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I have a problem answering a question in my lab report.

The question is about the empirical van't Hoff's rule - what does it say and what are its limitations?

I managed to find online that the reaction speed doubles when you increase the temperature by 10°C, and that somewhat matches our experiment.

Now, the question is when does it not work. I found only a few answers but they seemed complex (they were either written in a confusing way or it's because I'm not a native English speaker).

I found this: "The rule is an approximation that works best when temperatures approximate those under which the reaction normally occurs." #
But what does it mean, "temperature under which the reaction normally occurs"?

Also, related to the last question, should van't Hoff's rule work for a reaction between sodium thiosulphate and sulphuric acid?

Alien Lobster
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    https://chemistry.stackexchange.com/questions/22153; https://chemistry.stackexchange.com/questions/66233 – andselisk Feb 26 '23 at 19:49

1 Answers1

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The van't Hoff's empirical rule says:

The reaction rate typically raises 2-4 times with the temperature increase by $\pu{10 ^\circ C}$.

The range of the rule corresponds to typical range of reaction activation energies. The higher the reaction activation energy is, the higher is the coefficient of the rule.

The rule you have quoted is from a medical dictionary. Biochemical reactions have in average rather lower activation energies and for this specific domain the value $2$ may generally about fit. But note that the value for a particular reaction is a real number.

It is closely related to the Arrhenius equation (the simplest equation for the dependence of a reaction kinetic constant on temperature): $$k = A \exp{( - \frac{E_\mathrm{a}}{RT})}.$$

We can see in the below equation the relation of the van't Hoff rule coefficient and the reaction activation energy: $$C = \frac{ \exp{(- \frac{{E_\mathrm{a}}}{{R(T+\Delta T)}}})}{\exp{(- \frac{{E_\mathrm{a}}}{{RT}}} )}=\exp{\left(\frac{E_\mathrm{a}}{RT} - \frac{E_\mathrm{a}}{R(T+\Delta T)}\right)}=\exp{\left(\frac{E_\mathrm{a}\Delta T}{RT(T+\Delta T)}\right)}$$

  • $C$ is the ratio of reaction kinetic constants.
  • $E_\mathrm{a}$ is the reaction activation energy.
  • $T$ is the lower absolute temperature of measurement.
  • $\Delta T$ is the measurement temperature difference.

If $\Delta T = \pu{10 K}$, $C$ is equal to the van't Hoff rule empirical coefficient.

$E_\mathrm{a}$ can be computed from $C$, $T$ and $\Delta T$ this way:

$$\ln{C} = \frac{E_\mathrm{a}\Delta T}{RT(T+\Delta T)} $$

$$E_\mathrm{a} = RT\ln{(C)} \left(\frac {T+\Delta T}{\Delta T}\right)$$

For $T= \pu{298.15 K}$ and $\Delta T = \pu{10 K}$: $E_\mathrm{a} \approx \pu{76.4 \ln{(C)} kJ mol-1}$

For $C = 2$: $E_\mathrm{a} \approx \pu{76.4 \ln{(2)} kJ mol-1} \approx \pu{52.9 kJ mol-1}$

For $C = 4$: $E_\mathrm{a} \approx \pu{76.4 \ln{(4)} kJ mol-1} \approx \pu{106 kJ mol-1}$

Therefore, the coefficient of the van't Hoff rule is about 2 for reaction activation energies around $\pu{50 kJ mol-1}$.

Poutnik
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  • So the rule is that the reaction time raises around 2 times when the temperature increases by 10°C, but, the higher the activation energy of the reaction, the larger the increase? That actually makes sense. How would one calculate the theoretical activation energy of a reaction between sodium thiosulphate and sulphuric acid? – Alien Lobster Feb 26 '23 at 20:11
  • See the answer update. – Poutnik Feb 27 '23 at 06:35
  • Ok, but if I know the reactants and the products, how would I determine the activation energy of the reaction? The question I need to answer is if van't Hoff's rule works for a reaction between $\ce{Na2S2O3}$ and $\ce{H2SO4}$. I guess I need to see if that reaction's activation energy is around 50 kJ/mol. Based on the experiment I would guess it's lower than 50 kJ/mol. – Alien Lobster Feb 27 '23 at 09:33
  • The activation energy is an experimental value. I doubt you find it written, so you have to determine it experimentally, obtaining the reaction speed ratio at 2 different temperatures. But you need not to calculate the activation energy at all. Enough is to compare reaction speed at temperatures different by 10 K. – Poutnik Feb 27 '23 at 09:55
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    Oh ok, that's what I did. I wasn't sure if that was enough. I will mark your answer as correct. Thank you for taking your time to help. – Alien Lobster Feb 27 '23 at 10:01