# Question 496a7

##### 1 Answer
Dec 31, 2017

Here's what I got.

#### Explanation:

For starters, the units you have for the change in entropy is incorrect. The change in entropy is usually measured in joules per Kelvin, ${\text{J K}}^{- 1}$, so I assume that

$\Delta S = - \text{80 kJ}$

is actually

DeltaS = - 80 color(red)(cancel(color(black)("J"))) "K"^(-1) * "1 kJ"/(10^3color(red)(cancel(color(black)("J")))) = -"0.080 kJ K"^(-1)

As you know, we can use the change in Gibbs energy, $\Delta G$, to assess the spontaneity of a chemical reaction at a given temperature.

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{\Delta G = \Delta H - T \cdot \Delta S}}}$

Here

• $\Delta H$ is the change in enthalpy of the reaction
• $T$ is the absolute temperature at which the reaction takes place
• $\Delta S$ is the change in entropy of the reaction

Now, in order for a reaction to be spontaneous at a given temperature, you need to have

$\Delta G < 0$

In your case, the reaction is exothermic because its change in enthalpy is negative.

$\Delta H < 0 \implies \text{exothermic reaction}$

The change in entropy is negative, which means that the disorder of the system is actually decreasing.

$\Delta S < 0 \implies \text{decreasing disorder}$

This tells you that your reaction is enthalpy driven, which is what happens when an exothermic reaction overcomes a decrease in entropy.

In such cases, the spontaneity of the reaction depends on the temperature at which the reaction is taking place. In your case, you have

$\Delta G = {\overbrace{\Delta H}}^{\textcolor{b l u e}{< 0}} - {\overbrace{T \cdot \Delta S}}^{\textcolor{b l u e}{< 0}}$

For $\Delta S < 0$, this is equivalent to

$\Delta G = \Delta H + T \cdot | \Delta S |$

so in order for the reaction to be spontaneous, you need

$\Delta G < 0 \implies T \cdot | \Delta S | < | \Delta H |$

$T < \frac{| \Delta H |}{| \Delta S |}$

At equilibrium, you have

$\Delta G = 0 \implies T \cdot | \Delta S | = | \Delta H |$

which gets you

$T = \frac{\Delta H}{\Delta S}$

Plug in your values to find

T = (-50color(red)(cancel(color(black)("kJ"))))/(-0.080color(red)(cancel(color(black)("kJ")))"K"^(-1)) = "625 K"#

This means that your reaction is

• $\text{spontaneous " => " T < 626 K}$
• $\text{nonspontaneous " => " T > 625 K}$
• $\text{at equilibrium " => " T = 625 K}$

So you can say that when the temperature increases and reaches

$\text{625 K" = "625 K" - 273.15 = 352^@"C}$

your reaction goes from being spontaneous to being nonpontaneous.