Question

Consider the following gas-phase reaction: 3 NO(g) NO2(g) + N2O(g) Using data from Appendix C of your textbook calculate the temperature, To, at which this reaction will be at equilibrium under standard conditions (Go = 0) and choose whether >Go will increase, decrease, or not change with increasing temperature from the pulldown menu.

Answer 1

3 NO(g) ------------------> NO₂(g) + N₂O(g)

ΔH^o_rxn = ΔH^o_products – ΔH^o_reactants

ΔH^o_NO2 = 33.10 kJ/mol

ΔH^o_N2O = 82.05 kJ/mol

ΔH^o_NO = 90.29 kJ/mol

Total ΔH^o_NO = 3 x 90.29 = 270.87 kJ/mol

ΔH^o_rxn = (33.10 + 82.05) – (270.87)

= 115.15 – 270.87

ΔH^o_rxn = -155.72 kJ/mol

ΔS^o_rxn = ΔS^o_products – ΔS^o_reactants

ΔS^o_NO2 = 240.04 J/mol

ΔS^o_N2O = 219.96 J/mol

ΔS^o_NO = 210.76 J/mol

Total ΔS^o_NO = 3 x 210.76 = 632.28 J/mol

ΔS^o_rxn = (240.04 + 219.96) - 632.28

ΔS^o_rxn = 460 - 632.28

ΔS^o_rxn = - 172.28 J/mol

ΔG^o_rxn = 0

ΔG^o_rxn = ΔH^o_rxn – T ΔS^o_rxn

Accordingly

0 = -155720 – T (- 172.28)

155720 = T x 172.28

T = 155720 / 172.28 = 903.877K

If temperature increases ΔG^o_rxn value will become negative and the reaction will be spontaneous.