Show how ΔtotS = 0 leads to a limiting temperature for a reaction.

Setting ΔsysS − ΔH/T = 0 gives ΔsysS = ΔH/T, so T = ΔH/ΔsysS at the temperature where the sign of ΔtotS changes.

The decomposition of calcium carbonate has ΔH = +178 kJ mol⁻¹ and ΔsysS = +160.4 J K⁻¹ mol⁻¹. Calculate the temperature above which it becomes feasible.

T = 178 000/160.4 = 1110 K to three significant figures, above which ΔtotS becomes positive.

Formation of N₂O₄ from NO₂ has ΔH = −57.2 kJ mol⁻¹ and ΔsysS = −175.8 J K⁻¹ mol⁻¹. Find the temperature at which ΔtotS changes sign and say where the reaction is feasible.

T = 57 200/175.8 = 325 K, and the reaction is feasible only below this temperature.

A reaction has a negative ΔH and a positive ΔsysS. Over what temperature range is it feasible?

It is feasible at all temperatures, because both contributions to ΔtotS are positive.

A reaction has a positive ΔH and a negative ΔsysS. What can you say about its feasibility?

It is not feasible at any temperature, because both contributions to ΔtotS are negative.

Why does a positive total entropy change not guarantee that a reaction will be seen to happen?

Feasibility says nothing about rate; a large activation energy can make the reaction immeasurably slow.

OCR A-Level Chemistry B H433

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The limiting temperature, and why a feasible reaction may not happen

1 thing to know
Components 01, 02 and 03

Overview

Recall questions on setting the total entropy change to zero to find a limiting temperature, and why a feasible reaction may still not be observed.

What you need to know

Answer each one before you reveal it.

Your recallNot started

Numbers

Why does the decomposition of calcium carbonate (ΔH +178 kJ mol⁻¹, ΔsysS +160.4 J K⁻¹ mol⁻¹) become feasible only at high temperature, where is forming N₂O₄ from NO₂ (ΔH −57.2 kJ mol⁻¹, ΔsysS −175.8 J K⁻¹ mol⁻¹) feasible, and why does a positive ΔtotS not guarantee a reaction is seen?

Questions from this subtopic

Each one with the model answer underneath.

1
Show how ΔtotS = 0 leads to a limiting temperature for a reaction.
Model answer
Setting ΔsysS − ΔH/T = 0 gives ΔsysS = ΔH/T, so T = ΔH/ΔsysS at the temperature where the sign of ΔtotS changes.
2
The decomposition of calcium carbonate has ΔH = +178 kJ mol⁻¹ and ΔsysS = +160.4 J K⁻¹ mol⁻¹. Calculate the temperature above which it becomes feasible.
Model answer
T = 178 000/160.4 = 1110 K to three significant figures, above which ΔtotS becomes positive.
3
Formation of N₂O₄ from NO₂ has ΔH = −57.2 kJ mol⁻¹ and ΔsysS = −175.8 J K⁻¹ mol⁻¹. Find the temperature at which ΔtotS changes sign and say where the reaction is feasible.
Model answer
T = 57 200/175.8 = 325 K, and the reaction is feasible only below this temperature.
4
A reaction has a negative ΔH and a positive ΔsysS. Over what temperature range is it feasible?
Model answer
It is feasible at all temperatures, because both contributions to ΔtotS are positive.
5
A reaction has a positive ΔH and a negative ΔsysS. What can you say about its feasibility?
Model answer
It is not feasible at any temperature, because both contributions to ΔtotS are negative.
6
Why does a positive total entropy change not guarantee that a reaction will be seen to happen?
Model answer
Feasibility says nothing about rate; a large activation energy can make the reaction immeasurably slow.

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