Question

The value of ΔGo for the reaction, N2(g) + 3 H2(g) <=> 2 NH3(g) is -32.90 kJ at 298 K. Calculate the value of ΔG in kJ at 298 K if the partial pressures of N2, H2 and NH3 are 7.765, 8.219, and 14 atm respectively.

Answer 1

ΔG = -35.3 kJ/mol

Step-by-step explanation:

Step 1: Data given

ΔG° = -32.90 kJ at 298 K

PArtial pressure of N2 = 7.765 atm

Partial pressure of H2 = 8.219 atm

Partial pressure of NH3 = 14 atm

Step 2: The balanced equation

N2(g) + 3 H2(g) <=> 2 NH3(g) ΔG° = -32.90 kJ

Step 3: Calculate Q

Q = p(NH3)² / p(N2) * p^(H2)³

Q = (14)²/ (7.765 * 8.219³)

Q = 0.3737

Step 4: Calculate ΔG

ΔG = ΔG° + RT ln Q

⇒with ΔG° = -32.90 kJ/mol = -32900 J/mol

⇒with R = 8.314 J/mol

⇒with T = the temperature = 298 K

⇒with ln Q = ln (0.3737) = -0.9843

ΔG = -32900 J/mol + (8.314 * 298) * -0.9843

ΔG = -35338.7 J/mol

ΔG = -35.3 kJ/mol