Is the Haber Process here proceeding at positive Gibbs free energy change? $\ce{3H2 + N2 -&gt; 2NH3}$ 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 $\Delta G &lt; 0$, 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.
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.