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Question

At 300K, the standard enthalpies of formation of C6H5COOH(s),CO2(g), and H2O(l) are -408, -393 and -286 kJ mol respectively. Calculate the heat of combination of benzoic acid at constant volume.

A
+3201 kJ
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B
+3199.75 kJ
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C
-3201 kJ
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D
-3199.75 kJ
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Solution

The correct option is B +3199.75 kJ
C6H5COOH+152H2O7CO2+3H2O
According to the following combustion reaction:
ΔHcombustion of C6H5COOH=(ΔHformation of C6H5COOH)7(ΔHformation of CO2)3(ΔHformation of H2O)
=4087(393)3(286)=+3199.75 kJ

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