Question

Question #496a7

Answer 1

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, `"J K"^(-1)`, so I assume that

`DeltaS = - "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, `DeltaG`, to assess the spontaneity of a chemical reaction at a given temperature.

`color(blue)(ul(color(black)(DeltaG = DeltaH - T * DeltaS)))`

Here

• `DeltaH` is the change in enthalpy of the reaction
• `T` is the absolute temperature at which the reaction takes place
• `DeltaS` 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

`DeltaG < 0`

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

`DeltaH < 0 implies "exothermic reaction"`

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

`DeltaS < 0 implies "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

`DeltaG = overbrace(DeltaH)^(color(blue)(<0)) - overbrace(T * DeltaS)^(color(blue)(<0))`

For `DeltaS < 0`, this is equivalent to

`DeltaG = DeltaH + T * |DeltaS|`

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

`DeltaG < 0 implies T * |DeltaS| < |DeltaH|`

`T < (|DeltaH|)/(|DeltaS|)`

At equilibrium, you have

`DeltaG = 0 implies T * |DeltaS| = |DeltaH|`

which gets you

`T = (DeltaH)/(DeltaS)`

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

• `"spontaneous " => " T < 626 K"`
• `"nonspontaneous " => " T > 625 K"`
• `"at equilibrium " => " T = 625 K"`

So you can say that when the temperature increases and reaches

`"625 K" = "625 K" - 273.15 = 352^@"C"`

your reaction goes from being spontaneous to being nonpontaneous.