Question #b08e1
1 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,
#color(blue)(|bar(ul(color(white)(a/a)DeltaG = DeltaH - T * DeltaScolor(white)(a/a)|)))#
Here
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 = 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
This is why the reaction will never be spontaneous in the forward direction. The forward reaction is endothermic, since
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_"rev" = underbrace(overbrace(DeltaH_ "rev")^(color(blue)(<0)) - overbrace(T * DeltaS_ "rev")^(color(purple)(>0)))_(color(darkgreen)(<0))#
This time the term