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  3. Chemistry
  4. If equilibrium constant is 2.6 × 10-8 then find the value of Δ G° ?

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

JIPMERJIPMER 2019Equilibrium

Solution:

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