Question

The standard free energy change of a reaction is $\Delta G^{\circ}$ = - $115\,kJ$ at $298 \,K$. Calculate the equilibrium constant $k_p$ in $\log k_p$ $(R = 8.314 \,Jk^{-1} mol^{-1}).$

• A. 20.16
• B. 2.303
• C. 2.016
• D. 13.83

Answer 1

Answer: (A)

20.16

Answer 2

$\Delta G^{\circ}= - 115 \times 103 ^3\,J,$ $T = 298 \,K, R = - 8.314\, JK^{-1} mol ^{-1}$ $- \Delta G^ {\circ} = 2.303RT\, \log_{10}K_ p$ $-(- 115 \times 10 ^3) = 2.303\times 8.314 \times 298 \log _{10} K_p$ $\log_{10} K_p = \frac{115000}{ 2.303 \times 8.314 \times 298 }$ $\log_{10} K _p = 20.16$