Question

Calculate the change of Gibbs energy when 3.0 mol H2 mixes with 1.0 mol N2 at the same pressure with the volume of the vessel adjusted accordingly.

Answer 1

Solution.

$\chi (H2) = \frac{3}{3+1} = 0.75$

$\chi (N2) = \frac{1}{3+1} = 0.25$

n = 4 mol

$\Delta G = nRT(\chi(H2) \times ln(\chi(H2)) + \chi(N2) \times ln(\chi(N2)))$

$\Delta G = 4 \times 8.31 \times 298(0.75 \times ln(0.75) + 0.25 \times ln(0.25)) = -5570.22 J/mol$

Answer:

$\Delta G = -5570.22 J/mol$