Question

For a gaseous reaction, standard conditions are 298 K and a partial pressure of 1 atm for all species.

For the reaction

N2(g)+3H2(g)↽−−⇀2NH3(g)

the standard change in Gibbs free energy is Δ°=−32.8 kJ/mol
. What is ΔG for this reaction at 298 K when the partial pressures are N2=0.350 atm
, H2=0.300 atm
, and NH3=0.750 atm
?

Answer 1

We can use the following equation to calculate the Gibbs free energy change at non-standard conditions:

ΔG = Δ°G + RT ln(Q)

where Δ°G is the standard Gibbs free energy change, R is the gas constant (8.314 J/mol-K), T is the temperature in Kelvin, and Q is the reaction quotient.

First, we need to calculate Q for the given partial pressures:

Q = (P(NH3))^2 / (P(N2) * P(H2)^3)
= (0.750 atm)^2 / (0.350 atm * 0.300 atm^3)
= 4.08

Next, we can substitute the values into the equation:

ΔG = -32.8 kJ/mol + (8.314 J/mol-K * 298 K) * ln(4.08)
= -32.8 kJ/mol + (2471 J/mol) * 1.407
= -32.8 kJ/mol + 3476 J/mol
= -29.3 kJ/mol

Therefore, the Gibbs free energy change for the reaction at 298 K and the given partial pressures is -29.3 kJ/mol.