Chemical Equilibrium

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Question

For a reaction,
$A \Leftrightarrow B, K = 10^{- 3}$
$The value of ∆ G^{0} at 27^{0} C$ is (Note : Instead of using 2.303 in the equation, use 2.0 to get nearest integer) in Cal/Mole

Moderate

Solution

$∆ G^{0} = - 2.303 RT \log_{10} 10^{- 3} = 2 \times 3 \times 300 \times 2 cal / mol = 3600$

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