MCQ
The difference between the reaction enthalpy change $\left( {{\Delta _r}H} \right)$ and reaction internal energy change $\left( {{\Delta _r}U} \right)$ for the reaction

$2C_6H_6 (l) + 15O_2 (g) \longrightarrow 12CO_2 (g) + 6H_2O(l)$

at $300\, K$ is ....$J\,mol^{-1}$ ($R = 8.314\, J\, mol^{-1}\, K^{-1}$)

  • A
    $0$
  • B
    $2490$
  • C
    $-2490$
  • $-7482$

Answer

Correct option: D.
$-7482$
d
$\Delta H = \Delta U + \Delta {n_g}RT$

For the reaction  $\Delta {n_g} = 12 - 15 =  - 3$

$\Delta H - \Delta U =  - 3 \times 8.314 \times 300$

$ =  - 7482\,J\,mo{l^{ - 1}}$

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