Please click on one of the following topics:

- Predicting Spontaneity:How Its Done
- Gibbs-Helmholtz Equation
- Determining Delta G By Using Free Energies of Formation at 25C
- Determining Delta G at Any Given Temperature
- Temperature Effects On Spontaneity.

Predicting the Spontaneity of a Reaction can be a tricky business. On the surface you would think that the Change in Enthalpy (Delta H_{rx}) would predict whether a reaction is spontaneous or not simply by applying the First Law of Thermodynamics. If Delta H were negative that would mean that the chemical system would lose energy to the surroundings and that should make the products more stable. According to the First Law all systems tend toward lower (more stable) energy state. However the first law considers the total energy of the system. It doesn't take into account that maybe some of this energy change would involve an entropy increase big enough to make a reaction that might be predicted to be non-spontaneous to actually be spontaneous. Indeed there are some reactions that are spontaneous at low temperatures whereas others are spontaneous at only high temperatures. Therefore Delta H is not a fool proof way of predicting spontaneity.

Others might consider using the Change in Entropy of a reaction system to be a good indicator or predictor of spontaneity. According to the Second Law systems tend toward higher entropy so it might be concluded that if the change in Entropy is positive this would indicate spontaneity. The problem with that is that the second law is referring to "isolated" systems that have no input from the outside environment. If that happens and it is more than likely it will, then Delta S would not always predict spontaneity. There are systems that have a negative Delta S such as The precipitation of a slightly soluble salt from a chemical reaction which occurs spontaneously. What happens when you pour a solution of NaCl together with a solution of Silver Nitrate?

So what will be a good predictor? A good predictor of spontaneity would be one that considers the Delta H, the Delta S, and the temperature since there are some reactions as noted above that seem to respond to changes of temperature even though we find the Delta H and Delta S would be approximately the same at any temperature.

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Professor Gibbs and Helmholtz came up with a relationship that took all three of these factors into consideration. This is known as the Gibbs Helmholtz Equation:

Let's identify each term in this very important equation.

- The Delta H
_{rx}represents the total energy exchange that takes place between the system and its environment. - The T(Delta S
_{rx}) term represents the energy eused to take care of the intermolecular activity. This is wasted energy and has to be subtracted from the total energy. An analogy can be drawn here between a mechanical engine and a chemical reaction (engine). When a mechanical engine performs useful work not all of the energy output of the engine goes toward that end. Some of the energy output is wasted due to friction of moving parts. That is why we never have a perpetual motion engine. The friction of the moving parts will be wasted energy and must be subtracted from the total energy output to get the net useful energy capable of performing a task or work. - Delta G represents that net useful energy of a chemical system (engine). Due to the fact that molecules in motion will rub up against each other more or less depending on how independent they are of one another (the higher the entropy the more independent they are). This molecular friction (T Delta S) will siphon off from the total energy (Delta H) to result in net energy capable of performing a task (Delta G).

A third possibility is if Delta G was equal to zero. At that value the system is in a state of equilibrium. If one knows the Delta H and Delta S then one can determine the equilibrium temperature for the system by setting Delta G equal to 0 and solving for T. This equilibrium temperature would be the break even point for systems that might be positive or negative depending upon the temperature.

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Depending upon the conditions, Delta G for a reaction can be determined one of two ways:

- Calculate Delta G under standard conditions using the Standard Gibbs Free Energies of Formation
- Calculating Delta G under
**any**conditions using the Gibbs Helmholtz Equation.

If we are at standard conditions of 25 C, 1 atm pressure for any gaseous component and 1 Molar ion concentration, then we can determine the Gibbs Free Energy change by using Standard Free Energies of Formation. It must be emphasized that this method can only be used if the conditions are standardized. Any non-standard conditions will require the second method above. The Standard Free Energies of Formation are the Change in Free Energy upon the formation of the species from its elements under standard conditions. We can use Hess's Law of Summations similar to what we did in determining the
Delta H_{rx} and the
Delta S_{rx}. According to Hess's Law of Summations:

Delta G_{rx} = Sum of the Free Energies of Formation of Products - Sum of the the Free Energies of Formation of Reactants.

Let's take an example illustrating this method.

Determine the Delta G under standard conditions using Gibbs Free Energies of Formation found in a suitable Thermodynamics table for the following reaction:

4HCN(l) + 5O_{2}(g) ---> 2H_{2}O(g) + 4CO_{2}(g) + 2N_{2}(g)

- Check to make sure the equation is balanced
- Look up the Standard Free Energy of Formation of H2O(g) and multiply by its coefficient(2) in the equation.
2 moles ( -237.2 kj/mole) = -474.4 kj = Standard Free Energy of Formation for two moles H

_{2}O(l) - Look for the Standard Free Energy of Formation of CO
_{2}(g) and multiply by its coefficient (4)4 moles ( -394.4 kj/mole) = -1577.6 kJ = Standard Free Energy of Formation for four moles CO

_{2} - Look up the Standard Free Energy of Formation of N
_{2}(g) and multiply by its coefficient(2)2 moles(0.00 kj/mole) = 0.00 kJ = Standard Free Energy of Formation for 2 moles of N

_{2}(g) - Add the results of steps 2,3, and 4 to get the Standard Free Energy for the products
(-474.4 kJ) + (-1577.6 kJ) + 0.00 = -2052 kJ = Standard Free Energy for Products

- Look up the Standard Free Energy of Formation for HCN(l) and multiply by its coefficient(4)
4 moles ( 121 kj/mole) = 484 kJ = Standard Free Energy for 4 moles HCN(l)

- Look up the Standard Free Energy of Formation of O
_{2}(g) and multiply by its coefficient(5)5 moles (0.00 kJ/mole) = 0.00 kJ = Standard Free Energy of 5 moles of O2(g)

- Add the results of steps 7 and 8 to get the Standard Free Energy of the Reactants
(484) + (0.00) = 484 kJ = Standard Free Energy for Reactants

- Subtract the result of step 9 from the result of step 5 to get the Standrad Free Energy Change for the Reaction
Sum of Free Energy of Products - Sum of Free Energy of Reactants = (-2052 kJ) - (484 kJ) = -2536 kJ = Standard Free Energy Change for the Reaction.

Calculate the Free Energy Change for the following reaction using the Standard Free Energies of Formations found in the thermodynamics table:

2CH_{3}OH(l) + 3O_{2}(g)-----> 2CO_{2}(g) + 4H_{2}O(l)

When you have finished with your answer, you might check for the correct answer

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The Gibbs Free Energy can be determined at any temperature using the Gibbs Helmholtz Equation. One has to be given the Delta H for the reaction as well as the Delta S and the Temperature, or one have to be able to calculate Delta Hrx and Delta Srx using the previous methods.

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There are some systems that always seem to be spontaneous no matter what temperature they are run under. Other reaction systems look good on paper, but they will not go spontaneously at any temperature. Then there are those reactions that do not take off until you reach a threshold temperature which seems to set them off. And finally there are those systems that do not go spontaneously until you have reduced the temperature sufficiently. What makes these differences in how reaction systems respond to temperature changes? This is explained in the relationship between Delta H and Delta S of a system.

There are four scenarios that we can explore with the Gibbs Helmholtz Equation as the backdrop.

- Delta H > 0 and Delta S < 0
- Delta H < 0 and Delta S > 0
- Delta H < 0 and Delta S < 0
- Delta H > 0 and Delta S > 0

If Delta H was of a different sign to Delta S then it wouldn't make any difference what the temperature would be, the system would always be spontaneous if Delta H was negative and Delta S was positive (scenario 2 above). On the other hand if Delta H was positive and Delta S was negative (scenario 1 above) then the system would always be positive no matter what the temperature was.

It is only when the two thermodynamic properties have the same sign that temperature will make a difference. If they both are negative (scenario 3 above) then the TDelta S term will be positive since we are subtracting TDelta S from Delta H (see equation above). The question is how large must that term be in order for it to exceed the negative Delta H term and cause the reaction to have a non-spontaneous positive Delta G. Under those conditions any temperature BELOW the equilibrium temperature would make such a reaction spontaneous.

On the other hand, if Delta H and Delta S are both positive (scenario 4 above) then the TDelta S term will be negative because of the negative sign in the equation. Again the question becomes how large must the TDelta S term be so its negative value can overcome the positive Delta H value and cause the system to be spontaneous. Under these conditions any temperature ABOVE the equilibrium temperature would result in a spontaneous process.

Let's illustrate this method by the following example:

For the following reaction using the thermodynamics table:

CoCl_{2}(g) ---> CO(g) + Cl_{2}(g)

- Calculate at 127 C the Delta G
- Calculate the Temperature when the above reaction is at equilibrium (Delta G = 0)
- At what temperature will this reaction be spontaneous?

- Check to see if the equation is balanced
- Determine the Delta H of the Reaction using Hess Summation Law and Standard Enthalpies of Formation
Delta H = Sum of Delta H

_{f}of products - Sum Delta H_{f}of ReactantsDelta H = [ 1(-110.5) + 1(0.00)] - [ 1(-220)]

Delta H = -110.5 - (-220) = +110.5 kJ

- Determine Delta S for the reaction using Standard Molar Entropies and Hess Law of Summation
Delta S = Sum Standard Molar Entropies of Products - Sum of Standard Molar Entropies of Reactants

Delta S = [ 1 mole(197.5 J/mole-K) + 1 mole(223) J/mole-K] - [ 1 mole(283.7 J/mole-K)]

Delta S = 420.5 J/K - 283.7 J/K = 136.8 J/K

- Convert Delta S from J/K to kJ/K
136.8 J/K X 1 kJ / 1000 J = 136.8 / 1000 = .1368 kJ/K

- Convert 127 C to K
K = C + 273 = 127 + 273 = 400 K

- Plug results of step 2 and 4 into Gibbs Helmholtz Equation along with Kelvin Temperature to get Delta G of the Reaction
Delta G = Delta H - T(Delta S)

Delta G = 110.5 kJ - 400 K(.1368 kj/K)

Delta G = 110.5 - 54.72 kJ = + 55.78 kJ

- Set Delta G =0 and plug in Delta H and Delta S
Delta G = 0 = 110.5 kj - T(.1368 kJ/K)

- Solve for T
0 = 110.5 - .1368T

.1368T = 110.5

T = 110.5 / .1368 = 807.7 K or 534.7 C when the reaction is at equilibrium.

Since Delta H is positive and Delta S is positive the temperature would have to be large enough to make the T(Delta S) term large enough for the Delta G to become negative. The equilibrium temperature is the breakeven point where Delta G is 0 so any temperature ABOVE that temperature would make this reaction become spontaneous. Any temperature above 534.7 C would cause the above reaction to occur spontaneously.

Here is one for you to try.

For the following reaction at 327 C.

N_{2}(g) + 3H_{2}(g) ---> 2NH_{3}(g)

- Calculate the Delta G
- Calculate the temperature when this system will be at equilibrium (Delta G = 0)
- At what temperature will this reaction be spontaneous (if any)?

For further practice problems dealing with this topic, may I suggest the following Web site Thermodynamics Problem Set (problem sets developed by S.E. Van Bramer for chemistry and environmental science courses at Widener University.)

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R. H. Logan, Instructor of Chemistry, Dallas County Community College District, North Lake College.

Send Comments to R.H. Logan:

All textual content copyrighted (c) 1997 R.H. Logan, Instructor of Chemistry, DCCCD All Rights reserved

Revised: 11/30/98

Calculate the Free Energy Change for the following reaction using the Standard Free Energies of Formations found in the thermodynamics table:

2CH_{3}OH(l) + 3O_{2}(g)-----> 2CO_{2}(g) + 4H_{2}O(l)

- Check to make sure that the equation is balanced.
- Look up the Standard Free Energies of Formation of CO
_{2}(g) and multiply by its coefficient(2)2 moles(-394.4 kJ/mole) = -788.8 kJ = Standard Free Energy of 2 moles of CO

_{2}(g) - Look up the Standard Free Energy of Formation of H
_{2}O(l) and multiply by its coefficient(4)4 moles(-237.2 kJ/mole) = -948.8 kJ = Standard Free Energy for Formation of 4 moles of H

_{2}O(l) - Add the results of steps 2 and 3 to get the Standard Free Energy of the Products
(-788.8 kJ) + (-948.8 kJ) = -1737.6 kJ = Standard Free Energy for Products

- Look up the Standard Free Energy of Formation of CH
_{3}OH(l) and multiply by its coefficient(2)2 moles(-166.2 kJ/mole) = -332.4 kJ = Standard Free Energy of Formation of two moles of CH

_{3}OH(l) - Look up the Standard Free Energy of Formation of O
_{2}(g) and multiply it by its coefficient (3)3 moles (0.00 kJ/mole) = 0.00 kJ = Standard Free Energy for the formation of 3 moles of O

_{2}(g) - Add the results of steps 5 and 6 for the Standard Free Energy of the Reactants.
(-332.4 kJ) + (0.00 kJ) = -332.4 kJ = Standard Free Energy for Reactants

- Subtract the results from step 7 from the results of step 4 to get the Standard Free Energy for the reaction
(-1734.6 kJ) - (-332.4 kJ) = -1402.2 kJ = Standard Free Energy of the Reaction

For the following reaction at 327 C.

N_{2}(g) + 3H_{2}(g) ---> 2NH_{3}(g)

- Calculate the Delta G
- Calculate the temperature when this system will be at equilibrium (Delta G = 0)
- At what temperature will this reaction be spontaneous (if any)?

- Check to make sure the equation is balanced
- Calculate the Delta H using Standard Enthalpies of Formation and Hess Summation Equation
Delta H = Sum of Standard Enthalpies of Products - Sum of Standard Enthalpies of Reactants

Delta H = [2 moles (-45.9 kj/mole)] - [ 1 mole (0.00 kj/mole) + 3 moles (0.00 kJ/mole)]

Delta H = -91.8 kJ - 0.00 = -91.8 kJ

- Calculate the Delta S from the Standard Molar Entropies and the Summation Equation
Delta S = Sum Standard Entropies of Products - Sum Standard Entropies of Reactants

Delta S = [ 2 moles( 193 J/mole-K)] - [ 1 mole (191.5 J/mole-K) + 3 moles(130.6 J/mole-K)]

Delta S = 386 J/K - [ 191.5 J/K + 391.8 J/K]

Delta S = 386 J/K - 583.3 = - 197.3 J/K

- Convert J/K to kJ/K
-197.3 J/K X 1 kJ / 1000 J = -.1973 kJ/K

- Convert Celsius Temperature given to Kelvin
K = C + 273 = 327 + 273 = 600 K

- Using the Gibbs Helmholtz Equation Plug in results of steps 2 and 4 above along with the Kelvin Temperature to determine the Delta Grx
Delta G = Delta H - T(Delta S)

Delta G = -91.8 kJ - 600 K ( -.1973 kJ/K)

Delta G = -91.8 kJ + 118.4 kJ = +26.58 kJ

- Set Delta G = 0 for equilibrium conditions and plug in Delta H and Delta S
Delta Grx = 0 = -91.8 kJ - (-.1973)T

- Solve for T
0 = -91.8 kJ + .1973T

91.8 = .1973T

T = 91.8 / .1973 = 465.3 K or 192.3 C = Equilibrium Temp

Since Delta H is negative and Delta S is also negative then it does not matter what temperature is placed in the Gibbs Helmholtz equation it will always result in a negative Delta G. Therefore this reaction will be spontaneous at any temperature.