Question #1fad2

For a reaction to be feasible the total entropy change of the system and surroundings must be positive.

#N_(2(g))+2O_(2(g))rarr2NO_(2(g))#

To work out the entropy change in the system we need to find the total entropy of the products and subtract this from the total entropy of the reactants.

#DeltaS_(sys)=Sigma["entropy products"]-Sigma["entropy reactants"]#

You can get these data from standard tables:

#S^(0)(N_2)=191.5"J/K/mol"#

#S^(0)(O_2)=205.3"J/K/mol"#

#S^(0)(NO_2)=240.4"J/K/mol"#

#DeltaS_(sys)=(2xx240.4)-[191.5+(2xx205.3)]#

#DeltaS_(sys)=480.8-[602.1]#

#DeltaS_(sys)=-121.3"J/K/mol"#

Now we need to work out the entropy change in the surroundings which would result if this reaction were to happen.

This is given by:

#DeltaS_(surr)=-(DeltaH)/(T)#

#DeltaS_(surr)=-(66.4xx10^(3))/(298)=-222.8"J/K/mol"#

Now we need to add these two together:

#DeltaS_("total")=DeltaS_(sys)+DeltaS_(surr)#

#DeltaS_("total")=-121.3+(-222.8)#

#DeltaS_("total")=-344.1"J/K/mol"#

The negative value of #DeltaS_("total")# tells us that the reaction cannot be spontaneous at this temperature.