ThermochemistryHard

Question

Benzene burns in oxygen according to the following reactions.

C6H6(l) + 15/2 O2(g) → 3H2O(l) + 6CO2(g)

If the standard enthalpies of formation of C6H6(l), H2O(l) and CO2(g) are 11.7, −68.1 and −94 kcal/ mole, respectively, the amount of heat that will liberate by burning 780 g of benzene is

Options

A.7800 kcal
B.780 kcal
C.78 kcal
D.608.4 kcal

Solution

$\Delta_{c}{H^{o}}_{\text{Benzene}} = \Delta_{r}H$

$= \sum_{}^{}{\Delta_{f}H_{\text{Products}}} - \sum_{}^{}{\Delta_{f}H_{\text{Reactants}}} = 7800\text{ kcal}$

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