Question #b08e1

1 Answer
Jul 10, 2016

Answer:

Here's what I got.

Explanation:

The reaction given to you

#2"A"_ ((g)) + 3"B"_ ((g)) -> 2"C"_ ((g))#

for which

#DeltaH = "100 kJ mol"^(-1)" "# and #" " DeltaS = -"200 J mol"^(-1)"K"^(-1)#

will never proceed spontaneously in the forward direction, and that regardless of the temperature at which it takes place.

As you know, the spontaneity of a chemical reaction that takes place at constant pressure and constant temperature is given by the change in Gibbs free energy, #DeltaG#

#color(blue)(|bar(ul(color(white)(a/a)DeltaG = DeltaH - T * DeltaScolor(white)(a/a)|)))#

Here

#DeltaH# - the enthalpy change of reaction
#DeltaS# - the entropy change of reaction
#T# - the absolute temperature at which the reaction takes place

In order for a process to be spontaneous, you need to have

#DeltaG < 0#

As you can see by inspecting the values given to you, #DeltaG# will never be #<0# for a positive #DeltaH# and a negative #DeltaS#, since

#DeltaG = underbrace(overbrace(DeltaH)^(color(blue)(>0)) - overbrace(T * DeltaS)^(color(purple)(<0)))_(color(darkgreen)(>0))#

Keep in mind that the absolute temperature, i.e. the temperature expressed in Kelvin, is always positive.

The term #-T * DeltaS# will always be positive for a negative #DeltaS#, and so the value of #T# is not relevant here.

This is why the reaction will never be spontaneous in the forward direction. The forward reaction is endothermic, since #DeltaH>0#, and it leads to a decrease in the entropy of the system, since #DeltaS<0#, which is why it won't be spontaneous regardless of the temperature at which it takes place.

Interestingly enough, it will always be spontaneous in the reverse direction

#2"C"_ ((g)) -> 2"A"_ ((g)) + 3"B"_ ((g))#

That happens because

#DeltaH_"reverse" = -DeltaH_"forw"" "# and #" "DeltaS_"rev" = -DeltaS_"forw"#

In this case, you have

#DeltaH_"rev" = -"100 kJ mol"^(-1)#

#DeltaS_"rev" = -(-"200 J mol"^(-1)"K"^(-1)) = +"200 J mol"^(-1)"K"^(-1)#

The #DeltaG# for the reverse reaction will be

#DeltaG_"rev" = underbrace(overbrace(DeltaH_ "rev")^(color(blue)(<0)) - overbrace(T * DeltaS_ "rev")^(color(purple)(>0)))_(color(darkgreen)(<0))#

This time the term #-T * DeltaS_"rev"# is always negative, which is why #DeltaG_"rev"# is always #<0#, regardless of the temperature at which the reaction takes place.

http://www.eoht.info/page/Gibbs+free+energy