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) + 3H_2 (g) reversible arrow 2NH_3 (g). Assuming that delta H_0 and delta S_0 for this reaction do not change with temperature, Predict the change in delta G for the reaction as temperature increases, (don't do any calculations, but explain the change in delta G based on the Gibbs-Helmholtz equation). Now, with your calculator, calculate the values for delta G at 25degree C and 500degree C. Do your calculations in b) agree with your prediction in a)?

Explanation / Answer

1a)We know the Gibbs- Helmholtz equation is DElta G = Delta H - T . delta S

The given equilibrium for the production of ammonia is an exothermic reaction in the forward direction, thus delta H is negative .

The number of moles on the produt side are less than the reactant side, making the delta S always negative.

Thus at low temperatures when delta H > Tdelta S numerically , the delta G is negative value but as the temperature increases, the TDelta s Value wich is a positive value overcomes the negative delta H value, thus the delta G becomes positive.

Thus as temperature increases, the delta G value becomes less negative , becomes zero at equilibrium temperature and then becomes positive beyond that temperature.

b) Delta H for elements is zero and delta H of ammonia is -46.1kJ/mol.

standard molar entropies of N2, H2 and NH3 are 191, 130.68, 111.3 J/Kmol respectively.

Thus delta H reaction = -2x46.1 kJ and delta s reaction = 2x111.3 -(191 + 3x 130.68) = -360.4 J /K.mol

Then delta G at 25C = -(2x46.1x1000) J -298 [ -360.4] = 15199.2 J

Delta G at 500C = -(2x46.1x1000) J -773 [ -360.4] = 186389.2J

As the temperature increases the delta G is becoming more positive as predicted.

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.