Q52 / 200 · WBJEE 2013 · Chemistry

The standard Gibbs free energy change $\left(\Delta G^{o}\right)$ at $25^{\circ} \mathrm{C}$ for the dissociation of $\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{g})$ to $\mathrm{NO}_{2}(g)$ is (given, equilibrium constant $=0.15, R=8.314 \mathrm{JK} / \mathrm{mol})$

  • A
    $1.1 \mathrm{KJ}$
  • Correct answer
    B
    $4.7 \mathrm{KJ}$
  • C
    $8.1 \mathrm{KJ}$
  • D
    $38.2 \mathrm{KJ}$

Solution

$$ \begin{array}{l} =-2.303 \times 8.314 \times 298 \times \log 0.15 \\ =-2.303 \times 8.314 \times 298 \times(-0.82) \\ =4678.7 \mathrm{J}=4.67 \mathrm{kJ} \end{array} $$

Hence, the correct answer is option: 'B'.

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