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.