Consider the reaction below, which is not spontaneous at 298 K. CO(NH2)2(s) + H2
ID: 994037 • Letter: C
Question
Consider the reaction below, which is not spontaneous at 298 K. CO(NH2)2(s) + H2O(l) CO2(g) + 2NH3(g) H298 = +134 kJ
a. For the reaction, indicate whether the standard entropy change, S298, is positive, negative, or zero. Justify your answer.
b. Which factor, the change in enthalpy, H298, or the change in entropy, S298, provides the principle driving force for the reaction at 298 K? Explain.
c. For the reaction, how is the value of the standard free energy change, G, affected by an increase in temperature? Explain.
Explanation / Answer
i) entropy change is positive. You are getting 3 moles of gas from 2 moles of solid/liquid.
ii) entropy change must drive the reaction because the -T S term will be negative. The only way to have a spontaneous reaction is for the entropy term to be more positive than the enthalpy term.
iii) increasing the temperature will make the reaction spontaneous. The increase in temperature will make the -TdS term more negative. It will eventually become large enough to overcome the change in ethalpy and make
dG < 0.