Question

Calculate the change in G for a reaction mixture that consists of 1 atm N2, 3atm H2, and .5 atm NH3
(N2 +3H2 2NH3)

Answer 1

Solution-

Given

P(N2) = 1.0 atm

P(H2) = 3.0atm

P(NH3) = 0.5 atm

R = 8.3145 J/K.mol = 0.0083145 kJ/K.mol

Here we consider reaction is at room temperature (25 °), T = 298 K

Reaction is

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

Free energy on pressure (ΔG) is given by the formula

ΔG = ΔG° + RT ln Q
Here

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

= (0.5 atm)^2/1.0 atm*3.0atm = 0.083

Q = 0.083

When system is at equilibrium ΔG° = - 33.3 kJ/mol

Let’s calculate the ΔG
ΔG = ΔG° + RT ln Q = - 33.3 kJ/mol + 0.0083145 kJ/K.mol * 298 ln (0.083)

= -33.3 + 2.477 *(-2.488) = -33.3 – 6.16

ΔG = -39.46 kJ/mol

Answer change in G for a reaction mixture (ΔG) = -39.46 kJ/mol