At $300 \mathrm{~K}$, the equilibrium constant for a reaction is 10 . The standard free energy change (in…

At $300 \mathrm{~K}$, the equilibrium constant for a reaction is 10 . The standard free energy change (in $\mathrm{kJ} \mathrm{mol}^{-1}$ ) for the reaction is
  1. -57.4
  2. -115.2
  3. +57.4
  4. -57.4

Solution

From Gibbs free energy equation, $ \Delta G=-2.303 R T \log K $ Given, $K=10, T=300 \mathrm{~K}$ $ \begin{aligned} & R=8.314 \mathrm{~J} / \mathrm{K} / \mathrm{mol} \\ & \therefore \Delta G=-2.303 \times 8.314 \mathrm{~J} / \mathrm{mol} / \mathrm{K} \times 300 \mathrm{~K} \times \log 10 \\ & \Delta G=-2.303 \times 8.314 \mathrm{~J} / \mathrm{mol} / \mathrm{K} \times 300 \mathrm{~K} \\ & {[\because \log 10=1]} \\ & \Delta G=-5744.14 \mathrm{~J} / \mathrm{mol}, \Delta G=-5.74 \mathrm{~kJ} / \mathrm{mol} \end{aligned} $ Thus, option (d) is correct

Asked in: AP EAMCET 2019 (20 Apr Shift 2)

Practice more Chemical Thermodynamics questions on Aicharya