Question

Question #b08e1

Answer 1

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.