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S  0h ``  X*  0  `   Z*  0Í `   Z*z  bA޽h @ ?Parchment ̙33 Default Design 0 zr@ ( )   0 P    P*    0     R*  d  c $ ?    0xʈ  @  RClick to edit Master text styles Second level Third level Fourth level Fifth level!     S  6 `P   P*    6݈ `   R*  H  0޽h ? ̙3380___PPT10.:TE 0X(   $+ X X 0\ P    X*  X 0`     Z*  X 6`e `P   X*  X 6i `   Z* H X 0޽h ? ̙3380___PPT10.:T 0L0 P(B(     0 ~ .ENTHALPY CHANGES A guide for A level students b.(28(26f<fff & 0 OKNOCKHARDY PUBLISHING(2fJ ' C "A KTREj x S _ ( 0"`>B 2008 SPECIFICATIONS<G$G   B  s *޽h ? ̙33 0L0 )!`(    00O LENTHALPY CHANGES (2$fS  0ap0P INTRODUCTION This Powerpoint show is one of several produced to help students understand selected topics at AS and A2 level Chemistry. It is based on the requirements of the AQA and OCR specifications but is suitable for other examination boards. Individual students may use the material at home for revision purposes or it may be used for classroom teaching if an interactive white board is available. Accompanying notes on this, and the full range of AS and A2 topics, are available from the KNOCKHARDY SCIENCE WEBSITE at... www.knockhardy.org.uk/sci.htm Navigation is achieved by... either clicking on the grey arrows at the foot of each page or using the left and right arrow keys on the keyboard (2 2(2 2  75, dB  <D?  dB  <D)@0A  0Pp@ 0 0B  s *޽h ? ̙33N 0L0 p|(    0) LENTHALPY CHANGES (2$fJ  0P0% `CONTENTS Thermodynamics Enthalpy changes Standard enthalpy values Standard enthalpy of formation Standard enthalpy of combustion Enthalpy of neutralisation Bond dissociation enthalpy Hess s law Calculating enthalpy changes using bond enthalpies Calculating enthalpy changes using enthalpy of formation values Calculating enthalpy changes using enthalpy of combustion values Practical measurement Check listL (2 2 dB  <D)@0A  0Pp@ 0 0dB  @ <D)@X@   0X0@ 0 0   6g ? ,@ 0 0  <g ?e@ 0 0  <g ?2 @  0 0  <g ?  ~@ 0 0  <g ?X 1@ 0 0  <g ?  @ 0 0  <g ?P 3 @& 0 0  <g ?$  @- 0 0  <g ? e @7 0 0  <g ? a: @> 0 0  <g ?l  @E 0 0  <g ?1  @L 0 0  <g ? @U 0 0B  s *޽h ? ̙33 0L0 80 ((  ( ( 0ĖH ZBefore you start it would be helpful to& be able to balance equations be confident with simple arithmetical operationsH*(2R 2)R ( 0x LENTHALPY CHANGES (2$fdB ( <D)@0A ( 0Pp@ 0 0dB  (@ <D)@X@  ( 0X0@ 0 0B ( s *޽h ? ̙33 0L0 %( w   0s JTHERMODYNAMICS (2fdB  <D)@0A  0Pp@ 0 0dB @ <D)@X@  0X0@ 0 0  0  First Law Energy can be neither created nor destroyed but It can be converted from one form to another Energy changes all chemical reactions are accompanied by some form of energy change changes can be very obvious (e.g. coal burning) but can go unnoticed Exothermic Energy is given out Endothermic Energy is absorbed Examples Exothermic combustion of fuels respiration (oxidation of carbohydrates) Endothermic photosynthesis thermal decomposition of calcium carbonateh; 23@@Z)@.@$@/@ af f f  fL f?  B  s *޽h ? ̙33 0L0 ?7 ( w   0s !THERMODYNAMICS - ENTHALPY CHANGESD"(2fdB  <D)@0A  0Pp@ 0 0dB @ <D)@X@  0X0@ 0 0  0 ,LEnthalpy a measure of the heat content of a substance at constant pressure you cannot measure the actual enthalpy of a substance you can measure an enthalpy CHANGE written as the symbol DH ,  delta H  Enthalpy change (DH) = Enthalpy of products - Enthalpy of reactants @PK@ 5  ' B  s *޽h ? ̙33  0L0 ) !   (  dB  <D)@0A  0Pp@ 0 0dB @ <D)@X@  0X0@ 0 0^  0h, <Enthalpy a measure of the heat content of a substance at constant pressure you cannot measure the actual enthalpy of a substance you can measure an enthalpy CHANGE written as the symbol DH ,  delta H  Enthalpy change (DH) = Enthalpy of products - Enthalpy of reactants Enthalpy of reactants > products DH = - ive EXOTHERMIC Heat given out x@PT@P,@"@A @ 5   f*f &v  & 8 5J  5J V   C &Aexogrg5     0% J  LREACTION CO-ORDINATE(2    6@*A+  @ENTHALPY (2    0-s !THERMODYNAMICS - ENTHALPY CHANGESD"(2fB  s *޽h ? ̙33e 0L0   0 ( w dB  <D)@0A  0Pp@ 0 0dB @ <D)@X@  0X0@ 0 0x  0X8, Enthalpy a measure of the heat content of a substance at constant pressure you cannot measure the actual enthalpy of a substance you can measure an enthalpy CHANGE written as the symbol DH ,  delta H  Enthalpy change (DH) = Enthalpy of products - Enthalpy of reactants Enthalpy of reactants > products Enthalpy of reactants < products DH = - ive DH = + ive EXOTHERMIC Heat given out ENDOTHERMIC Heat absorbed@PT@PM@9@A>@ 5   fKf    @    ? N   C &Aexogrg5     0Z J  LREACTION CO-ORDINATE(2   6^A+  @ENTHALPY (2  8 m 5  5m  X   C (Aendogrg5    0bz    LREACTION CO-ORDINATE(2   6gm A+  @ENTHALPY (2    0Tes !THERMODYNAMICS - ENTHALPY CHANGESD"(2fB  s *޽h ? ̙33  0L0   0' (  0dB 0 <D)@0A 0 0Pp@ 0 0dB 0@ <D)@X@ 0 0X0@ 0 0  0 0u RWhy a standard? enthalpy values vary according to the conditions a substance under these conditions is said to be in its standard state ... Pressure:- 100 kPa (1 atmosphere) A stated temperature usually 298K (25C) " as a guide, just think of how a substance would be under normal lab conditions " assign the correct subscript [(g), (l) or (s) ] to indicate which state it is in " any solutions are of concentration 1 mol dm-3 " to tell if standard conditions are used we modify the symbol for DH . Enthalpy Change Standard Enthalpy Change (at 298K) >@(@( f0f FAf&     0 0s USTANDARD ENTHALPY CHANGES (2fP 0 C (Adhsigngy t B 0 s *޽h ? ̙33d  0L0   @8 ( w 8dB 8 <D)@0A 8 0Pp@ 0 0dB 8@ <D)@X@ 8 0X0@ 0 0  8 0(s ZSTANDARD ENTHALPY OF FORMATION (2fp 8 0  Definition The enthalpy change when ONE MOLE of a compound is formed in its standard state from its elements in their standard states. Symbol DHf Values Usually, but not exclusively, exothermic Example(s) C(graphite) + O2(g)    > CO2(g) H2(g) + O2(g)    > H2O(l) 2C(graphite) + O2(g) + 3H2(g)    > C2H5OH(l) Notes Only ONE MOLE of product on the RHS of the equation Elements In their standard states have zero enthalpy of formation. Carbon is usually taken as the graphite allotrope. >N  2       C= M B 8 s *޽h ? ̙33 0L0 G? 4 ( w 4dB 4 <D)@0A 4 0Pp@ 0 0dB 4@ <D)@X@ 4 0X0@ 0 0  4 0t Z Definition The enthalpy change when ONE MOLE of a substance undergoes complete combustion under standard conditions. All reactants and products are in their standard states. Symbol DHc Values Always exothermic Example(s) C(graphite) + O2(g)    > CO2(g) H2(g) + O2(g)    > H2O(l) C2H5OH(l) + 3O2(g)    > 2CO2(g) + 3H2O(l) Notes Always only ONE MOLE of what you are burning on the LHS of the equation To aid balancing the equation, remember... you get one carbon dioxide molecule for every carbon atom in the original and one water molecule for every two hydrogen atoms When you have done this, go back and balance the oxygen.N@,@:          I9+:    4 0זs [STANDARD ENTHALPY OF COMBUSTION (2 fB 4 s *޽h ? ̙33 0L0  < ( w <dB < <D)@0A < 0Pp@ 0 0dB <@ <D)@X@ < 0X0@ 0 0  < 0D.td  Definition The enthalpy change when ONE MOLE of water is formed from its ions in dilute solution. Values Exothermic Equation H+(aq) + OH(aq)    > H2O(l) Notes A value of -57kJ mol-1 is obtained when strong acids react with strong alkalis. See later slides for practical details of measurementz-6 gff ffff F6@        < 0(Gts VENTHALPY OF NEUTRALISATION (2fB < s *޽h ? ̙33r  0L0   @ ( w @dB @ <D)@0A @ 0Pp@ 0 0dB @@ <D)@X@ @ 0X0@ 0 0  @ 0Ot! "&Definition Energy required to break ONE MOLE of gaseous bonds to form gaseous atoms. Values Endothermic - Energy must be put in to break any chemical bond Examples Cl2(g)    > 2Cl(g) O-H(g)    > O(g) + H(g) Notes " strength of bonds also depends on environment; MEAN values quoted " making bonds is exothermic as it is the opposite of breaking a bond " for diatomic gases, bond enthalpy = 2 x enthalpy of atomisation " smaller bond enthalpy = weaker bond = easier to break @JIf      @ 0kts VBOND DISSOCIATION ENTHALPY (2fB @ s *޽h ? ̙33& 0L0  T(  dB  <D)@0A  0Pp@ 0 0dB @ <D)@X@  0X0@ 0 0.  0,st! Definition Energy required to break ONE MOLE of gaseous bonds to form gaseous atoms. Values Endothermic - Energy must be put in to break any chemical bond Examples Cl2(g)    > 2Cl(g) O-H(g)    > O(g) + H(g) Notes " strength of bonds also depends on environment; MEAN values quoted " making bonds is exothermic as it is the opposite of breaking a bond " for diatomic gases, bond enthalpy = 2 x enthalpy of atomisation " smaller bond enthalpy = weaker bond = easier to break Mean Values H-H 436 H-F 562 N-N 163 C-C 346 H-Cl 431 N=N 409 C=C 611 H-Br 366 NN 944 CC 837 H-I 299 P-P 172 C-O 360 H-N 388 F-F 158 C=O 743 H-O 463 Cl-Cl 242 C-H 413 H-S 338 Br-Br 193 C-N 305 H-Si 318 I-I 151 C-F 484 P-H 322 S-S 264 C-Cl 338 O-O 146 Si-Si 176L@JIf   DD"i    d  a       0Фts VBOND DISSOCIATION ENTHALPY (2f"  6t?   Average (mean) values are quoted because the actual value depends on the environment of the bond i.e. where it is in the molecule 2f   6$t?I [ pUNITS = kJ mol-1D 2B  s *޽h ? ̙33 0L0 NFP(  PdB P <D)@0A P 0Pp@ 0 0dB P@ <D)@X@ P 0X0@ 0 0n  P 0tf   The enthalpy change is independent of the path taken How The enthalpy change going from A to B can be found by adding the values of the enthalpy changes for the reactions A to X, X to Y and Y to B. DHr = DH1 + DH2 + DH3 7@"@7ufffffffffffff&  !   P 0ts PHESS S LAW (2 fV P C .Ahessring1g DB P s *޽h ? ̙33  0L0 ` X  (  dB  <D)@0A  0Pp@ 0 0dB @ <D)@X@  0X0@ 0 0  0Tt  The enthalpy change is independent of the path taken How The enthalpy change going from A to B can be found by adding the values of the enthalpy changes for the reactions A to X, X to Y and Y to B. DHr = DH1 + DH2 + DH3 If you go in the opposite direction of an arrow, you subtract the value of the enthalpy change. e.g. DH2 = - DH1 + DHr - DH3 The values of DH1 and DH3 have been subtracted because the route involves going in the opposite direction to their definition.7@"@7ufffffffffffffsffffffffff@       0ts PHESS S LAW (2 fV  C .Ahessring1g DB  s *޽h ? ̙33h 0L0  (  dB  <D)@0A  0Pp@ 0 0dB @ <D)@X@  0X0@ 0 0~  0v   The enthalpy change is independent of the path taken Use applying Hess s Law enables one to calculate enthalpy changes from other data, such as... changes which cannot be measured directly e.g. Lattice Enthalpy enthalpy change of reaction from bond enthalpy enthalpy change of reaction from DHc enthalpy change of formation from DHf07G@@D@@7_AfSffff%ffff `    0(vs PHESS S LAW (2 fB  s *޽h ? ̙33  0L0 S K  p (  pdB p <D)@0A p 0Pp@ 0 0dB p@ <D)@X@ p 0X0@ 0 0. p 08BvO Theory Imagine that, during a reaction, all the bonds of reacting species are broken and the individual atoms join up again but in the form of products. The overall energy change will depend on the difference between the energy required to break the bonds and that released as bonds are made. energy released making bonds > energy used to break bonds ... EXOTHERMIC energy used to break bonds > energy released making bonds ... ENDOTHERMIC)PPOPPP@"B A     p 0Nvs )Enthalpy of reaction from bond enthalpies *(2*f) V p C .Ahessring2gGA   p 0Ov&  Step 1 Energy is put in to break bonds to form separate, gaseous atoms Step 2 The gaseous atoms then combine to form bonds and energy is released its value will be equal and opposite to that of breaking the bonds Applying Hess s Law DHr = Step 1 + Step 2DG2F7Hf3f  ff&   B p s *޽h ? ̙33B 0L0  |@X(  |dB | <D)@0A | 0Pp@ 0 0dB |@ <D)@X@ | 0X0@ 0 0  | 0gvs )Enthalpy of reaction from bond enthalpies *(2*f)  | 6lv?S  R DH = S bond enthalpies  S bond enthalpies of reactants of productsz_;< | 0xv -  ,Alternative view Step 1 Energy is put in to break bonds to form separate, gaseous atoms. Step 2 Gaseous atoms then combine to form bonds and energy is released; its value will be equal and opposite to that of breaking the bonds DHr = Step 1 - Step 2 Because, in Step 2 the route involves going in the OPPOSITE DIRECTION to the defined change of bond enthalpy, it s value is subtracted.f@Mf fff3f37 &    8  OJ  |  Xu  | 60v * u+SUM OFTHE BOND ENTHALPIES OF THE REACTANTS.,(2+ f   | 6vK w A REACTANTS (2  ^  | S .A cyclebeY  | 6v   @PRODUCTS (2   | 6vF O =ATOMS(2  | 6v* FDH((2  | 64v&J, d*SUM OFTHE BOND ENTHALPIES OF THE PRODUCTS+ 2+ 2 | BܪvGpHI3%2S B B | s *޽h ? ̙334 0L0 0 tb( " tdB t <D)@0A t 0Pp@ 0 0dB t@ <D)@X@ t 0X0@ 0 0 t <̙? x @ 0 0 t 0 v =Calculate the enthalpy change for the hydrogenation of ethene >H>&7     t 0vs )Enthalpy of reaction from bond enthalpies *(2*f) V  t C .Ahesscalc1gAtB t s *޽h ? ̙33 0L0 z (  dB  <D)@0A  0Pp@ 0 0dB @ <D)@X@  0X0@ 0 0  <̙? x @ 0 0  0v =Calculate the enthalpy change for the hydrogenation of ethene >H>&7     0vs )Enthalpy of reaction from bond enthalpies *(2*f)    0v9  NDH2 1 x C=C bond @ 611 = 611 kJ 4 x C-H bonds @ 413 = 1652 kJ 1 x H-H bond @ 436 = 436 kJ Total energy to break bonds of reactants = 2699 kJ T*@(4{4V   C .Ahesscalc1gAtB  s *޽h ? ̙33*  0L0   P X (  dB  <D)@0A  0Pp@ 0 0dB @ <D)@X@  0X0@ 0 0  <̙? x @' 0 0  0v =Calculate the enthalpy change for the hydrogenation of ethene >H>&7     0vs )Enthalpy of reaction from bond enthalpies *(2*f)    0, { DH2 1 x C=C bond @ 611 = 611 kJ 4 x C-H bonds @ 413 = 1652 kJ 1 x H-H bond @ 436 = 436 kJ Total energy to break bonds of reactants = 2699 kJ DH3 1 x C-C bond @ 346 = 346 kJ 6 x C-H bonds @ 413 = 2478 kJ Total energy to break bonds of products = 2824 kJ Applying Hess s Law DH1 = DH2  DH3 = (2699  2824) =  125 kJTT*@(4-)@(3xS@{4< 1  V   C .Ahesscalc1gAtB  s *޽h ? ̙33B 0L0 @!\ `(  \dB \ <D)@0A \ 0Pp@ 0 0dB \@ <D)@X@ \ 0X0@ 0 0  \ 0{1Pj  QIf you formed the products from their elements you should need the same amounts of every substance as if you formed the reactants from their elements. Enthalpy of formation tends to be an exothermic processF@9H  8  V \ C .A hessring4gJg !\ 0hvs 1Enthalpy of reaction from enthalpies of formation 2(22f1 B \ s *޽h ? ̙33t 0L0  P(  dB  <D)@0A  0Pp@ 0 0dB @ <D)@X@  0X0@ 0 0$   0'{   Step 1 Energy is released as reactants are formed from their elements. Step 2 Energy is released as products are formed from their elements. DHr = - Step 1 + Step 2 or Step 2 - Step 1 In Step 1 the route involves going in the OPPOSITE DIRECTION to the defined enthalpy change, it s value is subtracted.C@xIfF3f ff3f3ff3f*"&   =F v 3    X  C (A dhfring {J  0@{v ]c  |2SUM OFTHE ENTHALPIES OF FORMATION OF THE REACTANTS.3(22 f   0F{   A REACTANTS (2    0J{ d" @PRODUCTS (2    0&{> 3 @ELEMENTS (2    00R{ :j  FDH((2   0V{i U  l2SUM OFTHE ENTHALPIES OF FORMATION OF THE PRODUCTS3 23   0tZ{s 1Enthalpy of reaction from enthalpies of formation 2(22f1 2  B>{GpHI3%^ B $  BTT{? d< r DH = S DHf of products  S DHf of reactants :   ,B  s *޽h ? ̙33 0L0 ;3` (  dB  <D)@0A  0Pp@ 0 0dB @ <D)@X@  0X0@ 0 0   0z{ BSample calculation Calculate the standard enthalpy change for the following reaction, given that the standard enthalpies of formation of water, nitrogen dioxide and nitric acid are -286, +33 and -173 kJ mol-1 respectively; the value for oxygen is ZERO as it is an element 2H2O(l) + 4NO2(g) + O2(g)    > 4HNO3(l)x@:H 0   N   0,{ G JBy applying Hess s Law ... The Standard Enthalpy of Reaction DHr will be... PRODUCTS REACTANTS [ 4 x DHf of HNO3 ] minus [ (2 x DHf of H2O) + (4 x DHf of NO2) + (1 x DHf of O2) ] DHr = 4 x (-173) - 2 x (-286) + 4 x (+33) + 0 ANSWER = - 252 kJN@5@@=  ffffff fffff fffff ffffff  L d       0{s 1Enthalpy of reaction from enthalpies of formation 2(22f1 $  B({?y r DH = S DHf of products  S DHf of reactants :   ,B  s *޽h ? ̙33 0L0 pT( w TdB T <D)@0A T 0Pp@ 0 0dB T@ <D)@X@ T 0X0@ 0 0V T C .A hessring3g T 0D{sg  2Enthalpy of reaction from enthalpies of combustion 3(23f2  T 0{?  If you burned all the products you should get the same amounts of oxidation products such a CO2 and H2O as if you burned the reactants. Enthalpy of combustion is an exothermic process @0H@^  "  0   B T s *޽h ? ̙33 0L0 me (  dB  <D)@0A  0Pp@ 0 0dB @ <D)@X@  0X0@ 0 0$   B{? HyK r DH = S DHc of reactants  S DHc of products :    ,   0sg  2Enthalpy of reaction from enthalpies of combustion 3(23f2 .   0tyP %  \Step 1 Energy is released as reactants undergo combustion. Step 2 Energy is released as products undergo combustion. DHr = Step 1 - Step 2 Because, in Step 2 the route involves going in the OPPOSITE DIRECTION to the defined change of Enthalpy of Combustion, it s value is subtracted.(/@=f;f3f3f&{   ?8  X   pa   6aX {3SUM OFTHE ENTHALPIES OF COMBUSTION OF THE REACTANTS,4 23 f   6T 6 A REACTANTS (2    6l! r& @PRODUCTS (2    6" V  HOXIDATION PRODUCTS   6&pdK FDH((2   6,h C\I  3SUM OFTHE ENTHALPIES OF COMBUSTION OF THE PRODUCTS24(23  ^  S .AdhcringM-' 2  H3GpHI3%L%  B B  s *޽h ? ̙33Q 0L0 `X@ ( `@`@ dB  <D)@0A  0Pp@ 0 0dB @ <D)@X@  0X0@ 0 0   03sg  2Enthalpy of reaction from enthalpies of combustion 3(23f2 D   0=  Sample calculation Calculate the standard enthalpy of formation of methane; the standard enthalpies of combustion of carbon, hydrogen and methane are -394, -286 and -890 kJ mol-1 . C(graphite) + 2H2(g)    > CH4(g)@/H     ,   0K!  By applying Hess s Law ... The Standard Enthalpy of Reaction DHr will be... REACTANTS PRODUCTS [ (1 x DHc of C) + (2 x DHc of H2) ] minus [ 1 x DHc of CH4] DHr = 1 x (-394) + 2 x (-286) - 1 x (-890) ANSWER = - 76 kJ mol-1O@.@@=   fffff fffff f ffffff D N  %  $  Bdo?c` r DH = S DHc of reactants  S DHc of products :    ,B  s *޽h ? ̙33y___PPT10Y+D=' = @B + 0L0  `(  `dB ` <D)@0A ` 0Pp@ 0 0dB `@ <D)@X@ ` 0X0@ 0 0  ` 0!  8Calorimetry involves the practical determination of enthalpy changes usually involves heating (or cooling) known amounts of water water is heated up reaction is EXOTHERMIC water cools down reaction is ENDOTHERMIC Calculation The energy required to change the temperature of a substance can be calculated using... q = m x c x DT where q = heat energy kJ m = mass kg c = Specific Heat Capacity kJ K -1 kg -1 [ water is 4.18 ] DT = change in temperature K@|@ {)( [ "$       ` 04s VMEASURING ENTHALPY CHANGES (2fB ` s *޽h ? ̙33  0L0 G ? (  dB  <D)@0A  0Pp@ 0 0dB @ <D)@X@  0X0@ 0 0  0! _Calorimetry involves the practical determination of enthalpy changes usually involves heating (or cooling) known amounts of water water is heated up reaction is EXOTHERMIC water cools down reaction is ENDOTHERMIC Calculation The energy required to change the temperature of a substance can be calculated using... q = m x c x DT where q = heat energy kJ m = mass kg c = Specific Heat Capacity kJ K -1 kg -1 [ water is 4.18 ] DT = change in temperature K Example On complete combustion, 0.18g of hexane raised the temperature of 100g water from 22C to 47C. Calculate its enthalpy of combustion. Heat absorbed by the water (q) = 0.1 x 4.18 x 25 = 10.45 kJ Moles of hexane burned = mass / Mr = 0.18 / 86 = 0.00209 Enthalpy change = heat energy / moles = 10.45 / 0.00209 ANS = 5000 kJ mol -1r@|@@ZPxB}         0ds VMEASURING ENTHALPY CHANGES (2fB  s *޽h ? ̙33M 0L0  x{(  xdB x <D)@0A x 0Pp@ 0 0dB x@ <D)@X@ x 0X0@ 0 0 x 0\Հ^!  Example 1 - graphical The temperature is taken every half minute before mixing the reactants. Reactants are mixed after three minutes. Further readings are taken every half minute as the reaction mixture cools. Extrapolate the lines as shown and calculate the value of DT. ^@@    x 0߀s VMEASURING ENTHALPY CHANGES (2fR  x C *Adhgraphg/:  B x s *޽h ? ̙33 0L0 (  dB  <D)@0A  0Pp@ 0 0dB @ <D)@X@  0X0@ 0 0;  0^q Example 1 - graphical The temperature is taken every half minute before mixing the reactants. Reactants are mixed after three minutes. Further readings are taken every half minute as the reaction mixture cools. Extrapolate the lines as shown and calculate the value of DT. Example calculation When 0.18g of hexane underwent complete combustion, it raised the temperature of 100g (0.1kg) water from 22C to 47C. Calculate its enthalpy of combustion. Heat absorbed by the water (q) = m C DT = 0.1 x 4.18 x 25 = 10.45 kJ Moles of hexane burned = mass / Mr = 0.18 / 86 = 0.00209 Enthalpy change = heat energy / moles = 10.45 / 0.00209 = 5000 kJ mol -1@@*      0s VMEASURING ENTHALPY CHANGES (2fR  C *Adhgraphg/:  B  s *޽h ? ̙33  0L0   d, (  ddB d <D)@0A d 0Pp@ 0 0dB d@ <D)@X@ d 0X0@ 0 0  d 0e B 2 d  d 0p  Example 2 25cm3 of 2.0M HCl was added to 25cm3 of 2.0M NaOH in an insulated beaker. The initial temperature of both solutions was 20C. The highest temperature reached by the solution was 33C. Calculate the Molar Enthalpy of Neutralisation. [The specific heat capacity (c) of water is 4.18 kJ K -1 kg -1] NaOH + HCl   > NaCl + H2O | @,@P24@PF                      d 00s VMEASURING ENTHALPY CHANGES (2fB d s *޽h ? ̙33F  0L0   t (  dB  <D)@0A  0Pp@ 0 0dB @ <D)@X@  0X0@ 0 0  06e B 2   0:  L Example 2 25cm3 of 2.0M HCl was added to 25cm3 of 2.0M NaOH in an insulated beaker. The initial temperature of both solutions was 20C. The highest temperature reached by the solution was 33C. Calculate the Molar Enthalpy of Neutralisation. [The specific heat capacity (c) of water is 4.18 kJ K -1 kg -1] NaOH + HCl   > NaCl + H2O Temperature rise (DT) = 306K  293K = 13K Volume of resulting solution = 25 + 25 = 50cm3 = 0.05 dm3 Equivalent mass of water = 50g = 0.05 kg (density is 1g per cm3) Heat absorbed by the water (q) = m x c x DT = 0.05 x 4.18 x 13 = 2.717 kJ  @,@P24@PFv@2r@@        P  H '             )   0es VMEASURING ENTHALPY CHANGES (2fB  s *޽h ? ̙33 0L0 D( w dB  <D)@0A  0Pp@ 0 0dB @ <D)@X@  0X0@ 0 0  0Lte B 2 |  0v  &Example 2 25cm3 of 2.0M HCl was added to 25cm3 of 2.0M NaOH in an insulated beaker. The initial temperature of both solutions was 20C. The highest temperature reached by the solution was 33C. Calculate the Molar Enthalpy of Neutralisation. [The specific heat capacity (c) of water is 4.18 kJ K -1 kg -1] NaOH + HCl   > NaCl + H2O Temperature rise (DT) = 306K  293K = 13K Volume of resulting solution = 25 + 25 = 50cm3 = 0.05 dm3 Equivalent mass of water = 50g = 0.05 kg (density is 1g per cm3) Heat absorbed by the water (q) = m x c x DT = 0.05 x 4.18 x 13 = 2.717 kJ Moles of HCl reacting = 2 x 25/1000 = 0.05 mol Moles of NaOH reacting = 2 x 25/1000 = 0.05 mol Moles of water produced = 0.05 mol Enthalpy change per mol (DH) = heat energy / moles of water = 2.717 / 0.05 = 54.34 kJ mol -1 @,@P24@PFv@2r@x@*@l@        P  H ' @             2  8     0Hs VMEASURING ENTHALPY CHANGES (2fB  s *޽h ? ̙33  0L0 \ T   (    0q JREVISION CHECK (2  0 ZWhat should you be able to do? (2   0   Explain the difference between an endothermic and exothermic reaction Understand the reasons for using standard enthalpy changes Recall the definitions of enthalpy of formation and combustion Write equations representing enthalpy of combustion and formation Recall and apply Hess s law Recall the definition of bond dissociation enthalpy Calculate standard enthalpy changes using bond enthalpy values Calculate standard enthalpy changes using enthalpies of formation and combustion Know simple calorimetry methods for measuring enthalpy changes Calculate enthalpy changes from calorimetry measurements pb 2f@f f1ff8ff?ff ff0f f6f fHff;f f0ff, H    0X؄   ~,CAN YOU DO ALL OF THESE? YES NO6- 2!    0 H @P 0 0   0f h @U 0 0dB   <D)@0A  0Pp@ 0 0dB @ <D)@X@  0X0@ 0 0H  0޽h ? ̙33 0L0  ;(  9  0  YYou need to go over the relevant topic(s) again Click on the button to return to the menudG(282/$f f) f  s * x  P @S 0 0dB  <D)@0A  0Pp@ 0 0dB @ <D)@X@   0X0@ 0 0H  0޽h ? ̙33 0L0  D(    0΄xK (WELL DONE! 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