Q77 / 200 · NEET 2011 (Screening) · Chemistry

If the enthalpy change for the transition of liquid water to steam is $30 \mathrm{~kJ} \mathrm{~mol}^{-1}$ at $27^{\circ} \mathrm{C}$, the entropy change for the process would be

  • A
    $1.0 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$
  • B
    $0.1 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$
  • Correct answer
    C
    $100 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$
  • D
    $10 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$

Solution

$$ \begin{aligned} & \Delta G^{\circ}=\Delta H^{\circ}-T \Delta S^{\circ} \\ & \text { Given, } \Delta H_{\text {vap. }}=30 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ & \Delta G^{\circ}=0 \text { at equilibrium, } \\ & \Delta S_{\text {vap }}=\frac{\Delta H_{\text {vap }}}{T} \\ & =\frac{30 \times 10^3 \mathrm{~J} \mathrm{~mol}^{-1}}{300 \mathrm{~K}} \\ & =100 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1} \\ & \end{aligned} $$

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

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