Question

3CH4(g)→C3H8(g)+2H2(g)

Calculate ΔG at 298 K if the reaction mixture consists of 41 atm of CH4, 0.013 atm of C3H8, and 2.0×10⁻² atm of H2.

Answer 1

given

3CH4(g)→C3H8(g) + 2H2(g)

P_C3H8= 0.013 atm

P_H2 = 2.0×10⁻² atm

P_CH4 = 41 atm

we know that

ΔG = ΔG₀ + RTlnQ

lets calculate ΔG₀of the reaction

ΔG₀rxn = ΔG₀ f(products) - ΔG₀ f(reactants)

= (ΔG₀ f(C3H8(g)) + 2*ΔG₀ f(H2(g)) ) – (3*ΔG₀ f(CH4(g)))

Now

ΔG₀ f(C3H8(g)) = -24.40 kJ/mol

ΔG₀ f(CH4(g)) = -50.80 kJ/mol

ΔG₀ f(H2(g)) = 0 kJ/mol

ΔG₀rxn = (-24.40 kJ/mol – 2*0) – (3*-50.80 kJ/mol)

= -24.40 kJ/mol + 152.40 kJ/mol

= 128 kJ/mol

So

ΔG0 = 128 kJ/mol

Q = P_products/P_reactants

= (P_C3H8) * (P_H2)² / (P_CH4)³

= (0.013)* (2.0×10⁻²)² / 41³

= 0.013 * 4*10^-4 / 68921

= 7.5 * 10^-11

ln(Q) = -23.3

now

ΔG = ΔG₀ + RTlnQ

= 128 kJ/mol + (8.314 J/mol K * 298 K * -23.3)

= 128 kJ/mol – 57.727 kJ/mol

= 70 kJ/mol