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JEE Main levelSingle correct optionDifficulty 2 of 3ce137

Take $R = 8.314 J mo l^{- 1} K^{- 1}$ . For a gas-phase reaction at $400 K$ , $K = 50$ . The reaction quotient of a mixture is $Q = 5.0$ . The Gibbs energy change $Δ G$ of the mixture is

A $+ 7.66 kJ mo l^{- 1}$ (backward reaction spontaneous)
B $- 13.0 kJ mo l^{- 1}$ (forward reaction spontaneous)
C $- 3.33 kJ mo l^{- 1}$ (forward reaction spontaneous)
D $- 7.66 kJ mo l^{- 1}$ (forward reaction spontaneous)

Answer: D

Δ G = Δ G^{\circ} + RT ln Q = RT ln (Q / K) = (8.314) (400) ln (5.0/50) /1000 = - 7.66 kJ mo l^{- 1}

. This is negative because

Q < K

, so the forward reaction is spontaneous. The value -13.0 is only

Δ G^{\circ}

, and -3.33 uses

lo g_{10}

.

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