Question

Use standard thermodynamic data (in the Chemistry References) to calculate G at 298.15 K for the following reaction, assuming that all gases have a pressure of 14.50 mm Hg. N2(g) + 3H2(g)2NH3(g)

Answer 1

N2(g) + 3H2(g) <----> 2NH3(g)

Given that;

all gases have a pressure of 14.50 mm Hg or 0.0191 atm

The dependence of free energy on pressure is given by the formula;

ΔG = ΔG° + RT ln Q

where Q = P^2(NH3) / P(N2) x P^3(H2)

Q = (0.0191)^2 / (0.0191)(0.0191)^3
Q = 2741.2

ΔG° = - 33.3 kJ/mol
T = 298 K
R = 8.3145 J/K.mol = 0.0083145 kJ/K.mol

ΔG = - 33.3 + (0.0083145)(298.15) ln(2741.2)
ΔG = - 33.3 + (0.0083145)(298.15) (7.92)
ΔG = - 33.3 +19.6

= - 13.7 kJ/mol