Consider the following data :Heat of formation of CO2(g) = –393.5 kJ mol–1Heat of formation of H2O(l) = – 286.0 kJ mol–1Heat of combustion of benzene = –3267.0 kJ mol–1The heat of formation of benzene is _______ kJ mol–1.(Nearest integer)

ΔHf[CO2(g)]=−393.5kJ/mole ΔHf[H2O(ℓ)]=−286.0kJ/mole ΔHc[C6H6]=−3267.0kJ/mole ΔHfC6H6=(?) C6H6+152O2(g)⟶6CO2(g)+3H2O(ℓ) ΔHR=ΔHC=ΣΔHf(P)−ΣΔHf(R) −3267=6×(−393.5)+3(−286)−ΔHf(C6H6) ΔHf(C6H6)=48kJ/mole

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Consider the following data :
Heat of formation of CO₂(g) = –393.5 kJ mol^–1
Heat of formation of H₂O(l) = – 286.0 kJ mol^–1
Heat of combustion of benzene = –3267.0 kJ mol^–1
The heat of formation of benzene is _______ kJ mol^–1.
(Nearest integer)

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answer is 48.

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Detailed Solution

$\begin{aligned} {ΔH}_{f} [{CO}_{2} (g)] = - 393.5 kJ / mole \\ {ΔH}_{f} [H_{2} O (ℓ)] = - 286.0 kJ / mole \\ {ΔH}_{c} [C_{6} H_{6}] = - 3267.0 kJ / mole \\ {ΔH}_{f} C_{6} H_{6} = (?) \end{aligned}$
$\begin{aligned} C_{6} H_{6} + \frac{15}{2} O_{2} (g) ⟶ 6 {CO}_{2} (g) + 3 H_{2} O (ℓ) \\ {ΔH}_{R} = {ΔH}_{C} = {ΣΔH}_{f} (P) - {ΣΔH}_{f} (R) \\ - 3267 = 6 \times (- 393.5) + 3 (- 286) - {ΔH}_{f} (C_{6} H_{6}) \\ {ΔH}_{f} (C_{6} H_{6}) = 48 kJ / mole \end{aligned}$