# Question b08e1

Jul 10, 2016

Here's what I got.

#### Explanation:

The reaction given to you

$2 {\text{A"_ ((g)) + 3"B"_ ((g)) -> 2"C}}_{\left(g\right)}$

for which

$\Delta H = \text{100 kJ mol"^(-1)" }$ and ${\text{ " DeltaS = -"200 J mol"^(-1)"K}}^{- 1}$

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, $\Delta G$

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \Delta G = \Delta H - T \cdot \Delta S \textcolor{w h i t e}{\frac{a}{a}} |}}}$

Here

$\Delta H$ - the enthalpy change of reaction
$\Delta S$ - 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

$\Delta G < 0$

As you can see by inspecting the values given to you, $\Delta G$ will never be $< 0$ for a positive $\Delta H$ and a negative $\Delta S$, since

$\Delta G = {\underbrace{{\overbrace{\Delta H}}^{\textcolor{b l u e}{> 0}} - {\overbrace{T \cdot \Delta S}}^{\textcolor{p u r p \le}{< 0}}}}_{\textcolor{\mathrm{da} r k g r e e n}{> 0}}$

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

The term $- T \cdot \Delta S$ will always be positive for a negative $\Delta S$, 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 $\Delta H > 0$, and it leads to a decrease in the entropy of the system, since $\Delta S < 0$, 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

$2 {\text{C"_ ((g)) -> 2"A"_ ((g)) + 3"B}}_{\left(g\right)}$

That happens because

$\Delta {H}_{\text{reverse" = -DeltaH_"forw"" }}$ and $\text{ "DeltaS_"rev" = -DeltaS_"forw}$

In this case, you have

$\Delta {H}_{\text{rev" = -"100 kJ mol}}^{- 1}$

$\Delta {S}_{\text{rev" = -(-"200 J mol"^(-1)"K"^(-1)) = +"200 J mol"^(-1)"K}}^{- 1}$

The $\Delta G$ 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 $- T \cdot \Delta {S}_{\text{rev}}$ is always negative, which is why $\Delta {G}_{\text{rev}}$ is always $< 0$, regardless of the temperature at which the reaction takes place.