How to Convert Entropy Units in Gibbs Free Energy Calculations

A reaction has an enthalpy change of \(-40.0\ \text{kJ mol}^{-1}\) and an entropy change of \(+90.0\ \text{J mol}^{-1}\text{K}^{-1}\). You need to calculate its Gibbs free energy change at 298 K.

You know the equation:

\[
\Delta G^\circ=\Delta H^\circ-T\Delta S^\circ
\]

So you substitute the numbers:

\[
\Delta G^\circ=-40.0-298(90.0)
\]

But there’s a problem. You’re trying to subtract a value measured in joules from one measured in kilojoules. The calculator will happily give you an answer, but it won’t be a physically meaningful one.

The fix is simple: convert entropy from J to kJ by dividing by 1000 before substituting.

\[
90.0\ \text{J mol}^{-1}\text{K}^{-1}
=0.0900\ \text{kJ mol}^{-1}\text{K}^{-1}
\]

Now:

\[
\begin{aligned}
\Delta G^\circ
&=-40.0-(298\times0.0900)\\
&=-40.0-26.82\\
&=-66.82\ \text{kJ mol}^{-1}
\end{aligned}
\]

The answer is \(-66.8\ \text{kJ mol}^{-1}\).

That’s the essential rule: make sure \(\Delta H\) and \(T\Delta S\) use the same energy units before subtracting them.

01The correct unit conversion

The Gibbs free energy equation combines three quantities:

\[
\Delta G^\circ=\Delta H^\circ-T\Delta S^\circ
\]

Each symbol represents something different:

• \(\Delta G^\circ\): standard Gibbs free energy change, usually in \(\text{kJ mol}^{-1}\).
• \(\Delta H^\circ\): standard enthalpy change, usually in \(\text{kJ mol}^{-1}\).
• \(T\): absolute temperature, measured in kelvin (K).
• \(\Delta S^\circ\): standard entropy change, usually in \(\text{J mol}^{-1}\text{K}^{-1}\).

Notice the mismatch.

Enthalpy is commonly provided in kilojoules, while entropy is commonly provided in joules.

Since \(1\text{ kJ}=1000\text{ J}\), you need to convert one of these quantities before calculating.

Method 1: Convert entropy into kilojoules

This is usually the easiest method for HSC calculations.

Divide the entropy change by 1000:

\[
\boxed{
\Delta S^\circ(\text{kJ mol}^{-1}\text{K}^{-1})
=
\frac{\Delta S^\circ(\text{J mol}^{-1}\text{K}^{-1})}{1000}
}
\]

Then substitute into the Gibbs equation.

The units now cancel correctly:

\[
\text{K}\times
\frac{\text{kJ}}{\text{mol K}}
=
\frac{\text{kJ}}{\text{mol}}
\]

Therefore, both \(\Delta H^\circ\) and \(T\Delta S^\circ\) are in \(\text{kJ mol}^{-1}\), allowing you to subtract them.

Method 2: Convert enthalpy into joules

You can also do the calculation entirely in joules.

Multiply the enthalpy change by 1000:

\[
\Delta H^\circ(\text{J mol}^{-1})
=
\Delta H^\circ(\text{kJ mol}^{-1})\times1000
\]

Then leave entropy in its original units.

Calculate \(\Delta G^\circ\) in joules per mole, and divide the final answer by 1000 if you want it in kilojoules per mole.

Both methods give exactly the same result.

A useful exam habit: Unless the question specifically asks for joules, convert entropy into kilojoules and keep your entire calculation in kJ.

02Don’t forget to convert Celsius into kelvin

There is another unit mistake worth catching before you start.

The Gibbs free energy equation requires absolute temperature, not temperature in degrees Celsius.

Convert using:

\[
T(\text{K})=T(^\circ\text{C})+273.15
\]

For example:

\[
25^\circ\text{C}=298.15\text{ K}
\]

If a question gives you \(25^\circ\text{C}\), substituting 25 directly into the Gibbs equation will give an incorrect answer.

One distinction is worth remembering: a temperature difference of \(1^\circ\text{C}\) has the same numerical size as a temperature difference of \(1\text{ K}\). However, the temperature \(T\) in the Gibbs equation must be in kelvin.

05Why the signs matter when calculating Gibbs free energy

Unit conversion is only half the calculation. You also need to handle positive and negative signs correctly.

The most common mistake occurs when entropy is negative.

For example:

\[
\Delta G^\circ=\Delta H^\circ-T(-\Delta S)
\]

Subtracting a negative number is equivalent to adding a positive number.

This means a negative entropy change makes the entropy contribution less favourable as temperature increases.

The following table summarises the four possible combinations.

These predictions assume enthalpy and entropy remain approximately constant with temperature.

Feasibility is not the same as reaction rate

A negative Gibbs free energy change tells you that a reaction is thermodynamically spontaneous in the forward direction under the stated conditions.

It does not tell you how quickly the reaction will occur.

Imagine a reaction with a very negative \(\Delta G^\circ\), but a large activation energy. The reaction may be thermodynamically favourable and still proceed extremely slowly.

A catalyst can lower the activation energy and increase the reaction rate. However, it does not change the overall Gibbs free energy change for the same initial and final states at a fixed temperature and pressure.

For HSC Chemistry, keep these two ideas separate:

• Gibbs free energy determines thermodynamic spontaneity or feasibility.
• Activation energy and reaction kinetics help determine reaction rate.

A reaction being spontaneous does not mean it happens immediately.

06A quick method to use in an exam

When you see a Gibbs free energy calculation, follow this sequence.

1. Write down the Gibbs equation: \(\Delta G^\circ=\Delta H^\circ-T\Delta S^\circ\).
2. Check the units of enthalpy and entropy.
3. Convert entropy from J to kJ by dividing by 1000, if enthalpy is in kJ.
4. Convert any temperature in Celsius into kelvin.
5. Calculate \(T\Delta S^\circ\), keeping the negative signs.
6. Subtract the entropy term from enthalpy.
7. State the sign of \(\Delta G^\circ\) and explain what it means for the forward reaction.

One extra detail: if the question gives you individual standard molar entropy values for the reactants and products, calculate the reaction entropy first. Use the balanced equation and its stoichiometric coefficients, then convert the resulting \(\Delta S^\circ\) into consistent units.

07Syllabus scope: 2017 Chemistry

The Gibbs free energy calculation covered here belongs to the 2017 Chemistry Stage 6 Syllabus, Year 11 Module 4: Drivers of Reactions.

This module includes the relationship between enthalpy, entropy, and Gibbs free energy, using:

\[
\Delta G^\circ=\Delta H^\circ-T\Delta S^\circ
\]

It also examines how temperature affects reaction spontaneity.

NESA’s separate Chemistry 11-12 Syllabus (2025) is scheduled for implementation from 2028, with its first HSC examination in 2029. Students should check the content for their particular syllabus edition rather than assuming that every 2017 syllabus requirement carries over.

You can consult the official documents here:

• NESA: Chemistry Stage 6 Syllabus (2017)
• NESA: Chemistry 11-12 Syllabus (2025)

08Questions and solutions

Try each question before reading its solution. The questions move from straightforward unit conversion towards more challenging applications involving temperature, reaction scaling, and thermodynamic reasoning.

Question 1

A chemical reaction has the following thermodynamic data at 298 K:

\[
\Delta H^\circ=-32.0\ \text{kJ mol}^{-1}
\]

\[
\Delta S^\circ=+75.0\ \text{J mol}^{-1}\text{K}^{-1}
\]

Calculate \(\Delta G^\circ\) and state whether the forward reaction is thermodynamically spontaneous under standard-state conditions.

Solution 1

The forward reaction is thermodynamically spontaneous, with:

\[
\boxed{\Delta G^\circ=-54.4\ \text{kJ mol}^{-1}}
\]

First, convert entropy into kilojoules:

\[
\Delta S^\circ=\frac{75.0}{1000}
=0.0750\ \text{kJ mol}^{-1}\text{K}^{-1}
\]

Substitute into the Gibbs equation:

\[
\begin{aligned}
\Delta G^\circ
&=-32.0-(298\times0.0750)\\
&=-32.0-22.35\\
&=-54.35\ \text{kJ mol}^{-1}
\end{aligned}
\]

To three significant figures:

\[
\Delta G^\circ=-54.4\ \text{kJ mol}^{-1}
\]

The negative result indicates that the forward reaction is thermodynamically spontaneous under standard-state conditions.

Both contributions favour the forward reaction: enthalpy is negative, and entropy is positive.

Question 2

A hypothetical decomposition reaction has:

\[
\Delta H^\circ=+42.0\ \text{kJ mol}^{-1}
\]

\[
\Delta S^\circ=+120\ \text{J mol}^{-1}\text{K}^{-1}
\]

Calculate \(\Delta G^\circ\) at:

a. \(20^\circ\text{C}\)

b. \(180^\circ\text{C}\)

c. Determine the temperature at which \(\Delta G^\circ=0\).

State the temperature conditions under which the forward reaction is thermodynamically spontaneous.

Assume \(\Delta H^\circ\) and \(\Delta S^\circ\) remain constant over the temperature range.

Solution 2

The forward reaction is thermodynamically spontaneous at \(180^\circ\text{C}\), but not at \(20^\circ\text{C}\). The crossover temperature is 350 K.

First, convert entropy:

\[
\Delta S^\circ=\frac{120}{1000}
=0.120\ \text{kJ mol}^{-1}\text{K}^{-1}
\]

Part a: At \(20^\circ\text{C}\)

Convert Celsius to kelvin:

\[
T=20+273.15=293.15\text{ K}
\]

Substitute:

\[
\begin{aligned}
\Delta G^\circ
&=42.0-(293.15\times0.120)\\
&=42.0-35.178\\
&=+6.82\ \text{kJ mol}^{-1}
\end{aligned}
\]

The positive value means the forward reaction is not thermodynamically spontaneous under standard-state conditions at this temperature.

Part b: At \(180^\circ\text{C}\)

Convert to kelvin:

\[
T=180+273.15=453.15\text{ K}
\]

Substitute:

\[
\begin{aligned}
\Delta G^\circ
&=42.0-(453.15\times0.120)\\
&=42.0-54.378\\
&=-12.4\ \text{kJ mol}^{-1}
\end{aligned}
\]

The negative value means the forward reaction is thermodynamically spontaneous under standard-state conditions.

Part c: Find the crossover temperature.

At the crossover:

\[
\Delta G^\circ=0
\]

Therefore:

\[
T=\frac{\Delta H^\circ}{\Delta S^\circ}
\]

\[
\begin{aligned}
T&=\frac{42.0}{0.120}\\
&=350\text{ K}
\end{aligned}
\]

The forward reaction becomes thermodynamically spontaneous above 350 K, under the stated approximation.

The important idea is that both enthalpy and entropy are positive. Increasing temperature makes the positive entropy contribution larger, and eventually it outweighs the positive enthalpy change.

Question 3

A student is calculating Gibbs free energy for a reaction at 298 K.

The supplied data are:

\[
\Delta H^\circ=-48.0\ \text{kJ mol}^{-1}
\]

\[
\Delta S^\circ=-110\ \text{J mol}^{-1}\text{K}^{-1}
\]

The student writes:

\[
\Delta G^\circ=-48.0-298(-110)
\]

They obtain a positive answer and conclude that the forward reaction is thermodynamically non-spontaneous.

Identify the error, calculate the correct Gibbs free energy change, and explain why the student’s conclusion is incorrect.

Solution 3

The correct answer is:

\[
\boxed{\Delta G^\circ=-15.2\ \text{kJ mol}^{-1}}
\]

The forward reaction is thermodynamically spontaneous under standard-state conditions at 298 K.

The student’s mistake is failing to convert the entropy value into the same energy units as enthalpy.

The enthalpy is in kilojoules, but the entropy is in joules.

Step 1: Convert entropy into kilojoules.

\[
\Delta S^\circ=\frac{-110}{1000}
=-0.110\ \text{kJ mol}^{-1}\text{K}^{-1}
\]

Step 2: Calculate the entropy contribution.

\[
\begin{aligned}
T\Delta S^\circ
&=298\times(-0.110)\\
&=-32.78\ \text{kJ mol}^{-1}
\end{aligned}
\]

Step 3: Calculate Gibbs free energy.

\[
\begin{aligned}
\Delta G^\circ
&=-48.0-(-32.78)\\
&=-48.0+32.78\\
&=-15.22\ \text{kJ mol}^{-1}
\end{aligned}
\]

Therefore:

\[
\boxed{\Delta G^\circ=-15.2\ \text{kJ mol}^{-1}}
\]

The negative entropy change makes the entropy contribution less favourable. However, at 298 K, it is not large enough to overcome the favourable negative enthalpy change.

The student has also made a sign error in interpreting the calculation. Subtracting a negative entropy term adds a positive contribution to Gibbs free energy, but that does not automatically make the final result positive.

Question 4

Consider the reaction:

\[
\ce{2A(g) + B(g) -> 2C(g)}
\]

The following thermodynamic data are supplied at 400 K:

\[
\Delta H^\circ=-86.0\ \text{kJ mol}^{-1}
\]

\[
\Delta S^\circ=-140\ \text{J mol}^{-1}\text{K}^{-1}
\]

The values refer to one mole of reaction as written.

a. Calculate \(\Delta G^\circ\) for the reaction.

b. A student doubles the balanced equation:

\[
\ce{4A(g) + 2B(g) -> 4C(g)}
\]

Determine the new values of \(\Delta H^\circ\), \(\Delta S^\circ\), and \(\Delta G^\circ\).

c. Explain whether doubling the balanced equation changes the thermodynamic spontaneity of the forward reaction.

Solution 4

The forward reaction is thermodynamically spontaneous in both representations.

For the original equation:

\[
\boxed{\Delta G^\circ=-30.0\ \text{kJ mol}^{-1}}
\]

For the doubled equation:

\[
\boxed{\Delta G^\circ=-60.0\ \text{kJ mol}^{-1}}
\]

Part a: Calculate Gibbs free energy for the original reaction.

Convert entropy:

\[
\Delta S^\circ=\frac{-140}{1000}
=-0.140\ \text{kJ mol}^{-1}\text{K}^{-1}
\]

Substitute:

\[
\begin{aligned}
\Delta G^\circ
&=-86.0-(400\times-0.140)\\
&=-86.0-(-56.0)\\
&=-30.0\ \text{kJ mol}^{-1}
\end{aligned}
\]

The negative result means the forward reaction is thermodynamically spontaneous under standard-state conditions.

Part b: Double the balanced equation.

When you multiply a balanced chemical equation by two, all its reaction thermodynamic quantities also double.

Therefore:

\[
\Delta H^\circ_{\text{new}}=2(-86.0)
=-172.0\ \text{kJ mol}^{-1}
\]

Similarly:

\[
\Delta S^\circ_{\text{new}}=2(-140)
=-280\ \text{J mol}^{-1}\text{K}^{-1}
\]

Convert the new entropy value:

\[
\Delta S^\circ_{\text{new}}=-0.280\ \text{kJ mol}^{-1}\text{K}^{-1}
\]

Now calculate Gibbs free energy:

\[
\begin{aligned}
\Delta G^\circ_{\text{new}}
&=-172.0-(400\times-0.280)\\
&=-172.0-(-112.0)\\
&=-60.0\ \text{kJ mol}^{-1}
\end{aligned}
\]

Part c: Does doubling the equation change spontaneity?

No. The forward reaction remains thermodynamically spontaneous because the sign of Gibbs free energy remains negative.

The numerical value doubles because the new equation represents twice the reaction amount.

This is an important distinction: changing the stoichiometric scale changes the numerical thermodynamic quantities, but it does not change the preferred reaction direction.

Question 5

A hypothetical reaction has the following standard thermodynamic data:

\[
\ce{X2(g) -> 2X(g)}
\]

\[
\Delta H^\circ=+36.0\ \text{kJ mol}^{-1}
\]

\[
\Delta S^\circ=+90.0\ \text{J mol}^{-1}\text{K}^{-1}
\]

Assume the thermodynamic values remain constant with temperature.

A student makes three claims:

1. At 500 K, the reverse reaction must be more spontaneous because reversing a reaction changes the signs of both enthalpy and entropy.
2. Doubling the balanced equation makes the forward reaction twice as spontaneous because its Gibbs free energy change becomes twice as negative.
3. Adding a catalyst could make the forward reaction thermodynamically spontaneous at 300 K.

Evaluate all three claims.

In your response:

a. Calculate \(\Delta G^\circ\) for the forward reaction at 300 K and 500 K.

b. Determine the crossover temperature.

c. Calculate \(\Delta G^\circ\) for the reverse reaction at 500 K.

d. Calculate \(\Delta G^\circ\) at 500 K if the forward equation is doubled.

e. Explain why a catalyst cannot change the thermodynamic conclusion at 300 K.

Solution 5

The forward reaction is thermodynamically favoured at 500 K, while the reverse reaction is favoured at 300 K.

Doubling the equation does not change the preferred direction, and a catalyst cannot change the sign of Gibbs free energy at a fixed temperature and pressure.

Let’s examine each claim.

Part a: Calculate Gibbs free energy at both temperatures.

First, convert entropy:

\[
\begin{aligned}
\Delta S^\circ
&=\frac{90.0}{1000}\\
&=0.0900\ \text{kJ mol}^{-1}\text{K}^{-1}
\end{aligned}
\]

At 300 K:

\[
\begin{aligned}
\Delta G^\circ
&=36.0-(300\times0.0900)\\
&=36.0-27.0\\
&=+9.0\ \text{kJ mol}^{-1}
\end{aligned}
\]

The forward reaction is thermodynamically non-spontaneous under standard-state conditions at 300 K.

At 500 K:

\[
\begin{aligned}
\Delta G^\circ
&=36.0-(500\times0.0900)\\
&=36.0-45.0\\
&=-9.0\ \text{kJ mol}^{-1}
\end{aligned}
\]

The forward reaction is thermodynamically spontaneous under standard-state conditions at 500 K.

Part b: Determine the crossover temperature.

Set Gibbs free energy equal to zero:

\[
T=\frac{\Delta H^\circ}{\Delta S^\circ}
\]

\[
\begin{aligned}
T&=\frac{36.0}{0.0900}\\
&=400\text{ K}
\end{aligned}
\]

Therefore:

• Below 400 K, the forward reaction is not thermodynamically spontaneous under standard-state conditions.
• Above 400 K, the forward reaction is thermodynamically spontaneous under standard-state conditions.
• At 400 K, \(\Delta G^\circ=0\).

This makes physical sense. The reaction has positive enthalpy and positive entropy. Higher temperatures increase the favourable contribution from entropy until it outweighs the enthalpy requirement.

Part c: Evaluate claim 1.

The claim is incorrect.

Reversing a reaction changes the signs of both enthalpy and entropy, but it does not automatically make the reverse direction more spontaneous at every temperature.

For the reverse reaction:

\[
\ce{2X(g) -> X2(g)}
\]

The thermodynamic values become:

\[
\Delta H^\circ=-36.0\ \text{kJ mol}^{-1}
\]

\[
\Delta S^\circ=-0.0900\ \text{kJ mol}^{-1}\text{K}^{-1}
\]

At 500 K:

\[
\begin{aligned}
\Delta G^\circ_{\text{reverse}}
&=-36.0-(500\times-0.0900)\\
&=-36.0+45.0\\
&=+9.0\ \text{kJ mol}^{-1}
\end{aligned}
\]

The reverse reaction has a positive Gibbs free energy change at 500 K.

Therefore, the forward reaction is favoured at 500 K, not the reverse reaction.

At 300 K, the reverse reaction has:

\[
\Delta G^\circ_{\text{reverse}}=-9.0\ \text{kJ mol}^{-1}
\]

So the reverse direction is favoured at the lower temperature.

The key is that reversing a reaction changes the signs of the thermodynamic quantities, but the temperature dependence still determines which direction is favoured.

Part d: Evaluate claim 2.

The claim is also incorrect.

Doubling the equation gives:

\[
\ce{2X2(g) -> 4X(g)}
\]

All reaction thermodynamic quantities double:

\[
\Delta H^\circ=+72.0\ \text{kJ mol}^{-1}
\]

\[
\Delta S^\circ=+180\ \text{J mol}^{-1}\text{K}^{-1}
\]

Convert entropy:

\[
\Delta S^\circ=+0.180\ \text{kJ mol}^{-1}\text{K}^{-1}
\]

At 500 K:

\[
\begin{aligned}
\Delta G^\circ
&=72.0-(500\times0.180)\\
&=72.0-90.0\\
&=-18.0\ \text{kJ mol}^{-1}
\end{aligned}
\]

The Gibbs free energy change has doubled from \(-9.0\) to \(-18.0\ \text{kJ mol}^{-1}\).

However, this does not mean the reaction has become twice as spontaneous.

The doubled equation represents twice the reaction amount. All extensive reaction quantities double, but the thermodynamic preference for the forward direction remains unchanged.

The crossover temperature also remains 400 K:

\[
T=\frac{72.0}{0.180}=400\text{ K}
\]

This is the crucial distinction between changing the scale of a reaction equation and changing the actual thermodynamic conditions.

Part e: Evaluate claim 3.

The claim is incorrect.

A catalyst changes reaction kinetics, not thermodynamic spontaneity.

At 300 K, the forward reaction has:

\[
\Delta G^\circ=+9.0\ \text{kJ mol}^{-1}
\]

Adding a catalyst does not change this value.

The catalyst lowers the activation energy, allowing the reaction to proceed faster in both directions. It does not change the enthalpy or entropy difference between the initial and final states.

Therefore, the forward reaction remains thermodynamically non-spontaneous under standard-state conditions at 300 K, even when a catalyst is present.

The three claims fail for different reasons:

• Reversing a reaction changes its direction and the signs of its thermodynamic quantities, but temperature still determines which direction is favoured.
• Doubling a reaction equation doubles its thermodynamic quantities without changing its preferred direction.
• A catalyst changes reaction rate, not Gibbs free energy.

This question brings together unit conversion, temperature dependence, reaction scaling, and the difference between thermodynamic feasibility and kinetics.

Once you can handle the Gibbs calculation reliably, the next step is making sure the values you put into it are correct. If you need to calculate reaction entropy from standard molar entropy data, read Entropy and Enthalpy Changes: HSC Chemistry Explained. If your enthalpy value comes from an experiment, Calorimetry and q = mcΔT: HSC Chemistry Guide explains how to obtain it.

The calculation only works when the thermodynamic data, reaction equation, temperature, and units all refer to the same chemical process.