_{Up Learn – A Level Chemistry (aqa) – Thermodynamics}

_{Up Learn – A Level Chemistry (aqa) – Thermodynamics}

**Units of Gibbs Free Energy Change**

**Gibbs Free Energy change is usually reported in kJ mol-1**

### More videos on Thermodynamics:

Positive and Negative Entropy Changes

Predicting Entropy Changes of Reactions

Entropy change formula: Calculating Entropy Changes

Why are the units of entropy change ‘per mole’?

Gibbs Free Energy: Why do some Feasible Reactions not happen?

Units of Gibbs Free Energy Change

Calculating Gibbs Free Energy Change

Gibbs Free Energy: Feasibility

Finding Entropy and Enthalpy from Gibbs vs. Temperature Graph

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## Thermodynamics

2. Order and Disorder

3. What is Entropy?

4. How Does Temperature Affect Entropy?

5. How Does State Change Affect Entropy?

6. Comparing Entropy Between Substances?

6. How Does Dissolving a Substance Affect Entropy?

8. How Does the Number of Particles Affect Entropy?

9. Entropy Changes

10. Predicting the Entropy Change of a Reaction 1

11. Predicting the Entropy Change of a Reaction 2

2. A Simple System 1

3. A Simple System 2

4. Relating Configurations to Entropy

5. The Exact Mathematical Definition of Entropy

6. Relating Our Simple System to Atomic Systems

7. Why Does Temperature Affect Entropy?

8. Why Does Number of Particles Affect Entropy?

9. Why Does State Affect Entropy?

10. So Is Entropy Really a Measure of Disorder?

2. Measuring Entropy for Larger Systems

3. Entropy at Absolute Zero

4. Explaining Entropy at 0 K Mathematically

5. Entropy at Non-Zero Temperatures

6. Graphing Entropy

7. Standard Molar Entropies

8. Investigating the Trends in the Table of Absolute Entropies

9. Calculating the Entropy Change of a Reaction

10. Why Did We Bother Predicting Entropy Changes in the First Place?

11. Why Are the Units of Entropy Change ‘Per Mole’?

2. The Entropy Change of the Surroundings

3. Calculating the Entropy Change of the Surroundings

4. The Entropy Change of the Universe

5. What Reactions Can’t Happen?

6. Feasibility

7. Why Do Some Feasible Reactions Not Happen?

8. The 2nd Law of Thermodynamics

9. Gibbs Free Energy Change

10. The Units of Gibbs Free Energy Change

11. Calculating Gibbs Free Energy Change

12 .Assessing Feasibility

13. Assessing Feasibility – Making Ice

14. Assessing Feasibility – Thermal Decomposition of Calcium Carbonate

15. Exam Technique: Explaining Feasibility

16. Graphing Gibbs Free Energy Change

17. Using Graphs to Find Enthalpy and Entropy Changes

18. Assessing Feasiblility from Graphs

19. Finding the Temperature Where Reactions Become Feasible

20. The Limitations of Our Temperature-Finding Equation

21. Doesn’t Entropy Change…. Change With Temperature?

22. Calculating Gibbs Free Energy Change for Reverse Reactions

23. What About Reversible Reactions?

24. How Are Reversible Reactions Compatible With the Second Law of Thermodynamics?

Last time we saw that, in chemistry, we use the Gibbs free energy change to decide whether a reaction is feasible or not

And that reaction is feasible when…

And that a reaction is feasible when the Gibbs free energy change is negative

Also, we saw that we use this equation to calculate Gibbs

But, so far, we haven’t actually calculated anything.

Now, we’ll get to calculating shortly, but first we need to talk about units

So what units do we tend to use for each of these components [dS, dH, T]?

In calculations, we tend to use Kelvin for temperature, kilojoules per mole for enthalpy change, and joules per mole per kelvin for entropy change

So, that means the units of this part of the equation [dH] are kJ mol^{-1}

And the units of this part of the equation [TdS] are…

The units of this part of the equation are joules per mole

Now these units pose a problem

For instance, say we’re given this information.

[enthalpy change = -203kJ mol^{-1} entropy change = 123 J K^{-1 }mol^{-1} temp = 298K]

It’s tempting to just plug all of these values straight into the equation like this…

[dG = (-203) – (298)(123)]

…Expand brackets…

[dG = -203 – 36654]

…And evaluate to get our answer

[dG = -36857]

But actually we can’t do that.

The units of these two terms are different

[arrow to (-203) saying ‘kJ mol^{-1} and arrow to -36654 saying J mol^{-1}]

…so we can’t just add them together to get our answer.

That’d be like saying 3 metres plus 24 kilometers equals 27… [3m + 24km = 27 ???]

…It just doesn’t work

So, before we do any addition, we need to make sure these terms have the same units

And, since it’s convention to give Gibbs free energy change in kilojoules per mole, that means we need to change the units of this term.

To do that, instead of plugging in this value for entropy [dG =dH -T(123)] , we plug in this value for entropy [dG =dH -T(1231000)]

Or if we’re given this data [dS = -223 J K^{-1} mol^{-1} dG =dH -T(dS)], we plug this into our equation [dG =dH -T(-2231000)],

Or If we’re given this data, we plug…

If we’re given this data, we should plug this into our equation

In fact, whenever we’re given entropy data with these units [J K^{-1} mol^{-1}], we need to divide it by 1000 before we plug it into this equation.

This changes the units from joules per kelvin per mole into kilojoules per kelvin per mole

And, with that change, the units of these two terms are both the same, so we can now add them together and find delta G.

And we’ll look at that, next

But first, to sum up…

If we’re given this data for a reaction, we can’t just plug it in and evaluate to find Gibbs free energy change, because the units don’t match up.

So, before we plug in this value for entropy we first need to convert it from these units to these units

And to do that, we divide by a thousand.