# Change of first order rate constant with temperature [closed]

In the decomposition of $\ce{N2O5}$, the time taken to reduce the concentration of $\ce{N2O5}$ half of the original amount is 12 min at $\pu{50^\circ C}$, while 5 h at $\pu{25^\circ C}$ and 10 days at $\pu{0^\circ C}$.

How can we understand that the given assertion is correct because the temperature is different every time?

• Because the temperature affects the decomposition kinetics (as is generally expected for any kinetic process). What is your question? – Jon Custer Jul 25 '18 at 13:01

## 1 Answer

First of all, decomposition of $\ce{N2O5}$ is a first order reaction, not a zero order reaction. So, your question headline needs to be corrected.

You can justify the given assertion by simple mathematics. For a first order reaction, we know that, $$t_{1/2} = \frac{\ln \space 2}{k}$$ which is independent of initial concentration. It only depends on the rate constant which will change if you change the temperature.

From Arrhenius equation also, we know the relation, $$\ln\left(\frac{k_2}{k_1}\right) = \frac{E_a}{R}\left(\frac{1}{T_1} -\frac{1}{T_2}\right)$$

So here we have three rate constants for three different temperature. Thus we will have at least two sets of above equations. Dividing one by another we will have, $$\frac{ln\left(\frac{k_2}{k_1}\right)}{ln\left(\frac{k_3}{k_2}\right)} = \frac{\left(\frac{1}{T_1} -\frac{1}{T_2}\right)}{\left(\frac{1}{T_2} -\frac{1}{T_3}\right)}$$

Now, from the relation between half life and rate constants we can have

$$\frac{\ln\left(\frac{\left(t_{1/2}\right)_1}{\left(t_{1/2}\right)_2}\right)}{\ln\left(\frac{\left(t_{1/2}\right)_2}{\left(t_{1/2}\right)_3}\right)} = \frac{\left(\frac{1}{T_1} -\frac{1}{T_2}\right)}{\left(\frac{1}{T_2} -\frac{1}{T_3}\right)}$$

Now from the given assertion by putting the values you have to just verify whether both the sides are giving the same value or not. In this case both sides, gives the value ≈ 1.2. So, I guess this assertion is thus correct.