Question

The equilibrium constant of a reaction is $0.008$ at $298K$. The standard free energy change of the reaction at the same temperature is

• A. $+11.96kJ$
• B. $-11.96kJ$
• C. $-5.43kJ$
• D. $-8.46kJ$

Answer 1

Answer: (A)

Standard free energy change,

$\Delta G^{\circ }=-2.303RT\log K$

$=-2.303\times 8.314\times 298\times \log 0.008$

$=-2.303\times 8.314\times 298\times (3\log 2-3\log 10)$

$=-2.303\times 8.314\times 298\times (0.903-3.0)$

$=-2.303\times 8.314\times 298\times (-2.097)$

$=+11965.16J=+11.96kJ$