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Question

Given the following standard heats of reactions:
(a) heat of formation of water =68.3 kcal, (b) heat of combustion of C2H2=310.6 kcal and (c) heat of combustion of ethylene=337.2 kcal. Calculate the heat of the reaction for the hydrogenation of acetylene at constant volume and at 25oC.

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Solution

The required equation is
C2H2(g)+H2(g)C2H4(g);ΔHf=?
Given
(a) H2(g)+12O2(g)H2O(l);(ΔH=68.3kcal)....(i)
(b) C2H2(g)+52O2(g)2CO2(g)+H2O(l);(ΔH=310.6kcal)....(ii)
(c) C2H4(g)+3O2(g)2CO2(g)+2H2O(l)(ΔH=337.2kcal)....(iii)
The required equation can be achieved by adding eqs. (i) and (ii) and subtracting (iii)
C2H2(g)+H2(g)+3O2(g)C2H4(g)3O2(g)2CO2+2H2O(l)2CO2(g)2H2O(l)
or C2H2(g)+H2(g)C2H4(g)
ΔH=68.3310.6(337.2)=378+337.2=41.7kcal
We know that
ΔH=ΔU+ΔnRT
or ΔU=ΔHΔnRT
or Δn=(12)=1,R=2×103kcalmol1K1
and T=(25+273)=298K
Substituting the values in above equation
ΔU=41.7(1)(2×103)(298)
=41.104kcal

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