For A B at 300 K, Delta G deg = +5.00 kJ mol^-1. Find Delta G in kJ mol^-1 (two decimals) when the reaction quotient is Q = 0.0100. Use R = 8.314 J mol^-1K^-1.

JEE Main levelNumerical answerDifficulty 2 of 3td100

For $A B$ at $300 K$ , $Δ G^{\circ} = + 5.00 kJ mo l^{- 1}$ . Find $Δ G$ in $kJ mo l^{- 1}$ (two decimals) when the reaction quotient is $Q = 0.0100$ . Use $R = 8.314 J mo l^{- 1} K^{- 1}$ .

Numerical answer type, answer in kJ/mol, tolerance 0.01.

Δ G = Δ G^{\circ} + RT ln Q = 5000 + 8.314 \times 300 \times ln (0.0100) = 5000 - 11486 J = - 6.49 kJ mo l^{- 1}

. It is negative, so the forward reaction is spontaneous at this composition.