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:
Want to see the whole course?
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)]
[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.