$C_6H_{6(l)} + 7.5O_{2(g)} \to 6CO_{2(g)} + 3H_2O_{(g)}$; $\Delta H = -3267.7 \ kJ \ mol^{-1}$. Given that the standard enthalpies of formation of $CO_{2(g)}$ and $H_2O_{(g)}$ are $-393.5 \ kJ \ mol^{-1}$ and $-285.85 \ kJ \ mol^{-1}$ respectively,calculate the standard enthalpy of formation of benzene $(C_6H_{6(l)})$.
(A) The combustion reaction is: $C_6H_{6(l)} + 7.5O_{2(g)} \to 6CO_{2(g)} + 3H_2O_{(g)}$
$\Delta H_{comb} = [6 \times \Delta H_f(CO_2) + 3 \times \Delta H_f(H_2O)] - [\Delta H_f(C_6H_6) + 7.5 \times \Delta H_f(O_2)]$
Given $\Delta H_{comb} = -3267.7 \ kJ \ mol^{-1}$,$\Delta H_f(CO_2) = -393.5 \ kJ \ mol^{-1}$,$\Delta H_f(H_2O) = -285.85 \ kJ \ mol^{-1}$,and $\Delta H_f(O_2) = 0 \ kJ \ mol^{-1}$.
$-3267.7 = [6 \times (-393.5) + 3 \times (-285.85)] - \Delta H_f(C_6H_6)$
$-3267.7 = [-2361 - 857.55] - \Delta H_f(C_6H_6)$
$-3267.7 = -3218.55 - \Delta H_f(C_6H_6)$
$\Delta H_f(C_6H_6) = -3218.55 + 3267.7 = 49.15 \ kJ \ mol^{-1}$.
$\Delta H_{comb} = [6 \times \Delta H_f(CO_2) + 3 \times \Delta H_f(H_2O)] - [\Delta H_f(C_6H_6) + 7.5 \times \Delta H_f(O_2)]$
Given $\Delta H_{comb} = -3267.7 \ kJ \ mol^{-1}$,$\Delta H_f(CO_2) = -393.5 \ kJ \ mol^{-1}$,$\Delta H_f(H_2O) = -285.85 \ kJ \ mol^{-1}$,and $\Delta H_f(O_2) = 0 \ kJ \ mol^{-1}$.
$-3267.7 = [6 \times (-393.5) + 3 \times (-285.85)] - \Delta H_f(C_6H_6)$
$-3267.7 = [-2361 - 857.55] - \Delta H_f(C_6H_6)$
$-3267.7 = -3218.55 - \Delta H_f(C_6H_6)$
$\Delta H_f(C_6H_6) = -3218.55 + 3267.7 = 49.15 \ kJ \ mol^{-1}$.
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Assertion $(A)$: The enthalpy of formation of graphite is taken as zero.
Reason $(R)$: Graphite is the thermodynamically most stable allotrope of carbon.
The correct answer is:
Assertion $(A)$: The enthalpy of formation of graphite is taken as zero.
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The correct answer is:
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