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Question

Given below are some standard heat of reaction:


(a) Heat of formation of water =68.3kcal
(b) Heat of combustion of acetylene =310.6kcal
(c) Heat of combustion of ethylene =337.2kcal.

The heat of reaction for the hydrogenation of acetylene at the constant volume at 25C.

A
41.104kcal
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B
49.345kcal
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C
62.203kcal
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D
None of these
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Solution

The correct option is B 41.104kcal
H2(g)+12O2H2O(l);
H=68.3Kcal

C2H2(g)+212O2(g)2CO2(g)+H2O(1);
H=310.6Kcal

C2H4(g)+3O2(g)2CO2(g)+2H2O(1);
H=337.2Kcal

E of the following equation is to be calculated:

C2H2(g)+H2(g)C2H4(g);E=?

Sum up the equations (i) and (ii) and subtract equation (iii)-

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯C2H2(g)+H2(g)C2H4(g);H=41.7Kcal
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
H is the heat of reaction at constant pressure. Heat at constant volume can be calculated by using the following equation:

H=E+ngRT

ng= moles of gaseous products - moles of gaseous reactants
=12=1

R=0.002Kcal/mole

T=273+25=298K

E=41.7(1×0.002×298)

=41.104Kcal

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