For the combustion of 1 mole of a liquid benzene at 25oC, the enthalpy is given by: C6H6(l)+152O2(g)→6CO2(g)+3H2O(l)ΔH=−780980 cal
What would be the internal energy of the reaction?
A
−780kcal
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B
780kcal
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C
−580kcal
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D
580kcal
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Solution
The correct option is A−780kcal We know. ΔH=ΔU+ngRT
where H and U are the enthalpy and the internal energy of a reaction. Δng = difference in moles of gaseous products and reactants = 6−152 = −32 = -1.5 ΔU=ΔH−ngRT = −780980−(−1.5)×2×298 ΔU=−780086 cal = −780 kcal