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Grade 10 Science · Problem type

ΔS° from a table, then ΔG°

Use this for “The products’ standard entropies add to 214 J/(mol·K) and the reactants’ to 188; ΔH° = −50 kJ/mol. What are ΔS° and ΔG° at 298 K?”

Diagram

18018519019520020521021522022526 J/(mol·K)reactants 188products 214
From 188 J/(mol·K) to 214 J/(mol·K): 26 J/(mol·K) to the right.
This is an example. Type your own values to start a new problem.
ΔS° (J/(mol·K))
=
S°ₚ (J/(mol·K))
−
S°ᵣ (J/(mol·K))
ΔG° (kJ/mol)
=
ΔH° (kJ/mol)
−
T (K)
×
ΔS° (J/(mol·K))
1,000

Formulas

ΔS° = S°ₚ − S°ᵣ
26 = 214 − 188
ΔG°
=
ΔH°
−
T
×
ΔS°/1,000
−57.7
=
−50
−
298
×
26
1,000

Assumptions

•
Each sum is over one side of the balanced equation: every S° is multiplied by its coefficient first.
•
A change is products minus reactants, so ΔS° = S°ₚ − S°ᵣ (p for products, r for reactants); ΔG°f values from a table subtract the same way.
•
ΔG° = ΔH° − TΔS°, with ΔS° divided by 1,000 to match ΔH°’s kJ; tables are for 298 K.
•
At other temperatures this treats ΔH° and ΔS° as unchanged, an estimate.

Step-by-step

WE KNOW
S°ₚ = 214 J/(mol·K), S°ᵣ = 188 J/(mol·K), ΔH° = −50 kJ/mol, T = 298 K
FIND
standard entropy change (ΔS°), standard free energy change (ΔG°)
Step 1 · Find standard entropy change (ΔS°)
Use ΔS° = S°ₚ − S°ᵣ
Products minus reactants: a positive change means the products are more spread out.
ΔS° = 214 − 188
ΔS° = 26 J/(mol·K)
Step 2 · Find standard free energy change (ΔG°)
Use ΔG° = ΔH° − T × ΔS°/1,000
Take TΔS°, in kJ, from ΔH°.
ΔG°
=
−50
−
298
×
26
1,000
ΔG° = −50 − 7.748
ΔG° = −57.7 kJ/mol (below 0: spontaneous at 298 K)
Check
Put the numbers back in. Both sides should match.
26 = 214 − 188 ✓
−57.7
≈
−50
−
298
×
26
1,000
✓
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