Question

If equilibrium constant is $2.6\times 10^{-8}$ then find the value of $\Delta G^{\circ }$ ?

• A. +43.36 kJ
• B. -63.2 kJ
• C. -23.3 kJ
• D. + 40 kJ

Answer 1

Answer: (A)

$\Delta G^{\circ }=-2.303RT\log K$
$\Delta G^{\circ }=-2.303\times 8.314\times 298\log \left(2.6\times 10^{-8}\right)$
$=-2.303\times 8.314\times 298\times (-7.6)$
$=+43.36kJ$