What is the value of ΔG at 298 K?

Question

In the Haber process, ammonia is synthesized from nitrogen and hydrogen:

N₂(g) + 3 H₂(g) → 2 NH₃(g)

ΔG° at 298 K for this reaction is -33.3 kJ/mol. The value of ΔG at 298 K for a reaction mixture that consists of 1.0 atm N₂, 3.0 atm H₂, and 0.50 atm NH₃ is ________.

Answer 1

Before we start on the math of this question, I'd like to point out what we should expect to happen conceptually given the concentrations and equation. For a completely balanced, equilibrated reaction, we should have a 1:3:2 ratio of each component of the reaction. If any of them are shifted, we can hopefully predict which way the reaction will shift, and then double check that our solution follows our prediction.

Looking at the given reaction mixture, it seems to be in equilibrium until we hit the NH₃ concentration! 0.5 atm is much too low compared to the 1 and 3 atm of the reactants. Therefore, we can predict that the reaction will shift towards the products.

Now we can move onto the math. To determine which direction our reaction will shift, we can compare the K_eq and Q values of the equation. The given K_eq of the Haber Bosch process at 300K is about 2.7x10⁸ (value obtained from LibreTexts). If our calculated Q value is bigger than this, our reaction will shift back towards the reactants. If it's smaller, it will favor the production of products to get that number back up.

To calculate Q, use the same format of our equilibrium constant (product concentrations over reactant concentrations raised to their molar ratio). Our Q equation should look like this:

Q = ([0.5]²)/([1]¹[3]³) = 9.25x10^-3

Needless to say, our Q value is much smaller than our K value, which means our reactants will want to make more product (shifting right). To find ΔG, we can use our Gibbs Free Energy Equation using Q:

ΔG = G^o + RTlnQ

where G^o is the standard free energy, R is the gas constant, T is the temperature, and Q is our reaction quotient which we calculated above. One of the most difficult parts of using this equation is making sure you use the correct gas constant format. The gas constant considers many different units of measurement, so we need to make sure we pick the one for our needs. We are in units of Kelvin and atmospheres, which matches the R value of 0.082057 L atm per mol kelvin. Our temperature was given as 298 K, and we plug in our Q value from above, which should give us a Gibbs free energy of -33.3 + -114.5 = -147.8 kJ/mol.

The more negative the Gibbs Free Energy is, the more spontaneous and product-favored the reaction is. It's safe to say that with a lack of product, this reaction favors the products and is fighting to re-equilibrate, so our prediction at the beginning is supported by our math.