Question

the standard few energy changes for the reaction N2(g) + 3H2(g) 2Nh3(g) is grxn=53.8 kj/mol at 800k. Calculate G at 800k if the reaction mixture consists of 30 atm of H2, 20 atm of N2, and .5 atm of nh3

Answer 1

ΔG = -63.14 kJ/mol

Step-by-step explanation:

It is possible to obtain ΔG of a reaction at any point to predict the spontaneity of the reaction using:

ΔG = ΔG° + RT ln Q (1)

Where ΔG is Change in free energy in the point you want to study, ΔG° is the standard ΔG of the reaction, R is gas constant (8.314x10⁻³kJ/molK), T is absolute temperature (800K), and Q is reaction quotient

For the reaction:

N₂(g) + 3H₂(g) ⇄ 2NH₃(g)

ΔG° = 53.8kJ/mol at 800K

Reaction quotient is:

Q = P[NH₃]² / P[H₂]³ P[N₂]

Replacing:

Q = (0.5atm)² / (30atm)³(20atm)²

Q = 2.31x10⁻⁸

Replacing in (1):

ΔG = 53.8kJ/mol + 8.314x10⁻³kJ/molK×800K ln 2.31x10⁻⁸

ΔG = -63.14 kJ/mol