Combustion of 1 mole of benzene is expressed at C 6 H 6 ( l ) + 2 15 O 2 ( g ) → 6 CO 2 ( g ) + 3 H 2 O ( l ) . The standard enthalpy of combustion of 2 mol of benzene is − ′ x ′ kJ . x =

Question

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 $CO_{2} (g)$ , for the reaction $C$ (graphite) $+ O_{2} (g) \to CO_{2} (g)$ is $- 393.5 kJ mol^{- 1}$ . 3. Standard and 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 $- 286 kJ mol^{- 1}$ .

Answer:

6535

Solution:

$6 C (graphite) + 3 H_{2} (g) \to C_{6} H_{6} (ℓ); Δ H = 48.5 kJ / mol C (graphite) + O_{2} (g) \to CO_{2} (g); Δ H = - 393.5 kJ / mol H_{2}^{(g)} + \frac{1}{2} (g) ⟶ H_{2} O (ℓ); Δ H = - 286 kJ / mol equation - (1) \times 1 + (2) \times 6 + (3) \times 3 - 48.5 - 6 \times 393.5 - 3 \times 286 = - 3267.5 kJ for 1 mol = - 6535 kJ for 2 mol Ans. 6535 kJ$

TestHub · Accepted Answer

​ 6 C ( graphite ) + 3 H 2 ​ ( g ) → C 6 ​ H 6 ​ ( ℓ ) ; Δ H = 48.5 kJ / mol C ( graphite ) + O 2 ​ ( g ) → CO 2 ​ ( g ) ; Δ H = − 393.5 kJ / mol H 2 ( g ) ​ + 2 1 ​ ( g ) ⟶ H 2 ​ O ( ℓ ) ; Δ H = − 286 kJ / mol equation − ( 1 ) × 1 + ( 2 ) × 6 + ( 3 ) × 3 − 48.5 − 6 × 393.5 − 3 × 286 = − 3267.5 kJ for 1 mol = − 6535 kJ for 2 mol Ans. 6535 kJ ​

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