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JEE Main 2016 Chemistry question

ChemistryEquilibriumJEE Main 2016

For the reaction, , H and S are, respectively, kJ mol and kJ K mol at 298 K. The equilibrium constant for the reaction at 298 k is:

Answer: (3)

Solution

Compute the standard Gibbs energy change:

Δ G° = Δ H° - TΔ S° = -29.8 - (298)(-0.100) = -29.8 + 29.8 = 0\ kJ mol^{-1}.

The relation to the equilibrium constant is , so $ K

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Question ID VK-115530