Can I get help on this please?

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 Δ𝐺°=−32.8 kJ/mol . What is ΔG for this reaction at 298 K when the partial pressures are 𝑃N2=0.250 atm , 𝑃H2=0.300 atm , and 𝑃NH3=0.700 atm ? Δ𝐺= kJ/mol

Answer 1

∆G = ∆Gº + RT ln Q

∆G = ?

∆Gº = -32.8 kJ/mol

R = 8.314 J/Kmol = 0.008314 kJ/Kmol (convert to units consistent with those of ∆Gº)

T = 298K

Q = reaction quotient (see below)

N2 + 3H2 ==> 2NH3

0.250...0.700....0.300

Q = (NH3)² / (N2)(H2)³ = (0.300)² / (0.250)(0.700)³

Q = 0.09 / 0.08575

Q = 1.0496

∆G = ∆Gº + RT ln Q

Solve for ∆G....

∆G = -32.8 kJ/mol + (0.008314 J/Kmol)(298K) ln 1.0496

∆G = -32.7 kJ/mol

(be sure to check all of the math)