xB  ,B HDjJ?*    , s *P?a sThe order of reaction can be found by measuring the rate at different times during the reaction and plotting the rate against either concentration or time. The shape of the curve provides an indication of the order. PLOTTING RATE AGAINST CONCENTRATIONF@$H$3    , 0,Z?&  RATE OF REACTION / mol dm-3 s-1H @          , s *c?6c  rCONCENTRATION / mol dm-3*@       , s *h?   2ZERO ORDER  the rate does not depend on the concentration. The line is parallel to the x axis.D`@ U ` ( , s *o? p ZFIRST ORDER  the rate is proportional to the concentration so you get a straight line of fixed gradient. The gradient of the line equals the rate constant for the reaction.D@    , s *0v? {w NSECOND ORDER  the rate is proportional to the square of the concentration. You get an upwardly sloping curve.<r@ e  r H , 0޽h ? ̙33y___PPT10Y+D=' = @B +M 0L0 \T (    0 XORDER OF REACTION  GRAPHICAL DETERMINATION  -(2-fF "   bS xB  HDo?"xB  HDo?"H     0e0e    B CDE(F3 o 8c8c     ?A)BCD|E|| t:K0 H) 0"   @   "5 VN *   " xB  HDjJ?** xB  B HDjJ?*     s *{?a sThe order of reaction can be found by measuring the rate at different times during the reaction and plotting the rate against either concentration or time. The shape of the curve provides an indication of the order. PLOTTING RATE AGAINST CONCENTRATIONF@$H$3     0c?&  RATE OF REACTION / mol dm-3 s-1H @           s *?6c  rCONCENTRATION / mol dm-3*@        s *?   &ZERO ORDER  the rate does not depend on the concentration. The line is parallel to the x axis.8`@ U  `    s *蝉? p ZFIRST ORDER  the rate is proportional to the concentration so you get a straight line of fixed gradient. The gradient of the line equals the rate constant for the reaction.<@      s *줉? {w NSECOND ORDER  the rate is proportional to the square of the concentration. You get an upwardly sloping curve.<r@ e  r H  0޽h ? ̙33y___PPT10Y+D=' = @B +" 0L0 1)0(    0 XORDER OF REACTION  GRAPHICAL DETERMINATION  -(2-f   s *d?a jThe order of reaction can be found by measuring the rate at different times during the reaction and plotting the rate against either concentration or time. The shape of the curve provides an indication of the order. PLOTTING RATE AGAINST TIMEF@H3     0̸?"`X  RATE OF REACTION / mol dm-3 s-1H @           s *tƉ?"` bTIME / s* @        s * ?"`t GZERO ORDER A straight line showing a constant decline in concentration.8H@ <  H   s *<?"`j  KFIRST ORDER A slightly sloping curve which drops with a constant half-life.<L@ ?  L   s *щ?"`   BSECOND ORDER The curve declines steeply at first then levels out.<C@4  C TL *  # oR xB  HDjJ?** xB B HDjJ?*  vB  NDo?R   0e0e    B&C DE(F o 8c8c     ?A)BCD|E|| q!nM(@X & @   S" R M   0e0e    BC~DE(F o 8c8c     ?A)BCD|E|| ?;v4'_?WGb~ @   S" S Z pB  HD3jJ? xs pB  HDjJ?xspB  HDjJ?sH  0޽h ? ̙33y___PPT10Y+D=' = @B +  0L0 @ "+(    0މw  h*ORDER OF REACTION GRAPHICAL DETERMINATION "+8P2+f8    d  S 4Agrhalflife B  ZDo?B  ZDo?VZ\B  ZDo? X B  ZDo?f ~ f pb  HjJ?4#pb  HjJ?>D4h pb  HjJ?C pb  HjJ?  G " B? C  %Calculate the rate of reaction at 1.0, 0.75, 0.5 and 0.25 mol dm-3 Plot a graph of rate v [A] Calculate the time it takes for [A] to go from... 1.00 to 0.50 mol dm-3 0.50 to 0.25 mol dm-3 Deduce from the graph that the order wrt A is 1 Calculate the value and units of the rate constant, kr&@ b  i?H  0޽h ? ̙33y___PPT10Y+D=' = @B + 0L0 (    0D܉ ORATE DETERMINING STEP(2fdB  <D)@0A  0Pp@ 0 0dB @ <D)@X@  0X0@ 0 0   s *?n`8  CMany reactions consist of a series of separate stages. Each step has its own rate and rate constant. The overall rate of a multi-step process is governed by the slowest step (like a production line where overall output can be held up by a slow worker). This step is known as the RATE DETERMINING STEP. If there is more than one step, the rate equation may not contain all the reactants in its format.~@gH3i3e  B  s *޽h ? ̙33y___PPT10Y+D=' = @B +W  0L0 f^ 4( w 4 4 0L ORATE DETERMINING STEP(2fdB 4 <D)@0A 4 0Pp@ 0 0dB 4@ <D)@X@ 4 0X0@ 0 0g  4 s *?   )THE REACTION BETWEEN PROPANONE AND IODINE Iodine and propanone CH3COCH3 + I2 CH3COCH2I + HI react in the presence of acid The rate equation is... r = k [CH3COCH3] [H+] Why do H+ ions appear in the rate equation? Why does I2 not appear in the rate equation? *XnX*3  3 3333333 (   'f&6    vB  4 NDjJ?&B 4 s *޽h ? ̙33y___PPT10Y+D=' = @B +  0L0   `pK ( w p p 0X6 ORATE DETERMINING STEP(2fdB p <D)@0A p 0Pp@ 0 0dB p@ <D)@X@ p 0X0@ 0 0 p s *t<? "  \FTHE REACTION BETWEEN PROPANONE AND IODINE Iodine and propanone CH3COCH3 + I2 CH3COCH2I + HI react in the presence of acid The rate equation is... r = k [CH3COCH3] [H+] Why do H+ ions appear in The reaction is catalysed by acid the rate equation? [H+] affects the rate but is unchanged overall Why does I2 not appear The rate determining step doesn t involve I2 in the rate equation? *XnXw*3  3 3333333 J ,   9 f&6   e vB p NDjJ?&B p s *޽h ? ̙33y___PPT10Y+D=' = @B +h 0L0 w o pt ( w t t 0t ORATE DETERMINING STEP(2fdB t <D)@0A t 0Pp@ 0 0dB t@ <D)@X@ t 0X0@ 0 0x t s *?   THE REACTION BETWEEN PROPANONE AND IODINE Iodine and propanone CH3COCH3 + I2 CH3COCH2I + HI react in the presence of acid The rate equation is... r = k [CH3COCH3] [H+] Why do H+ ions appear in The reaction is catalysed by acid the rate equation? [H+] affects the rate but is unchanged overall Why does I2 not appear The rate determining step doesn t involve I2 in the rate equation? The slowest step of any multi-step reaction is known as the rate determining step and it is the species involved in this step that are found in the overall rate equation. Catalysts appear in the rate equation because they affect the rate but they do not appear in the stoichiometric equation because they remain chemically unchanged*XnXwO*3  3 3333333 J ,   9 Pf@6   s  3 vB t NDjJ?&B t s *޽h ? ̙33y___PPT10Y+D=' = @B + 0L0  xO(  x x 0 ORATE DETERMINING STEP(2fdB x <D)@0A x 0Pp@ 0 0dB x@ <D)@X@ x 0X0@ 0 0 x s *?h `FHYDROLYSIS OF HALOALKANES Haloalkanes (general formula RX) are RX + OH- ROH + X- hydrolysed by hydroxide ion to give alcohols. With many haloalkanes the rate equation is... r = k [RX][OH-] SECOND ORDER This is because both the RX and OH- must collide for a reaction to take place in ONE STEP HT`.@`(@3 $ --3 3330 -f3f3@       vB x NDjJ?,B x s *޽h ? ̙33y___PPT10Y+D=' = @B +  0L0     (    0 ORATE DETERMINING STEP(2fdB  <D)@0A  0Pp@ 0 0dB @ <D)@X@  0X0@ 0 0   s *?h) HYDROLYSIS OF HALOALKANES Haloalkanes (general formula RX) are RX + OH- ROH + X- hydrolysed by hydroxide ion to give alcohols. With many haloalkanes the rate equation is... r = k [RX][OH-] SECOND ORDER This is because both the RX and OH- must collide for a reaction to take place in ONE STEP but with others it only depends on [RX]... r = k [RX] FIRST ORDER The reaction has taken place in TWO STEPS... - the first involves breaking an R-X bond i) RX R+ + X- Slow - the second step involves the two ions joining ii) R+ + OH- ROH Fast The first step is slower as it involves bond breaking and energy has to be put in. The first order mechanism is favoured by tertiary haloalkanes because the hydroxide ion is hindered in its approach by alkyl groups if the mechanism involves the hydroxide ion and haloalkane colliding. HT`.@`(@Shd`23 $ -- 15*3 3  6  1 R 3 3t      >   w     vB   NDjJ?,vB  NDjJ?  vB  NDjJ?8 ]@ B  s *޽h ? ̙33y___PPT10Y+D=' = @B +  0L0    O (    0,2 ORATE DETERMINING STEP(2fdB  <D)@0A  0Pp@ 0 0dB @ <D)@X@  0X0@ 0 0@   s *B?u  DThe reaction H2O2 + 2H3O+ + 2I I2 + 4H2O takes place in 3 steps Step 1 H2O2 + I IO + H2O SLOW Step 2 IO + H3O+ HIO + H2O FAST Step 3 HIO + H3O+ + I I2 + 2H2O FAST The rate determining step is STEP 1 as it is the slowest E3333 3333 3333 333#        8vB   NDjJ?  vB   ND3jJ?. vB  NDjJ?D\ DvB  NDjJ?Z B  s *޽h ? ̙33y___PPT10Y+D=' = @B +E 0L0 TL|( w | | 0H ORATE DETERMINING STEP(2fdB | <D)@0A | 0Pp@ 0 0dB |@ <D)@X@ | 0X0@ 0 0 | s *?ud wThe reaction H2O2 + 2H3O+ + 2I I2 + 4H2O takes place in 3 steps Step 1 H2O2 + I IO + H2O SLOW Step 2 IO + H3O+ HIO + H2O FAST Step 3 HIO + H3O+ + I I2 + 2H2O FAST The rate determining step is STEP 1 as it is the slowest The reaction 2N2O5 4NO2 + O2 takes place in 3 steps Step 1 N2O5 NO2 + NO3 SLOW Step 2 NO2 + NO3 NO + NO2 + O2 FAST Step 3 NO + NO3 2NO2 from another Step 1 FAST The rate determining step is STEP 1 rate = k [N2O5]x3333 3333 3333 333#        83333 3333$      . 33333vB | NDjJ?  vB  | ND3jJ?. vB  | ND3jJ?& .& vB  | NDjJ?D\ DvB  | NDjJ?Z vB  | NDjJ?N PN vB | NDjJ? B vB | NDjJ? % B | s *޽h ? ̙33y___PPT10Y+D=' = @B +^  0L0     (    0 F OTHER TOPICS (2 fdB  <D)@0A  0Pp@ 0 0dB @ <D)@X@  0X0@ 0 0   s * ?N>  (Autocatalysis A small number of reactions appear to speed up, rather than slow down, for a time. This is because one of the products is acting as a catalyst and as more product is formed the reaction gets faster. One of the best known examples is the catalytic properties of Mn2+(aq) on the decomposition of MnO4(aq). You will notice it in a titration of KMnO4 with either hydrogen peroxide or ethanedioic (oxalic) acid. Molecularity The number of individual particles of the reacting species taking part in the rate determining step of a reaction e.g. A + 2B C + D molecularity is 3 - one A and two B s need to collide A 2B however has a molecularity of 1 - only one A is involved@ 3    1 = 3s  33 3 t         c   ! vB   NDjJ?w $w vB   NDjJ? k B  s *޽h ? ̙33 0L0 ~ @(   X @ @ <̙?8x@= 0 0 @ 0 ~ THE RATE EQUATION The Endr(28(26f<fff @ 0x5eT h. 2009 JONATHAN HOPTON & KNOCKHARDY PUBLISHING/(2/ fJ @ C "A KTRE: .H @ 0޽h ? ̙33r\@0"&w$,OG5>K`z{ȍg/@GRdϔp`E- M`k>z YJ0}Qc CYP{do 0 6_&/+ N$:l;p$&0G=h#ZL.9jHjm?Oh+'0T hp   No Slide TitleHOPTONJONATHAN HOPTON762Microsoft PowerPoint@0y6@@hNx@ {EGSg  )'    """)))UUUMMMBBB999|PP3f333f3333f3ffffff3f̙3ff333f333333333f33333333f33f3ff3f3f3f3333f33̙33333f333333f3333f3ffffff3f33ff3f3f3f3fff3ffffffffff3ffff̙fff3fffff3fff333f3f3ff3ff33f̙̙3̙ff̙̙̙3f̙3f333f3333f3ffffff3f̙3f3f3f333f3333f3ffffff3f̙3f3ffffffffff!___www4'A x(xKʦ """)))UUUMMMBBB999|PP3f3333f333ff3fffff3f3f̙f3333f3333333333f3333333f3f33ff3f3f3f3333f3333333f3̙33333f333ff3ffffff3f33f3ff3f3f3ffff3fffffffff3fffffff3f̙ffff3ff333f3ff33fff33f3ff̙3f3f3333f333ff3fffff̙̙3̙f̙̙̙3f̙3f3f3333f333ff3fffff3f3f̙3ffffffffff!___wwwýýýýýýýüýüýüýüýüýýýüýüýüýýýüýýýýýýýüýüýüýüýüýýýüýüýüýýýüýýýýýýýüýüýüýüýüýýýüýüýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýüýüýýýýýýýýýýýýüýýýüýýýý___________________________________ýýýüýýýýýýýüýüýüýýýýýý________________________________ýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýüýýýýýýýýýýüýýýýýüýýýüýýýýýýýýýýüýýýýýüýýýüýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýüýýýýýýýýýýýüýýýýýýýýýýýrsrýýýýýýýýýýüýýýýýýýýýýýýýýýKnýýýýýýýýýýýýýüýýýýýtnýýýýýýýüýýýýýýýýýýýnüýýüýüýýüýüýýýýüýüýýýüýýýüýýüýüýýrxKýýýýüýüýýýüýýýüýýüýüýýüýüýýýýüýüýýýýýýýýx-Onnýýýýýýýýýýýýýýýýýýýýýýýýýýýý-wnD-ýýýýýýýýýýýýýýýýýýýýýýýIsKEPOýýýýüýüýýýýýýýýýýüýýýýýýýýýýüýüÓtJVOýýýýýýýüýýýýýýýýýýüýüýýýýýýýýýýüýýxJ('Orýýýýýýýýýýýýýýýüýýýýýýýýý-xP'-OýýýýüýýýýýýýýýýýýýýýýýüxOO(OOüýýýýýýýýýýýüýüýüýýýýýýýüýýýýýýýý-OOP-ýüýüýýýýýýýüýýýýýýýýýýýüýüýüýýýýýýýrxOO'O'O'ýýýýýýýýýüýýýýýýýýýýüýýýýýýýýýüýx-ONOxrýýýýýüýýýýýýýýýüýýýýýýýýýýüýýO-O'-'O-ýýýýüýýýüýüýüýýýüýýýýýýýüýüýüýýýüýýýüýüýr--O'---OýýýýýýýüýüýüýýýüýýýüýüýüýýýüýýýýýýýüýýýýýýýO'-'-'-OýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýP---'--xýýýýýýýýýýýýýýýýýýýýýýýýýýýO-'-'-OýýýýýýýýüýýýüýýýýýýýüýüýýýüýýýýýýýýüýýsU-UOýýýýýýüýüýýýüýýýýýýýýüýýýüýýýýýýýüýüýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýüýýýýýüýýýüýýýýýýýýýýüýýýýýüýýýüýýýýýýýýýýüýýýýýüýýýüýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýüýýýýýýýýýýýýüýýýýýýýýýýýýüýýýýýýýýýýýýüýýýýýýýýýýýýüýýýýýýýýýýýýýýýýýýýýýüýýýýýýýýýýýýýýýýüýýýýýýýýýýýýýýýýüýýýýýýýýýüýüýýüýýýüýýüýüýýüýüýýýýüýüýýüýýýüýýüýüýýüýüýýýýüýüýýüýýýüýýüýüýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýüýýýýüýýýýýýýüýýýýüýüýýýýýüýýýýüýýýýýýýüýýýýüýüýýýýýüýýýýüýýýýýýýüýýýýüýýýýýýýýýýýýýýýüýýýýýýýýýýýýýýýýýýýýüýýýýýýýýýýýýýýýýýýýýüýýýýýýýýýýüýüýýýýýýýýýüýýýýýýýýýýüýüýýýýýýýýýüýýýýýýýýýýüýüýýýýýýýýýüýýýýýýýýýýýýýýýýýýýýüýýýýýýýýüýýýýýýýýýüýýýýýýýýüýýýýýýýýýüýýýýýýýýüýýýýüýýýýýýýüýüýüýüýüýýýüýüýüýýýüýýýýýýýüýüýüýüýüýýýüýüýüýýýüýýýýýýýüýüýüýüýüýýýüýüýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýüýüýýýýýýýýýýýýüýýýüýýýýýýýüýüýýýýýýýýýýýýüýýýüýýýýýýýüýüýýýýýýýýýýýýüýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýüýýýýýýýýýýüýýýýýüýýýüýýýýýýýýýýüýýýýýüýýýüýýýýýýýýýýüýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýüýýýýýýýýýýýüýýýýýýýýýýýýüýýýýýýýýýýýüýýýýýýýýýýýýüýýýýýýýýýýýýýýýýýýýýýýüýýýýýýýýýýýýýüýýýýýýýýýýýüýýüýüýýüýüýýýýüýüýýýüýüýýüýüýýüýüýýýýüýüýýýüýýýüýýüýüýýüýüýýýýüýüýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýüýüýýýýýýýýýýüýýýýýýýüýüýýýýýýýýýýüýýýýýýýýýýýýýýýýýýüýýýýýýýýýýýýýýýýýüýýýýýýýýýýýýýýýýýüüýýýýýýýýýýýüýüýüýýýýýýýüýýýýýýýýýýýüýüýüýýýýýýýüýýýýýýýýýýýüýüýüýýýýýýýýýýýýýýýýüýýýýýýýýýýüýýýýýýýýýüýýýýýýýýýýüýýýýýýýýýüýýýýýýýýýýüýýýýýýüýýýüýüýüýýýüýýýýýýýüýüýüýýýüýüýüýüýýýüýýýýýýýüýüýüýýýüýýýüýüýüýýýüýýýýýýýüýýýýý__ýýýýýýýýýýýýýýýýýýýýý_______________ý_____________________________ý___ýýýýýýýýýýýýýýýý______________________________________________________ýýýýýýýýüýýýüýýýýýýýüýü____ü_________________________ý________________________ýýýýýüýýýüýýýýýýýüýüýýýýý____________________________________________________ýýýýýýýýýýýýýýýýýýýýý_________________________________________________________ýýýýýýýýýýýýýýý________________________________________________________ýýüýýýýýüýýýüýýýý____________________________________________________ýýýýýüýýýüýýýýý_________________________________________________________ýýýýýýýýýýýýýýýý________ý_____ý________ý____________________ý___________ýýýýýýýýýýýýýýýýýýýýýýýýýüýýýýýýýýýýýýüýýýýýýýýýýýýüýýýýýýýýýýýýüýýýýýýýýýýýýüýýýýýýýýýýýýýýýýýýýýýüýýýýýýýýýýýýýýýýüýýýýýýýýýýýýýýýýüýýýýýýýýýüýüýýüýýýüýýüýüýýü____________________üý___________________________ýüýüýýüýýýüýýüýüýýýý__________________________________________________ýýýýýýýýýýýýýýýýýý________________ý________________________ýýýýýýýýýýýýýýý________________________________________ýüýýýýüýýýýýýýüýýýýüýü_____________________ýý_______________ý____________üýýýýüýýýýýýýüýýýýüýýý_____________________ý____________________ý________ýýýýýýýýüýýýýýýýýýý________________ý________________________ýýýýýüýýýýýýýýýý__________________________________________________________üýüýýýýýýýýýüýýýýýýý__________________________ý_______ý________________________üýýýýýýýýýüýýýýýýýýýýýýýýýýýýýýüýýýýýýýýüýýýýýýýýýüýýýýýýýýüýýýýýýýýýüýýýýýýýýüýýýýüýýýýýýýüýüýüýüýüýýýüýüýüýýýüýýýýýýýüýüýüýüýüýýýüýüýüýýýüýýýýýýýüýüýüýüýüýýýüýüýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýüýüýýýýýýýýýýýýüýýýüýýýýýýýüýüýýýýýýýýýýýýüýýýüýýýýýýýüýüýýýýýýýýýýýýüýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýýý՜.+,D՜.+,<    On-screen Show H C JTimes New Roman Arial BlackArialTahoma Courier NewSymbolDefault DesignSlide 1Slide 2Slide 3Slide 4Slide 5Slide 6Slide 7Slide 8Slide 9 Slide 10 Slide 11 Slide 12 Slide 13 Slide 14 Slide 15 Slide 16 Slide 17 Slide 18 Slide 19 Slide 20 Slide 21 Slide 22 Slide 23 Slide 24 Slide 25 Slide 26 Slide 27 Slide 28 Slide 29 Slide 30 Slide 31 Slide 32 Slide 33 Slide 34 Slide 35 Slide 36 Slide 37 Slide 38 Slide 39 Slide 40 Slide 41 Slide 42 Slide 43 Slide 44 Slide 45 Slide 46 Slide 47 Slide 48 Slide 49 Slide 50 Slide 51 Slide 52 Slide 53 Slide 54 Slide 55 Slide 56 Slide 57 Slide 58 Slide 59 Slide 60 Slide 61 Slide 62 Slide 63 Slide 64 Slide 65 Slide 66 Slide 67  Fonts UsedDesign Template Slide TitlesC 8@ _PID_HLINKSAB316,3,Slide 3316,3,Slide 3313,46,Slide 46351,5,Slide 5383,8,Slide 8352,7,Slide 7345,11,Slide 11361,18,Slide 18397,41,Slide 41350,58,Slide 58402,44,Slide 44'_яJONATHAN HOPTONJONATHAN HOPTON  !"#$%&'()*+,-./0123456789:;<=>?@ABCDEFGHIJKLMNOPQRSTUVWXYZ[\^_`abcdefghijklmnopqrstuvwxyz{|}~      !"#$%&'()*+,-./0123456789:;<=>?@ABCDEFGHIJKLMNOPQRSTUVWXYZ[\]^_`abcdefghijklmnopqrstuvwxyz{|}~      !"#$&'()*+,-./0123456789:;<=>?@ABCDEFGHIJKLMNOQRSTUVWYZ[\]^_fRoot EntrydO)PicturesCurrent UserXSummaryInformation(%TPowerPoint Document(]DocumentSummaryInformation8P