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given the following standard heats of reaction at constant pressure: (i) Heat of formation of water =-68.3 kcal (ii) Heat of combustion of acetylene=-310.6 kcal (iii) Heat of combustion of ethylene =-337.2 kcal Calculate the heat of reaction for the hydrogenation of acetylene at constant volume (25^(@)C). |
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Answer» Solution :Given that (i) `H_(2)(g)+(1)/(2)O_(2)(g) to H_(2)O(1),DELTAH=-68.3kcal` (ii) `C_(2)H_(2)(g)+2(1)/(2)O_(2)(g) to 2CO_(2)(g)+H_(2)O(1),DeltaH=-310.6kcal` (iii) `C_(2)H_(4)(g)+3O_(2)(g) to 2CO_(2)(g)+2H_(2)O(1),DeltaH=-337.2kcal.` we have to calculate `Deltamu` for the equation, (iv) `C_(2)H_(2)(g)+H_(2)(g) to C_(2)H_(4)(g),Deltamu=?` `C_(2)H_(2)` in eqns. (ii) and (iv), and `H_(2)` in eqns. (i) and (iv) are on the same sides, whereas `C_(2)H_(4)` in eqns. (iii) and (iv) is on opposite sides. Thus APPLYING [Eqn. (i) +Eqn. (ii) - Eqn. (iii)] we get `H_(2)(g)+(1)/(2)O_(2)(g)+C_(2)H_(2)(g)+2(1)/(2)O_(2)(g)-C_(2)H_(2)(g)-H_(2)(g) to ` `H_(2)O(1)+2CO_(2)(g)+H_(2)O(1)-2CO_(2)(g)-2H_(2)O(1),` `DeltaH=-68.3+(-310.6)-(-337.2)` or `C_(2)H_(2)(g)+H_(2)(g) to C_(2)H_(4)(g),DeltaH=-41.7kcal.` `DeltaH` is the heat of reaction at constant pressure and to calculate heat at constant volume, we determine `Deltamu` by using equation, `DeltaH=Deltamu+Deltan_(g)RT...(Eqn. 8)` `Deltan_(g)= "moles of gaseous product" - "moles of gaseous reactant" ` `=1-2=-1` `R=0.002kcal//K// "mole" ` `T=(273+(25)=298K` Thus we have, `Deltamu=-41.7-(-1xx0.002xx298)` `=-41.104kcal.` [Note: if the heat of reaction is given in KJ, R shouldbe taken as ` 8.314xx10^(-3)kJ//K// "mole"` ] |
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