The Haber process for the production of ammonia involves the equilibrium N_2(g)

Question

The Haber process for the production of ammonia involves the equilibrium N_2(g) + 3 H_2(g) equilibrium 2 NH_3(g) Assume that Delta H degree = -92.38 kJ and deltas degree = -198.3 J/K for this reaction do not change with temperature. Without doing calculations, predict the direction in which Delta G degree for the reaction changes with increasing temperature. Explain your prediction Calculate Delta G degree at 25 degree C and 500 degree C. At what temperature does the Haber ammonia process become nonspontaneous?

Explanation / Answer

del G changes its sign from negative to positive with increase of temperature.

As temperature increases the product (-T)*delS increases positively and given del H is negative and therefore del G becomes negative up to certain temperature and thereafter as the temperature increases del G changes its sign.

b) At T=25 oc =298 K

del G= del H - T*del S=-92380-298*(-198.3)=-33286 j

K=exp(-del G/RT) =exp(33286/8.314*298)=6.83625*105

At 500 oc =773K

We use Vontoff equation

ln(K1/K2)= - ((del H)/R) * ((1/T1)-(1/T2)) ---(1)

Here T1=298 K and K1=6.83625*105

T2=773 K and del H= -92380 J ,R=8.324 J/mole/K

Substituting in equation (1)

We have K2= 7.66*10-5

c) del G = del H - T*del S

Del G>=0 the reaction is non spontaneous

Keeping del G=0 in the above equation we have

T=del H/del S=-92380/(-198.3)=465 K

Therefore Temperatures above 465 K the reaction is non spaontaneous.

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