Le Chatelier’s Principle was first taught in secondary schools, and delved deeper in A levels Chemistry. While many know what the principle is about, few really understand how it works for the effect of a change in temperature.
The effect of temperature on a reversible reaction can be seen through a shift in the position of equilibrium and the value of the equilibrium constant, K (Kc or Kp if expressed in terms of concentration or partial pressure respectively).
Consider a reversible reaction:
k_{f} | |||
A + B | ⇌ | C | ΔH>0 |
k_{b} |
where kf and kb are the forward and backward rate constants respectively.
Consider an INCREASE in temperature on the position of equilibrium.
Many of us know that the position of equilibrium shifts to the right.
A level Chemistry Teachers’ Reason:
When heat is applied to increase the temperature, the reaction will shift forward (endothermic) to absorb the heat, according to Le Chatelier’s principle.
Alternative reason (kinetics perspective):
When temperature increases, the forward and backward rate constants increase, according the Arrhenius equation:
k = A_{o }exp (-Ea/RT)
where
k = rate constant; Ao = frequency factor; EA = activation energy; R = gas constant; T = temperature, exp = exponential (or e)
We sometimes hear or read from some literature that says that for every 10^{o}C change in temperature, the rate of reaction doubles. Is that true?
Well, this is NOT ALWAYS TRUE!
If we were to perform a ln on the Arrhenius equation, we obtain
ln k = ln Ao – (Ea/R)(1/T)
Plotting ln k against 1/T yields the graph with ln Ao as the y-intercept and -Ea/R as the gradient . Notice that EA affects the gradient of the graph? Reactions of higher EA values are more sensitive to temperature changes.
Thus, since an endothermic reaction has a high EA value than an exothermic one, a change in temperature affects endothermic reactions more than exothermic reactions!
Going back to the reversible reaction at the beginning of this blog, we notice that kf is k_{endo} and kb is k_{exo }. Thus an increase in temperature increases kf more than kb.
This means the NET reaction is in the Forward direction, which is the same conclusion as the typical reason our A level Chemistry teachers taught us!
Consider an INCREASE in temperature on the value of equilibrium constant, K
A Level Chemistry Teachers’ Explanation:
k_{f} | |||
A + B | ⇌ | C | ΔH>0 |
k_{b} |
Since
Kc = | [C] ——– [A][B] |
increasing temperature shifts position of equilibrium to the right (previous section).
This increase [C] and decreases [A] and [B], thus Kc increases.
Does this explanation sound convincing to you?
Have you asked yourself why a change of concentrations of any reactants or products (eg. A, B or C in the reversible reaction in the above example) does not change the value of K?
Alternative reason (thermodynamics perspective):
In 1884, J.B van’t Hoff proposed a relationship between the equilibrium constant, K of a reversible reaction to the reaction temperature, T:
d (ln K) ——– dT |
= |
ΔH |
How does the famous van’t Hoff equation lead us to the same conclusion as what our A level teachers have taught us, but in a more convincing manner?
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