# Is the Haber Process here proceeding at positive Gibbs free energy change?

$\ce{3H2 + N2 -> 2NH3}$ is the forward reaction used in the Haber process, the industrial production of ammonia. After studying Gibbs energy and how it is just another way of saying that the 2nd law of thermodynamics is obeyed if $\Delta G < 0$, I decided to apply it to some common reactions that I know, just to confirm that my understanding of the topic is correct.

When applying it to the Haber process forward reaction, I worked out change in entropy to be $\pu{-199 J K^{-1} mol^{-1}}$. The enthalpy change for the forward reaction is exothermic and has a value of $\pu{-92 kJ mol^{-1}}$. Looking up common reaction conditions for the Haber process, I found that a compromise temperature of $\pu{650K}$ is used.

Applying this to the Gibbs energy equation $-92 - 650(-199/1000)$, I calculated a positive change in Gibbs energy for the reaction of which had a value of $\pu{37.35 kJ mol^{-1}}$.

As a result of this, I am now quite confused as to how the forward reaction can be allowed to proceed by the 2nd law of thermodynamics at a temperature of $\pu{650K}$, and I was wondering where my understanding of the Gibbs energy and thermodynamics falls short. I need help to explain the pitfalls in my calculations, understanding or even whether there is another factor that I am not considering.

• This source has it as -32.7 kJ/mole N2: surfguppy.com/thermodynamics/… – Gert Dec 4 '17 at 20:49
• Did you adjust the values for $\Delta H$ for the higher temperature? – Gert Dec 4 '17 at 20:58
• @Gert That source uses 298K as the temperature which results in change in Gibbs energy being negative. However in industry, a higher compromise temperature of 650K is used for the forward reaction to achieve higher rates of reaction. At this higher compromise temperature, change in Gibbs energy is positive. I assumed that the value for ΔH does not change with temperature. I know that it does but I thought it would only be very negligible. chemguide.co.uk/physical/equilibria/haber.html This source takes ΔH = -92kJmol<sup>-1</sup> and states conditions used are 673K-723K. – Acids-Bases Dec 4 '17 at 21:00
• You need to do the same thing for the $\Delta S$. – Chet Miller Dec 4 '17 at 21:29
• Just because the change of standard free energy for a reaction is positive, that does not mean that the reaction will not go at all. It just means that the equilibrium conversion will be low. – Chet Miller Dec 5 '17 at 13:05

It's all about the pressure! The Haber process requires a pressure of about 200 atm. This is how the process becomes favorable, since LeChatelie's principle says that increasing pressure shifts toward the side with less moles of gas.

As long as Q is smaller than K, the reaction will go forward. For example, if nitrogen and hydrogen are supplied at stoichiometric ratio and 1% of a given reaction mixture is product, we would have the following partial pressures:

$$P_\ce{N2} = 0.2475 \times \pu{200 atm}$$

$$P_\ce{H2} = 0.7425 \times \pu{200 atm}$$

$$P_\ce{NH3} = 0.01 \times \pu{200 atm}$$

Accordingly, Q would be

$$Q = \frac{2^2}{50 \times 75^3} = \pu{1.9e-7}$$

In the Haber process, the concentration of product is kept small by continuously removing product (liquifying ammonia at low temperature) and feeding the remaining reactants back into the reaction mixture.

The tendency of a reaction to proceed in a certain direction depends on the state of the concentration of each reactant. If you calculate the standard Gibbs free energy of the reaction and it's a positive number, that only means that - if the composition of the reaction medium corresponds to all components at standard concentration ($$1 \ \textrm{mol/L}$$) - then the reaction will shift in the reverse direction (to which you supposed when you calculated $$\Delta G$$).

I am not sure though if this is the case for the Haber Process as you presented.

It must be noted that often one will increase the temperature of a reaction in a way that is non-intuitive from a thermodynamic perspective: one might conclude that the reaction is being made less spontaneous in the desired reaction. However, in a practical setting, the kinetics are also important, that is, how fast the reaction proceeds towards equilibrium. A temperature increase will always increase the reaction rate, which may get us to an acceptable yield earlier, even if the maximum achievable yield (dictated by thermodynamics) will be compromised.