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## AP®︎/College Chemistry

### Course: AP®︎/College Chemistry > Unit 9

Lesson 2: Gibbs free energy and thermodynamic favorability# Worked example: Determining the effect of temperature on thermodynamic favorability

When Δ

*H*° and Δ*S*° for a reaction have the same sign, the thermodynamic favorability of the reaction depends on temperature. In this video, we'll determine the thermodynamic favorability of a reaction with Δ*H*° < 0 and Δ*S*° < 0 at two different temperatures. We'll also calculate the temperature at which the reaction changes from being thermodynamically favored to thermodynamically unfavored. Created by Jay.## Want to join the conversation?

- 5:55shouldn't this be regular delta G and not delta G naught? Since it's not necessarily at 25degreesC anymore?(2 votes)
- True, when the temperature isn't at 25°C it isn't at standard conditions. However over a limited temperature range the changes between the standard and nonstandard free energy is small so that it can be used as an apt approximation.

Hope that helps.(3 votes)

- What is the meaning of ∆Go ?

I was under the impression that ∆Go referred to the difference in the Gibbs free energy of the reactants and products at that instant in time where in the reaction mixture, the reactants and products are in their standard states.

For example, for the reaction:

N2 (g) + 3 H2 (g) < ------- > 2NH3 (g) ,∆Go = -33 kJ

I thought that at that instant in time, where in the reaction mixture there is N2 at 1atm, H2 at 1atm and NH3 at 1atm, that:

∆Go = Gibbs Free Energy (Reactants) – Gibbs Free Energy (Products)

= Gibbs Free Energy (N2 at 1atm, H2 at 1atm) - Gibbs Free Energy (NH3 at 1atm)

= -33 kJ

However, I have also heard that ∆Go refers to the difference in Gibbs Free Energy in the reaction mixture when the products and the reactants are in their standard states and the equilibrium reaction mixture.

For example, for the reaction:

N2 (g) + 3 H2 (g) < ------- > 2NH3 (g) ,∆Go = -33 kJ

∆Go = Gibbs Free energy (standard state mixture) - Gibbs Free energy (equilibrium mixture)

= -33 kJ

Which one of these statements is correct?(1 vote)- ΔG° is the standard change in Gibbs free energy between the products and reactants of a reaction. Essentially it’s measuring the energy difference between the reactants and products. It’s also a measure of spontaneity, or whether a reaction will proceed in a certain direction naturally without any outside work of energy needed to help it along. A negative ΔG° indicates a reaction is spontaneous and a positive ΔG° indicates a reaction is nonspontaneous.

Differences like ΔG° are defined as the final state minus the initial state. So ΔG°rxn should actually be ΔG°products - ΔG°reactants. And actually the ΔG° for the reactants and products should actually be free energies of formation, or ΔG°f. The free energy of formation is the change in free energy when 1 mol of a compound in its standard state forms from its constituent elements in their standard states. Additionally, the stoichiometric coefficients of the chemical equation also factors into the calculation.

So the real equation should be: ΔG°rxn = ∑np ΔG°f(products) - ∑nrΔG°f(reactants), where np are the stoichiometric coefficients of the products, nr are the stoichiometric coefficients of the reactants, and the ∑ (sigma) symbols mean to add them together.

The degree symbol used for these definitions, °, means that everything is in its standard state. Standard states represent a convenient and mathematically simple reference point we so can compare thermodynamics values. So if we were to compare two free energy values at nonstandard conditions it wouldn’t be possible since the free energy values would also be affected by differences in temperature and pressure. Currently standard state refers to gases at a pressure of 1 bar, concentrations of solutes in solution to be 1 M, and for liquids and solids in their pure and most stable form at a pressure of 1 bar and a temperature of interest (usually 25°C or 298K).

So a ΔG° of -33 kJ/mol states that the reaction is spontaneous in the forward direction and that the concentration of the products will increase compared to their standard state 1M concentration.

Hope that helps.(1 vote)

## Video transcript

- [Lecturer] Let's do a work example where we calculate the standard change in free energy, delta G naught,
for a chemical reaction. And for our reaction, let's look at the synthesis of ammonia gas from nitrogen gas and hydrogen gas. At 25 degrees Celsius, delta H naught for this reaction is equal to negative 92.2 kilojoules
per mole of reaction. Since delta H naught is negative, this reaction is exothermic and at 25 degrees Celsius, delta S naught for this reaction is equal to negative 198.7 joules
per kelvin mole of reaction. Our goal is to calculate delta G naught for this reaction at 25 degrees Celsius. Remember if delta G naught is negative, the forward reaction is
thermodynamically favorable, but if delta G naught is positive, the forward reaction is
thermodynamically unfavorable. Since the change in enthalpy is negative, and the change in entropy is negative, whether or not delta G naught is negative depends on the temperature. To calculate the value for delta G naught, we're going to use the following equation. Delta G naught is equal to delta H naught minus T delta S naught, where T is the temperature in kelvin. Next, we need to plug
everything into our equation. So delta H nought is equal to negative 92.2 kilojoules
per mole of reaction. The temperature is 25 degrees Celsius. So 25 plus 273 is equal to 298 kelvin. So we're gonna plug 298
kelvin into our equation. And for delta S naught,
notice we have this in joules, whereas for delta H naught,
it was in kilojoules. So we need to convert delta S naught into kilojoules per
kelvin mole of reaction. One way to do that is just to move the decimal place three to the left, which gives us negative 0.1987 kilojoules per kelvin mole of reaction. So we're gonna plug all of
this in for delta S naught. Here's our equation with
everything plugged in. Notice that kelvin will cancel out, which gives us kilojoules per
mole of reaction as our units. And when we do the math, we find that delta G naught is equal to negative 33.0 kilojoules
per mole of reaction. Because delta G naught is negative, that means the reaction is
thermodynamically favorable in the forward direction. And the superscript naught
means that both the reactants and products are in their standard states. So what our calculation tells
us is if we had a mixture of nitrogen, hydrogen, and ammonia gas at 25 degrees Celsius, and all three gases had a partial pressure of one atmosphere, the reaction is
thermodynamically favorable in the forward direction, meaning nitrogen gas and
hydrogen gas would come together to synthesize or make more ammonia. So at 298 kelvin, delta G naught for this reaction is negative. Let's do another calculation
for the same reaction at a different temperature, a temperature of 1,000 kelvin. Also, even though the
value for delta G naught is highly dependent on temperature, the values for delta H
naught and delta S naught are not as dependent on the temperature. Therefore, we're gonna
assume that these values for delta H naught and
delta S naught don't change, and we're gonna use the same ones at the higher temperature of 1,000 kelvin. Here's the equation with
everything plugged in for the calculation of delta G naught at the higher temperature of 1,000 kelvin. And notice that the values
we're using for delta H naught and delta S naught are
the same as the ones that we used at the lower
temperature of 298 kelvin. Once again, kelvin cancels out and gives us kilojoules
per mole of reaction as our units for our final answer. And after we do the math, we find that delta G naught is equal to positive 106.5 kilojoules
per mole of reaction. Since delta G naught is positive, that means the forward reaction is thermodynamically unfavorable, which means the reverse reaction is thermodynamically favorable. So if we had a mixture of
nitrogen gas, hydrogen gas, and ammonia gas at a
temperature of 1,000 kelvin, and all three gases had a partial pressure of one atmosphere, the ammonia gas would
turn into nitrogen gas and hydrogen gas. And while we have both
calculations on the screen, let's analyze why one calculation
gives a negative value for delta G naught, and the other one gives a positive value. For the calculation at 298 kelvin, the temperature was low enough that the entropy term didn't overwhelm the negative value for the enthalpy term, and we ended up with a
negative overall value for delta G naught. However, for the
calculation at 1,000 kelvin, this time the temperature was high enough where the entropy term outweighed the negative value for the enthalpy term, and therefore the overall value for delta G naught
ended up being positive. We've just seen that for
the synthesis of ammonia from nitrogen gas and hydrogen gas at relatively low
temperatures like 298 kelvin, delta G naught is less than zero, and at relatively high
temperatures like 1,000 kelvin, delta G naught is greater than zero. So there must be some temperature between 298 kelvin and 1,000 kelvin where the forward reaction goes from being thermodynamically favorable to thermodynamically unfavorable. And this crossover point
occurs when the enthalpy term and the entropy term
are perfectly balanced. Therefore, delta G naught is equal to zero at this crossover point. So we plug in zero for delta G naught and we solve for the temperature
at the crossover point. After some algebra, we find that the temperature is equal to delta H naught divided
by delta S naught, which is equal to negative 92.2 kilojoules per mole of reaction divided
by negative 0.1987 kilojoules per kelvin mole of reaction. Kilojoules per mole of
reaction cancels out and gives us 464 kelvin for
the crossover temperature. So for this particular reaction, delta G naught is equal
to zero at 464 kelvin. At temperatures less than 464 kelvin, the forward reaction is
thermodynamically favorable. However, at temperatures
higher than 464 kelvin, the forward reaction is
thermodynamically unfavorable.