Without doing any calculations, determine the sign of Ssys and Ssurr for each of

Question

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

Explanation / Answer

Since a) 2CO(g)+O2(g)2CO2(g)

In this reaction, Hrxn= -566.0 kJ which means reaction is exothermic. So, heat flows from system to surroundings.So, Ssurr>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. Ssys<0

Since, Ssys + Ssurr > 0, therefore , Ssys<0, Ssurr>0

b) 2NO2(g)2NO(g)+O2(g)

For this reaction, Hrxn= +113.1 kJ i.e. the reaction is endothermic. So, heat flows from surroundings to system. Thus, Ssurr<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 Ssys>0.

Since, Ssys + Ssurr > 0,Therefore, Ssys>0, Ssurr<0

c) 2H2(g)+O2(g)2H2O(g)

For this reaction, Hrxn= -483.6 kJ i.e exothermic.

So, heat flows from system to surroundings.So, Ssurr>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. Ssys<0

Since, Ssys + Ssurr > 0, therefore , Ssys<0, Ssurr>0

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