Question

Write Arrhenius equation. Derive an expression for temperature variations.

Answer 1

Hint: We have to know that the Arrhenius equation is a formula for the temperature dependence of reaction rates. The equation was given by Svante Arrhenius in 1889. An important application of Arrhenius equation is in determining rate of chemical reactions and for calculation of energy of activation.

Complete step by step answer:
We have to know that the Arrhenius equation is helpful in calculating the rate of reaction and plays an important portion in chemical kinetics. We can write Arrhenius equation as,
$K = A{e^{ - {E_a}/RT}}$
Where, K is the rate constant
A is the pre-exponential factor
${E_a}$ is the activation energy
R is the gas constant
T is the temperature (in Kelvin)
With an increase in temperature, the rate of the reaction also increases. This means that temperature and rate are proportional. Kinetic energy also increases with respect to temperature. Therefore, when we raise the temperature, the number of molecules with kinetic energy is greater than the increase in activation energy. This leads to increase in the rate of the overall reaction with reduction in the activation energy.
Let us take the Arrhenius equation at temperature ${T_1}$ and ${T_2}$ . The rates of the reactions are ${K_1}$ and ${K_2}$.
At temperature ${T_1}$, we can write the expression for ${K_1}$ as,
${K_1} = A{e^{ - {E_a}/R{T_1}}}\,\,\,\,\,\,\,\,\,\left( 1 \right)$
At temperature ${T_2}$, we can write the expression for ${K_2}$ as,
${K_2} = A{e^{ - {E_a}/R{T_2}}}\,\,\,\,\,\,\,\,\,\left( 2 \right)$
Let us take log on both sides,
$\log \,{K_1} = \log \,A - \dfrac{{{E_a}}}{{2.303\,R{T_1}}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( 3 \right)$
$\log \,{K_2} = \log \,A - \dfrac{{{E_a}}}{{2.303\,R{T_2}}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( 4 \right)$
Let us subtract eqn (4) and eqn (3). Upon subtracting, we get the equation as,
$\log \,{K_2} - \log \,{K_1} = \dfrac{{ - {E_a}}}{{2.303R{T_2}}} + \dfrac{{ - {E_a}}}{{2.303R{T_1}}}$
$\log \left( {\dfrac{{{K_2}}}{{{K_1}}}} \right) = \dfrac{{{E_a}}}{{2.303R}}\left[ {\dfrac{1}{{{T_2}}} - \dfrac{1}{{{T_1}}}} \right]$
$\log \left( {\dfrac{{{K_2}}}{{{K_1}}}} \right) = \dfrac{{{E_a}}}{{2.303R}}\left[ {\dfrac{{{T_2} - {T_1}}}{{{T_1}{T_2}}}} \right]$
This is the integrated form of Arrhenius equation.


Note:
We must know that the basis of the Arrhenius equation is collision theory. Based on this theory, a reaction happens due to collision between two molecules to give an intermediate. The formed intermediate is unstable and exists for a shorter period of time. The unstable intermediate gets broken into two product molecules and the energy used in forming the intermediate is called activation energy. With the help of Arrhenius equation, we can find the temperature, frequency, presence of catalyst, effect of energy barrier and orientation of collisions.