Combustion of 1 mole of benzene is expressed at

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Solution

Q.
Combustion of 1 mole of benzene is expressed at $C_{6} H_{6} (l) + \frac{15}{2} O_{2} (g) \to 6 {CO}_{2} (g) + 3 H_{2} O (l)$

The standard enthalpy of combustion of 2 mol of benzene is – 'x' kJ. x = ____________ .

Given:

1.  Standard Enthalpy of formation of 1 mol of $C_{6} H_{6} (l),$ for the reaction 6 C (graphite) $+ 3 H_{2} (g) \to C_{6} H_{6} (l)$ is 48.5 kJ ${mol}^{- 1}$ .

2.  Standard Enthalpy of formation of 1 mol of $C O_{2} (g)$ , for the reaction C(graphite) $+ O_{2} (g) \to {CO}_{2} (g)$ is –393.5 kJ ${mol}^{- 1}$ .

3.  Standard Enthalpy of formation of 1 mol of $H_{2} O (l)$ , for the reaction

$H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (l)$ is $- 286 kJ {mol}^{- 1}$ .   [2024]

Ans.
(6535)

$C_{6} H_{6} (l) + \frac{15}{2} O_{2} (g) \to 6 {CO}_{2} (g) + 3 H_{2} O (l)$

$Δ H_{r} = Δ_{f} H (products) - Δ_{f} H (reactants)$

$Δ H_{r} = 6 Δ_{f} H ({CO}_{2}) + 3 Δ_{f} H (H_{2} O) - Δ_{f} H (C_{6} H_{6}) + \frac{15}{2} Δ_{f} H (O_{2})$

$= [6 \times (- 393.5) + 3 (- 286) - 48.5 + \frac{15}{2} \times 0] kJ {mol}^{- 1}$

$= - 3267.5 kJ {mol}^{- 1}$

$Δ H_{r} (2 mol) = 2 \times - 3267.5 kJ {mol}^{- 1} = 6535 kJ {mol}^{- 1}$

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