Worked example: Determining the effect of temperature on thermodynamic favorability | Khan Academy

Let's do a worked example where we calculate the standard change in free energy, ΔG⁰, for a chemical reaction. For our reaction, let's look at the synthesis of ammonia gas from nitrogen gas and hydrogen gas at 25 degrees Celsius.

ΔH⁰ for this reaction is equal to negative 92.2 kilojoules per mole of reaction. Since ΔH⁰ is negative, this reaction is exothermic. At 25 degrees Celsius, ΔS⁰ for this reaction is equal to negative 198.7 joules per kelvin mole of reaction.

Our goal is to calculate ΔG⁰ for this reaction at 25 degrees Celsius. Remember, if ΔG⁰ is negative, the forward reaction is thermodynamically favorable, but if ΔG⁰ 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 ΔG⁰ is negative depends on the temperature.

To calculate the value for ΔG⁰, we're going to use the following equation:

ΔG⁰ = ΔH⁰ - TΔS⁰, where T is the temperature in Kelvin.

Next, we need to plug everything into our equation. So ΔH⁰ 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. We're going to plug 298 Kelvin into our equation.

For ΔS⁰, notice we have this in joules, whereas for ΔH⁰ it was in kilojoules. So we need to convert ΔS⁰ 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 going to plug all of this in for ΔS⁰. 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. When we do the math, we find that ΔG⁰ is equal to negative 33.0 kilojoules per mole of reaction.

Because ΔG⁰ is negative, that means the reaction is thermodynamically favorable in the forward direction. The superscript knot 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, ΔG⁰ for this reaction is negative. Let's do another calculation for this same reaction at a different temperature, a temperature of 1000 Kelvin. Also, even though the value for ΔG⁰ is highly dependent on temperature, the values for ΔH⁰ and ΔS⁰ are not as dependent on the temperature. Therefore, we're going to assume that these values for ΔH⁰ and ΔS⁰ don't change, and we're going to use the same ones at the higher temperature of 1000 Kelvin.

Here's the equation with everything plugged in for the calculation of ΔG⁰ at the higher temperature of 1000 Kelvin. Notice that the values we're using for ΔH⁰ and ΔS⁰ 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.

After we do the math, we find that ΔG⁰ is equal to positive 106.5 kilojoules per mole of reaction. Since ΔG⁰ 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 1000 Kelvin, and all three gases had a partial pressure of one atmosphere, the ammonia gas would turn into nitrogen gas and hydrogen gas.

While we have both calculations on the screen, let's analyze why one calculation gives a negative value for ΔG⁰ and the other 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 ΔG⁰.

However, for the calculation at 1000 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 ΔG⁰ 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, ΔG⁰ is less than zero, and at relatively high temperatures like 1000 Kelvin, ΔG⁰ is greater than zero. So there must be some temperature between 298 Kelvin and 1000 Kelvin where the forward reaction goes from being thermodynamically favorable to thermodynamically unfavorable.

This crossover point occurs when the enthalpy term and the entropy term are perfectly balanced; therefore, ΔG⁰ is equal to zero at this crossover point. We plug in zero for ΔG⁰ and we solve for the temperature at the crossover point. After some algebra, we find that the temperature is equal to ΔH⁰ divided by ΔS⁰, 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, ΔG⁰ 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.