I got stuck on this Chemistry problem, please help

Question

The reaction below took place at a temperature of 500 K.  

N2 (g) + 3H2 (g) ↔ 2NH3 (g) ΔH = -92.4 kJ K = 1.45 x 10-5

(a) Identify one way that a chemist could increase the production of the ammonia gas. Support your answer with an explanation.

(b) Calculate the value of the ΔG, in kJ, for this reaction. Explain how this value supports the known equilibrium constant of the reaction.

Answer 1

N2(g) + 3H2(g) <==> 2NH3(g) ... ∆H = -92.4 kJ and K = 1.45x10^-5

a) One way to increase production of NH3 would be to lower the temperature. Because the ∆H is negative, this means that heat is given off by the reaction and one can view heat as a product. Removing a product (heat) would pull the reaction toward the product side thus making more NH3(g). This is according to Le Chatelier.

b) ∆G = -RT ln K

∆G = ?

R = gas constant = 8.314 J/Kmol = 0.008314 kJ/molK

T = temperature in K = 500K

K = equilibrium constant = 1.45x10^-5

Solving for ∆G, we have...

∆G = -(0.008314)(500)(-11.1)

∆G = +46.1 kJ

This value supports the known equilibrium constant of the reaction because they BOTH suggest that the reaction is NOT SPONTANEOUS. A positive ∆G tells us the reaction is not spontaneous and a K value <1 also tells us the same.