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Q.

For the combustion of 1 mole of liquid benzene at 27° C, the heat of reaction at constant pressure is given by :

C6H6(l)+152O2(g)⟶6CO2(g)+3H2O(l) 

ΔH=−78 kcal

What would the be heat of reaction at constant volume?

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a

-78.9 kcal

b

816.0 kcal 

c

-77.1 kcal

d

-78.0 kcal

answer is C.

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Detailed Solution

The change in the enthalpy is,

ΔH=ΔU+ΔngRT

ΔU=ΔH−ΔngRT

=−78−(−1.5)×21000×300

=−7832×21000×300=−78+0.9

=−77.1 kcal

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