Question

Calculate ∆G°’ for the reaction A + B ↔ C + D +H+ at 37°C and pH 7.3 when the equilibrium conditions are [A] = 25 μM, [B] = 10 μM, [C] = 60 μM, [D] =15 μM.

Answer 1

$\Delta G^{o\prime} = -RT\text{ln}K_{eq}^{\prime}$

If we neglect the activity coefficients, the equilibrium constant is: (remember that we should ignore the protons as the pH is kept constant)

$K_{eq}^{\prime} = \frac{[C][D]}{[A][B]}$

Then, the equilibrium constant is:

$K_{eq}^{\prime} = \frac{15\cdot10^{-6}\cdot60\cdot10^{-6}}{25\cdot10^{-6}\cdot10\cdot10^{-6}} = 3.6$

$\Delta G = -8,314 (\text{J mol}^{-1}\text{K}^{-1}) \cdot(37+273.15 (\text{K}))\cdot\text{ln}(3.6)$

$\Delta G = -3303 \text{ J mol}^{-1} = -3.303 \text{ kJ mol}^{-1}$

Answer: -3.303 kJ/mol