Question

Suppose that 2.0 mol H2 at 2.0 ATM and 25*C and 4.0 mol N2 at 3.0 atm and 25*C are mixed at constant volume. Calculate delta mix G. What would be the value of delta mix G had the pressure been identical initially

Answer 1

Solution.

$\Delta G = -RT \sum(ni \times ln(P/Pi))$

$\Delta G = -8.31 \times 298(2 \times ln(\frac{5}{2}) + 4 \times ln(\frac{5}{3})) = -9598.16 J/mol$

$V = \frac{n \times R \times T}{p}$

V(H2) = 0.0244 m3

V(N2) = 0.0326 m3

V(sum) = 0.0570 m3

$\Delta G = -RT \sum (ni \times ln(V/Vi))$

$\Delta G = -8.31 \times 298(2 \times ln(\frac{0.0570}{0.0244}) + 4 \times ln(\frac{0.0570}{0.0326})) = -9736.86 J/mol$

Answer:

$\Delta G (V=const) = -9598.16 J/mol$

$\Delta G(p = const) = -9736.86 J/mol$