The difference between heat of reactions at constant pressure and constant volume for the reaction C6H6(l)+152O2(g)→6CO2(g)+3H2O(l) at 25∘C in kJ is
A
-7.43
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B
3.72
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C
-3.72
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D
7.43
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Solution
The correct option is C -3.72 We have the relation for a reaction △H−△U=△ngRT 2C6H6(l)+15O2(g)→12CO2(g)+6H2O(l) at 25∘C
here Δng=6−152=−1.5
Putting the values in the formula we get, △H−△U=(−1.5)×8.3141000×298△H−△U=−3.716kJmol−1