Question

Without doing any calculations, determine the sign of ΔSsys and ΔSsurr for each of the chemical reactions below. Part A 2CO(g)+O2(g)⇌2CO2(g)ΔH∘rxn= -566.0 kJ -566.0 ΔSsys>0, ΔSsurr>0 ΔSsys<0, ΔSsurr>0 ΔSsys>0, ΔSsurr<0 ΔSsys<0, ΔSsurr<0 SubmitMy AnswersGive Up Part B 2NO2(g)→2NO(g)+O2(g)ΔH∘rxn= +113.1 kJ +113.1 ΔSsys>0, ΔSsurr>0 ΔSsys<0, ΔSsurr>0 ΔSsys>0, ΔSsurr<0 ΔSsys<0, ΔSsurr<0 SubmitMy AnswersGive Up Part C 2H2(g)+O2(g)→2H2O(g)ΔH∘rxn= -483.6 kJ -483.6 ΔSsys>0, ΔSsurr>0 ΔSsys<0, ΔSsurr>0 ΔSsys>0, ΔSsurr<0 ΔSsys<0, ΔSsurr<0

Answer 1

Since a) 2CO(g)+O₂(g)⇌2CO₂(g)

In this reaction, ΔH∘rxn= -566.0 kJ which means reaction is exothermic. So, heat flows from system to surroundings.So, ΔSsurr>0.Not only that 3 molecules of reactants give two molecules of product. Therefore, there is reduction in degree of freedom of the system  i.e. ΔSsys<0

Since, ΔSsys + ΔSsurr > 0, therefore , ΔSsys<0, ΔSsurr>0

b) 2NO₂(g)→2NO(g)+O₂(g)

For this reaction, ΔH∘rxn= +113.1 kJ i.e. the reaction is endothermic. So, heat flows from surroundings to system. Thus, ΔSsurr<0. Not only that 2 molecules of reactants give 3 molecules of product. Therefore, there is increase in degree of freedom of the system , that makes ΔSsys>0.

Since, ΔSsys + ΔSsurr > 0,Therefore, ΔSsys>0, ΔSsurr<0

c) 2H₂(g)+O₂(g)→2H₂O(g)

For this reaction, ΔH∘rxn= -483.6 kJ i.e exothermic.

So, heat flows from system to surroundings.So, ΔSsurr>0.Not only that 3 molecules of reactants give two molecules of product. Therefore, there is reduction in degree of freedom of the system  i.e. ΔSsys<0​

Since, ΔSsys + ΔSsurr > 0, therefore , ΔSsys<0, ΔSsurr>0