Explain temperature dependency of reaction rate constants and reaction rate from Arrhenius law ?
1 Answer

Activation Energy

The minimum amount of energy is requires to start the reaction is known as activation energy.

Arrhenius law

$$\mathrm {k =k_{o}e^{\frac{-E}{RT}}}$$


$\mathrm k=$ rate constant

$\mathrm k_{o} =$ Pre exponential factor

$\mathrm{E} =$ Activation energy, J/mol or cal/mol

$\mathrm R=$ gas constant

$\mathrm T=$ absolute temperature, $\mathrm{K}$

It has been verified emperically to give the temperature behaviour of most reaction rate constants within the experimental accuracy over fairly large temperature ranges:

The activation energy is the minimum energy that must be possessed by reacting molecules before the reaction will occur i.e energy possessed by reacting moleculer for getting them converted into product.

The activation energy is determined experimentally by carrying out the reaction at several different temperatures.

Taking the natural logarithm of Arrhenius equation, we get

$$ \mathrm {\ln k=\ln k_{o}-\frac{E}{R T}} ...(i) $$ It follows from equation $(i)$ that a plot of $\ln \mathrm{k}$ v/s 1 T should give a straight line with slope equal to $-E / R$.

So one must have data of reaction rate constant as a function of temperature for getting E known.

1) Temperature dependency of reaction rate constant

Alternatively, energy of activation can also be obtained by measuring the rate constants $k_{1}$ and $k_{2}$ at temperatures $T_{1}$ and $T_{2}$ respectively.

$$ \begin{aligned} \ln \mathrm k_{1} &=\mathrm {\ln k_{o}-\frac{E}{R T_{1}}} \\ \ln \mathrm k_{2} &=\mathrm {\ln k_{o}-\frac{\mathrm{E}}{\mathrm{RT}_{2}}} \\ \ln \mathrm {k_{2}-\ln k_{1} }&=\mathrm {\frac{-E}{R T_{2}}+\frac{E}{R T_{1}}} \\ \ln \mathrm {\left(k_{2} / k_{1}\right) }&=\mathrm{-\frac{E}{R}\left[\frac{1}{T_{2}}-\frac{1}{T_{1}}\right]} \\ \ln \mathrm {\left(k_{2}/k_{1}\right)} &=\mathrm {\frac{E}{R}\left[\frac{1}{T_{1}}-\frac{1}{T_{2}}\right]} \end{aligned} $$

2) Temperature dependency of reaction rate

Consider that the rate of reaction is given by

$$\begin {aligned} \mathrm r &=\mathrm {k C_{A}^{\alpha} C_{B}^{\beta}}\\ \mathrm r &=\mathrm{k_{o}e^{\frac{-E}{RT}} C_{A}^{\alpha} C_{B}^{\beta}} ...(ii) \end{aligned}$$

Consider equation (ii), The concentration function $\mathrm{C}_{A}^{\mathrm \alpha}, \mathrm{C}_{B}^{\mathrm{\beta}}$ would be constant for constant concentrations. Hence, it could be combined with $k_{0}$ to give a new constant $k_{0}$.

Then, equation (ii) becomes

$$ \begin{aligned} \mathrm r &=\mathrm k_{o} \mathrm{e}^{-\mathrm{\frac{E}{RT}}} \\ \ln \mathrm r &=-\frac{\mathrm{E}}{\mathrm{R}}\left(\frac{1}{\mathrm T}\right)+\ln \mathrm k_{\mathrm{o}} \end{aligned} $$

From above equation, it is clear that plot of $\ln \mathrm r$ v/s $1 / \mathrm T$ yields a straight line with slope equal to $\mathrm{-E/R}$ and hence we can calculnte $\mathrm{E}$.

If rate is known at two different temperatures $\mathrm T_{1}$ and $\mathrm T_{2}$ then we can calculate $E$. By using following equation, $$ \ln \frac{\mathrm{r}_{2}}{\mathrm{r}_{1}}=-\frac{\mathrm{E}}{\mathrm{R}}\left[\frac{1}{\mathrm{~T}_{2}}-\frac{1}{\mathrm{~T}_{1}}\right] $$

Activation energy is usually reported in cal/mol or J/mole. In all above equations we have to use temperature in Kelvin and R=1.987 (cal/mol.K ) or R=8.314 (J/mol.K)

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