Question

Kp for the reaction, MgCO₃ (s) → MgO (s) + CO₂ (g) is 9x10⁻¹⁰ . Calculate ΔG° for the reaction at 25°C. (51.6 kJ mol⁻¹)

Answer 1

Explanation:

g = -nfE0 and E0= -0.0529/n logk

Answer 2

Answer:

$G^{0} = 51580.72$ J.

Explanation:

Since we know the relation between equilibrium constant and Gibs free energy.

$\Delta G^{0} = -2.303RT log(K_{p})$..................(1)

Here R = 8.314 ,T = 25+273=298 K.

$K_{p} = 9\times 10^{-10}$

So,substitute all the given valves in the equation (1)

$G^{0} = -2.303\times 8.314\times 298\ log(9\times 10^{-10} )$

After solving the above equation we get

Gib's free energy $G^{0} = 51580.72$ J.