Q81 / 160 · JIPMER 2013 · Chemistry

$\frac{3}{2} \mathrm{O}_2(g) \longrightarrow \mathrm{O}_3(g) ; K_p$ for this reaction is $2.47 \times 10^{-29}$. At $298 \mathrm{~K}, \Delta_r G^{\circ}$ for conversion of oxygen to ozone will be

  • A
    $100 \mathrm{~kJ} \mathrm{~mol}^{-1}$
  • B
    $150 \mathrm{~kJ} \mathrm{~mol}^{-1}$
  • Correct answer
    C
    $163 \mathrm{~kJ} \mathrm{~mol}^{-1}$
  • D
    $2303 \mathrm{~kJ} \mathrm{~mol}^{-1}$

Solution

$\begin{aligned} \Delta_r G^{\circ}= & 2.303 \operatorname{log~Kp}\left(R=8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\right) \\ \Delta_r G^{\circ}= & -2.303 \times 8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1} \times 298 \mathrm{~K} \\ & \times \log 2.47 \times 10^{-29} \\ = & 163000 \mathrm{Jmol}^{-1}=163 \mathrm{~kJ} \mathrm{mol}^{-1}\end{aligned}$

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

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