Q12 / 160 · TS EAMCET 2018 (05 May Shift 2) · Chemistry

If equilibrium constant of a process is $3.8 \times 10^{-3}$ at $25^{\circ} \mathrm{C}$, standard free energy change of the process is $$ \left(R=8.314 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}, \log 0.0038=-2.42\right) $$

  • A
    $5.7 \mathrm{~kJ} \mathrm{~mol}^1$
  • B
    $9.9 \mathrm{~kJ} \mathrm{~mol}^1$
  • Correct answer
    C
    $13.8 \mathrm{~kJ} \mathrm{~mol}^1$
  • D
    $15.6 \mathrm{~kJ} \mathrm{~mol}^1$

Solution

$$ \begin{aligned} \Delta G^{\circ} & =-R T \ln K \\ \Delta G^{\circ} & =-2.303 R T \log K \\ \Delta G^{\circ} & =-2.303 \times 8.314 \mathrm{~mol}^{-1} \times 298 \log 3.8 \times 10^{-3} \\ \Delta G^{\circ} & =13.8 \times 10^3 \mathrm{~J} \mathrm{~mol}^{-1} \\ \Delta G^{\circ} & =13.8 \mathrm{~kJ} \mathrm{~mol}^{-1} \end{aligned} $$

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

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