Question

1. Consider the reaction: 2NH3(g) → N2(g) + 3 H2 (8) AG = +33.3 kJ a. Is this reaction spontaneous? Explain. b. Predict the sign of AS. Explain. C. Based on your answer to part b, is this reaction exothermic or endothermic? Explain. For the reaction N2(g) + 3H2(g) 2NH3 (8) a. Using values in Appendix Cin your book, calculate AHⓇ and AS. b. Assuming that AHºand ASº don't change with temperature, calculate the value for AG at 400K Is this reaction spontaneous at 400K? Explain. What is the equilibrium constant at 298 for this reaction? Assuming that AH° and ASº do not change with temperature, at what temperature will the spontaneity of the reaction change. f. What is AG for this reaction at 298 K if the pressure of NH3(g) is 3.0 atm while that of Nz and Hz are 0.75 atm? d.

Answer 1

1) 2NH3 ----> N2 + 3H2 , Delta G° = +33.3 KJ

A) Since, Delta G° is positive thus reaction is non spontaneous.

B) Delta S is positive for reaction as product side has more gaseous moles than reactants moles.

C) Reaction is endothermic because Delta H is positive as per below;

Delta G° + T * Delta S = Delta H

Positive + T * positive = positive

2) N2 + 3H2 ---> 2NH3

A) Delta H(rxn) = 2*Delta H(NH3) - 3*Delta H(H2) - Delta H(N2)

Delta H(rxn) = 2*(-46.11) - 3*(0) - 0

Delta H(rxn) = -92.22 KJ/mol = -92220 J ...Answer

Similarly,

Delta S(rxn) = 2*(192.45) - 3*(130.684) - (191.61)

Delta S(rxn) = -198.762 J/K ....Answer

B) Delta G = Delta H - T * Delta S

Delta G = -92220 - 400 * (-198.762)

Delta G = -12715.2 J = -12.72 KJ/mol ....Answer

C) Since, Delta G is negative. Reaction is spontaneous.

D) Delta G = - R * T * ln Kc

-12715.2 = - 8.314 * 298 * ln Kc

Kc = 169.38 ....Answer

E) Delta H = T * Delta S

-92220 = T * (-198.762)

T = 463.973 K ....Answer

Let me know if any doubts/answer is not matching.