Question #496a7

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.