Calculate the amount of heat evolved during the complete combustion of 100 ml of liquid benzene from the following data:
(i) 18 g graphite on complete combustion evolves 590 kJ of heat
(ii) 15889 kJ of heat is required to dissociate all the molecules of 1 liter water into H, and Or.
(iii) The heat of formation of liquid benzene is +50 kJ/mol.
(iv) Density of C₆H₆(l) = 0.87 g ml^-1

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answer is 3645.

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Detailed Solution

$\begin{array}{l} C (s) + O_{2} (g) \to {CO}_{2} (g) \\ {ΔH}_{f}^{\circ} ({CO}_{2}) = \frac{590}{18 / 12} kJ / mol = 393.33 kJ / mol \\ H_{2} (g) + 1 / 2 O_{2} (g) \to H_{2} O (l) \\ {ΔH}_{f}^{\circ} (H_{2} O (1)) = - \frac{15889}{1000 / 18} = - 286 kJ {mol}^{- 1} \\ C_{6} H_{6} (l) + 7.5 O_{2} (g) \to 6 {CO}_{2} (g) + 3 H_{2} O (l) \\ {ΔH}_{comb}^{\circ} = 6 \times (- 393.33) + 3 (- 286) - 50 \\ = - 2359.98 - 858 - 50 ≅ - 3268 kJ / mol . \\ Heat evolved = 3268 \times \frac{100 \times 0.87}{78} kJ = 3645 kJ . \end{array}$

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