Chemical Thermodynamics

We have seen in the previous post, that the Gibbs free energy of a reaction is a comprehensive measure of the spontaneity of the reaction that can be calculated based on the parameters of the system.

 

 

Today, we are going to see how the enthalpy, temperature, and entropy of the reaction affect the ΔGo. The first thing is you need to remember that a negative ΔGo indicates a spontaneous reaction, and a positive ΔGo indicates a nonspontaneous reaction:

 

 Δ< 0: A negative ΔG corresponds to a spontaneous process.
 ΔG > 0: A positive ΔG corresponds to a nonspontaneous process.
 Δ= 0: The system is at equilibrium. There is no net change.

 

Let’s first look at the big picture by isolating the two terms in the Gibbs free energy equation: the ΔHo and TΔSo terms:

 

 

To make the (more) process spontaneous, you want to make the ΔGo a negative large number. Therefore, a negative ΔHo, meaning an exothermic reaction, is a good contributor to making the reaction spontaneous. On the other hand, increasing the entropy makes the ΔGo smaller because of the negative sign and therefore, it also indicates a spontaneous reaction:

 

  

In general, there are four possibilities for the signs of the ΔHo and ΔSo which are summarized below:

 

 

Scenario 1. When ΔH is negative and ΔS is positive, ΔG is negative at all temperatures because regardless of the values, the net product is a negative number.

Therefore, we can state that exothermic reactions with a positive entropy change are always spontaneous.

 

Scenario 2. When ΔH is positive and ΔS is negative, ΔG is positive at all temperatures, and therefore, the reaction is nonspontaneous. This is expected because an endothermic reaction goes upwards energetically and decreasing the entropy makes the reaction less spontaneous too. To summarize, we can say that endothermic reactions with a negative entropy change are always nonspontaneous, regardless of the temperature.

 

Scenario 3. If both ΔH and ΔS are positive, then ΔG will be negative only when the T ΔS > H. Increasing the temperature will favor spontaneity as it makes the T ΔS a larger negative number because of the negative sign.

 

Scenario 4. If both ΔH and ΔS are negative, then ΔG will be negative only when the T ΔS < H. Decreasing the temperature will favor spontaneity as it makes the T ΔS a smaller positive number.

 

For example, given the values, ΔH°rxn = -18,000 J, and ΔS°rxn = -70 J/K, calculate the ΔG°rxn at 25 oC and determine if the reaction is spontaneous or not.

Solution:

ΔG°rxn = ΔH°rxn – TΔS°rxn = -18,000 J – (25 + 273) (-70 J/K) = 2,860 J

 

ΔG°rxn  > 0, therefore, the process is nonspontaneous at 25 oC.

 

The temperature at Which the Reaction Becomes Spontaneous

Now, you need to know that if the reaction is not spontaneous at a given temperature, it does not mean it won’t be spontaneous at a different temperature. Sometimes, you may get a question on a test asking to determine the temperature at which a given reaction becomes spontaneous.

 

To do that, you need to first, set ΔG = 0 which corresponds to the equilibrium, and determine the corresponding temperature. This is the crossover temperature at which a reaction switches between being spontaneous and nonspontaneous and we derive it from the Gibbs free energy equation.

 

For example, at what temperature the reaction in the previous example will become spontaneous?

 

Solution: To find the temperature at which the reaction becomes spontaneous, first, set ΔG = 0 which corresponds to the equilibrium, and determine the corresponding temperature.

 

ΔG = ΔHTΔS

0 = ΔHTΔS

 

\[T\, = \,\frac{{\Delta H}}{{\Delta S}}\]

                                                                        

\[T{\mkern 1mu}  = {\mkern 1mu} \frac{{\Delta H}}{{\Delta S}}\; = \;\frac{{18,000\,{\rm{J}}}}{{70\,{\rm{J/K}}}}\, = \;257\,{\rm{K}}\,\]

 

Next, identify the favorable term that makes ΔG negative. In this case, ΔH is the favorable term because it is negative and therefore, it decreases the ΔG. Because the entropy is negative, TΔSorxn is positive, we need to make the TΔSorxn a less significant contributor by decreasing the temperature. It means the process will become spontaneous below 257 K.

 

Note: If the ΔHorxn and ΔSorxn are not given in the problem, you need to calculate them first, using the data in the appendix, and then plug the values in the Gibbs free energy equation.

 

 

Practice

1.

Given the values of ΔH and ΔS, which of the following changes will be spontaneous at constant T and P?

a) ΔH = +35 kJ, ΔS = +10 J/K, T = 300. K

b) ΔH = +45 kJ, ΔS = +153 J/K, T = 312 K

c) ΔH = – kJ, ΔS = +8.0 J/K, T = 298 K

d) ΔH = – kJ, ΔS = -60. J/K, T = 250 K

a)
answer

Nonspontaneous

b)
answer

Spontaneous

c)
answer

Spontaneous

d)
answer

Nonspontaneous

Solution

ΔG = ΔHTΔS and the process will be spontaneous when ΔG < 0.

 

a) ΔH = +35 kJ, ΔS = +10 J/K, T = 300 K

Make sure the units of ΔH and ΔS match in terms of Joules.

ΔH = +35 kJ = 35 x 103 J

ΔG = 35 x 103 J – 300. K ΔS x 15.0 J/K = 30,500 J

ΔG > 0, therefore, the process is nonspontaneous.

 

b) ΔH = +45 kJ, ΔS = +153 J/K, T = 312 K

ΔH = +45 kJ = 45 x 103 J

ΔG = 45 x 103 J – 312 K x 153 J/K = -2736 J

ΔG < 0, therefore, the process is spontaneous.

 

c) ΔH = – kJ, ΔS = +8.0 J/K, T = 298 K

ΔH = -20. kJ = -20. x 103 J

ΔG = -20. x 103 J – 298 K x 8.0 J/K = -22,384 J

ΔG < 0, therefore, the process is spontaneous.

 

d) ΔH = – kJ, ΔS = -60. J/K, T = 250 K

ΔH = -15. kJ = -15. x 103 J

ΔG = -14. x 103 J – 250 K x (-60. J/K) = 1000 J

ΔG > 0, therefore, the process is nonspontaneous.

2.

At what temperatures will the following processes be spontaneous?

a) ΔH = +25.0 kJ, ΔS = +40.0 J/K

b) ΔH = -5.00 kJ, ΔS = -30.0 J/K

c) ΔH = -20.0 kJ, ΔS = +60.0 J/K

c) ΔH = +45.0 kJ, ΔS = -70.0 J/K

a)
answer

T > 625 K

b)
answer

T < 167 K

c)
answer

Spontaneous at any temperature

d)
answer

Not spontaneous at any temperature

Solution

We need to find the temperature at which the ΔG becomes negative. One strategy is to set ΔG = 0 which corresponds to the equilibrium and determine the corresponding temperature. Whether the temperature will need to be higher or lower than it is at equilibrium depends on the signs of ΔH and ΔS.

 

a) ΔH = +25.0 kJ, ΔS = +40. J/K

 

ΔG = ΔHTΔS

0 = ΔHTΔS

 

\[T\, = \,\frac{{\Delta H}}{{\Delta S}}\]

 

Before plugging the numbers, make sure the units of ΔH and ΔS match in terms of Joules.

 

\[T\, = \,\frac{{{\rm{25,000}}\,{\rm{J}}}}{{{\rm{40}}.0\,{\rm{J/K}}}}\; = \;625\,K\]

 

At this point, find the favorable term that makes ΔG negative. That is a negative ΔH, and a positive ΔS. In this case, ΔS is the favorable term, therefore, therefore, the process will become spontaneous when the temperature becomes higher than 625 K.

Write the Gibbs free energy expression to help visualize this:

 

ΔG = ΔHTΔS = +25,000 J – T x 40.0 J/K

 

Higher temperature is favorable for a process with positive a ΔH and a positive ΔS because of the negative sign in from of the “TΔS” term.

 

b) ΔH = -5.00 kJ, ΔS = -30.0 J/K

 

\[T\, = \,\frac{{\Delta H}}{{\Delta S}}\; = \,\frac{{ – 5.000\; \times \,{{10}^3}\,{\rm{J}}}}{{ – 30.0\;{\rm{J/K}}}}\; = \,167\;{\rm{K}}\]

 

ΔG = -5,000 – T (-30.0) = -5000 + 30.0 T

 

The favorable term is the ΔH because it is negative and therefore, we want to minimize the unfavorable impact of the entropy. It means the process will become spontaneous below 167 K.

 

c) ΔH = -20.0 kJ, ΔS = +60.0 J/K

When ΔH is negative and  ΔS is positive, ΔG will always be negative and therefore, the process is spontaneous at any temperature.

 

d) ΔH = +45.0 kJ, ΔS = – J/K

The process can never be spontaneous at any temperature when ΔH is positive and ΔS is negative because ΔG can never be negative.

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