How do you calculate gibbs

What happens when one of the potential driving forces behind a chemical reaction is favorable and the other is not? We can answer this question by defining a new quantity known as the Gibbs free energy G of the system, which reflects the balance between these forces.

The Gibbs free energy of a system at any moment in time is defined as the enthalpy of the system minus the product of the temperature times the entropy of the system. The Gibbs free energy of the system is a state function because it is defined in terms of thermodynamic properties that are state functions.

The change in the Gibbs free energy of the system that occurs during a reaction is therefore equal to the change in the enthalpy of the system minus the change in the product of the temperature times the entropy of the system.

If the reaction is run at constant temperature, this equation can be written as follows. The change in the free energy of a system that occurs during a reaction can be measured under any set of conditions.

If the data are collected under standard-state conditions, the result is the standard-state free energy of reaction G o. The beauty of the equation defining the free energy of a system is its ability to determine the relative importance of the enthalpy and entropy terms as driving forces behind a particular reaction.

The change in the free energy of the system that occurs during a reaction measures the balance between the two driving forces that determine whether a reaction is spontaneous. As we have seen, the enthalpy and entropy terms have different sign conventions. The entropy term is therefore subtracted from the enthalpy term when calculating G o for a reaction. Because of the way the free energy of the system is defined, G o is negative for any reaction for which H o is negative and S o is positive.

G o is therefore negative for any reaction that is favored by both the enthalpy and entropy terms. We can therefore conclude that any reaction for which G o is negative should be favorable, or spontaneous. Conversely, G o is positive for any reaction for which H o is positive and S o is negative.

Any reaction for which G o is positive is therefore unfavorable. Free energy calculations become important for reactions favored by only one of these factors. Calculate H and S for the following reaction:. Use the results of this calculation to determine the value of G o for this reaction at 25 o C, and explain why NH 4 NO 3 spontaneously dissolves is water at room temperature.

Click here to check your answer to Practice Problem 6. Click here to see a solution to Practice Problem 6. The balance between the contributions from the enthalpy and entropy terms to the free energy of a reaction depends on the temperature at which the reaction is run. Use the values of H and S calculated in Practice Problem 5 to predict whether the following reaction is spontaneous at 25C:. Click here to check your answer to Practice Problem 7. Click here to see a solution to Practice Problem 7.

The equation used to define free energy suggests that the entropy term will become more important as the temperature increases. Since the entropy term is unfavorable, the reaction should become less favorable as the temperature increases. Assume that the values of H o and S used in Practice Problem 7 are still valid at this temperature. Click here to check your answer to Practice Problem 8. Click here to see a solution to Practice Problem 8. G o for a reaction can be calculated from tabulated standard-state free energy data.

Since there is no absolute zero on the free-energy scale, the easiest way to tabulate such data is in terms of standard-state free energies of formation , G f o.

As might be expected, the standard-state free energy of formation of a substance is the difference between the free energy of the substance and the free energies of its elements in their thermodynamically most stable states at 1 atm, all measurements being made under standard-state conditions. We are now ready to ask the obvious question: What does the value of G o tell us about the following reaction? By definition, the value of G o for a reaction measures the difference between the free energies of the reactants and products when all components of the reaction are present at standard-state conditions.

G o therefore describes this reaction only when all three components are present at 1 atm pressure. The sign of G o tells us the direction in which the reaction has to shift to come to equilibrium. The fact that G o is negative for this reaction at 25 o C means that a system under standard-state conditions at this temperature would have to shift to the right, converting some of the reactants into products, before it can reach equilibrium.

Enthalpy change. Entropy change. Gibbs free energy. At what temperature will the reaction above become spontaneous? The fact that both terms are negative means that the Gibbs free energy equation is balanced and temperature dependent:. Because this reaction has a positive Delta G it will be non-spontaneous as written. CO 2 g CH 3 OH l We're assuming that we're at room temperature, or degrees Kelvin. That's I should just say, Kelvin.

I should get in the habit of not saying degrees when I say Kelvin. Which is 25 degrees Celsius, times our change in entropy. Now, this is going to be a minus. Now you might say, OK, minus , you might want to put that there. But you have to be very, very, very careful. This right here is in kilojoules. This right here is in joules. So if we want to write everything in kilojoules, since we already wrote that down, let's write this in kilojoules.

So it's 0. And so now our Gibbs free energy right here is going to be minus kilojoules minus so the minus and the minus, you get a plus. And that makes sense, that the entropy term is going to make our Gibbs free energy more positive. Which, as we know, since we want to get this thing below 0, this is going to fight the spontaneity. But let's see if it can overwhelm the actual enthalpy, the exothermic nature of it. And it seems like it will, because you multiply a fraction times this, it's going to be a smaller number than that.

But let's just figure it out. So divided by 1, 2, 3. That's our change in entropy times , that's our temperature, is minus So this term becomes-- and then we put a minus there-- so it's plus So this is the entropy term at standard temperature. It turns into that. And this is our enthalpy term. So we can already see that the enthalpy is a much more negative number than our positive term from our temperature times our change in entropy.

So this term is going to win out. Even though we lose entropy in this reaction, it releases so much energy that's going to be spontaneous. This is definitely less than 0, so this is going to be a spontaneous reaction. As you can see, these Gibbs free energy problems, they're really not too difficult. You just really need to find these values. And to find these values, it'll either be given, the delta h, but we know how to solve for the delta h. You just look up the heats of formations of all the products, subtract out the reactants, and of course you wait by the coefficients.

And then, to figure out the change in entropy, you do the same thing. You have to look up the standard molar entropies of the products' weight by the coefficients, subtract out the reactants, and then just substitute in here, and then you essentially have the Gibbs free energy.

And in this case, it was negative. Now, you could imagine a situation where we're at a much higher temperature. Like the surface of the sun or something, where all of a sudden, instead of a here, if you had like a 2, or a 4, there. Then all of a sudden, things become interesting.

If you could imagine, if you had a 40, Kelvin temperature here, then all of a sudden the entropy term, the loss of entropy, is going to matter a lot more. And so this term, this positive term, is going to outweigh this, and maybe it wouldn't be spontaneous at a very, very, very, very high temperature. Another way to think about it. A reaction that generates heat that lets out heat-- the heat being released doesn't matter so much when there's already a lot of heat or kinetic energy in the environment.