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J. W. Hill, R. H. Petrucci, T. W. McCreary, & S. S. Perry General Chemistry 4th edition

Problem 17.36

Would you expect each of the following reactions to be spontaneous at low temperatures, high temperatures, all temperatures, or not at all? Explain.

(a) PCl3(g) + Cl2(g) PCl5(g)     Ho = –87.9 kJ

(b) 2 NH3(g) N2(g) + 3 H2(g)     Ho = +92.2 kJ

(c) 2 N2O(g) 2 N2(g) + O2(g)      Ho = –164.1 kJ

(d) H2O(g) + ½O2(g)H2O2(g)    Ho = +105.5 kJ

(e) CH4(g) + O2(g)CO2(g) + 2 H2O(g)     Ho = –802.3 kJ

(f) 2 CO(g) + O2(g) 2 CO2(g)    Ho = –566.0 kJ






Answer:

(a) PCl3(g) + Cl2(g) PCl5(g) Ho = –87.9 kJ

So is predicted to be negative since there are fewer moles of gaseous products than reactants. Thus, at low temperatures enthalpy will dominate (TSo is small) and the reaction will be spontaneous (Ho < 0) while at high temperature the entropy dominates and the reaction becomes nonspontaneous (TSo becomes large and negative).



(b) 2 NH3(g) N2(g) + 3 H2(g) Ho = +92.2 kJ

So is predicted to be positive since there are more moles of gaseous products than reactants. Thus, at low temperatures enthalpy will dominate (TSo is small) and the reaction will be nonspontaneous (Ho > 0) while at high temperature the entropy dominates and the reaction becomes spontaneous (TSo becomes large and positive).

(c) 2 N2O(g) 2 N2(g) + O2(g) Ho = –164.1 kJ

So is predicted to be positive since there are more moles of gaseous products than reactants. Thus, at low temperatures enthalpy will dominate (TSo is small) and the reaction will be spontaneous (Ho < 0) while at high temperature the entropy dominates and the reaction is also spontaneous (TSo becomes large and positive). therefore, the reaction is expected to be spontaneous over all temperatures.



(d) H2O(g) + ½O2(g)H2O2(g) Ho = +105.5 kJ

So will be negative (1.5 moles gases in reactants but only 1 mole of gases in the product) and Ho is positive so Go will be positive at all temperatures. Thus, the reaction is never spontaneous.


(e) CH4(g) + O2(g)CO2(g) + 2 H2O(g) Ho = –802.3 kJ

So will be positive (2 moles gases in reactants but 3 moles of gases in the product) and Ho is negative so Go will be negative at all temperatures. Thus, the reaction is spontaneous at all temperatures.


(f) 2 CO(g) + O2(g) 2 CO2(g) Ho = –566.0 kJ

So will be negative (3 moles gases in reactants but only 2 moles of gases in the product) and Ho is negative so Go will be negative at low temperatures and positive at high temperatures. Thus, the reaction is spontaneous only at low temperature.