# 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 $\Delta S$. – Chester 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. – Chester Miller Dec 5 '17 at 13:05