Question

Solve Arrhenius equation and find expression for ${T_2}$ .

Answer 1

Temperature and rate constant relation having pre-exponential factor. Use Arrhenius equation step by step by taking initial conditions of a reaction a s temperature ${T_1}$ and it’s rate constant ${k_1}$ and for final condition temperature ${T_2}$ take rate constant ${k_2}$ . Solve step by step and find the value of ${T_2}$ in terms of other quantities.

Complete step by step answer:
We know that Arrhenius equation is $k = A\,\exp (\dfrac{{ - {E_a}}}{{RT}})$
Where, $k$ is the rate constant for given reaction
$A$ is the pre-exponential factor
${E_a}$ is the activation energy
$R$ is the universal gas constant
$T$ is absolute temperature
We know that as given by the collision theory, that when reactant molecules collide having a certain amount of energy they will convert into a product. That specific amount of energy is termed as threshold energy. Some molecules if we take an example of reaction $A + \,B \to \,C$ where the two reactants must collide to form a product. They must contain threshold energy, it means they should have a minimum amount of energy for converting into product. If reactant molecules don't have that minimum energy they are supplied extra energy from outside in terms of heat. This extra energy supplied to reactant molecules is called activation energy.
It was seen that on increasing the temperature, rate constant gets doubled it means double the number of reactants collide and changes into product. There is a Maxwell distribution curve which graphically explains the doubled area of a fraction of molecules.
Arrhenius gave an equation after his name called Arrhenius equation which symbolises the relation between temperature and rate constant. In the equation as we see above, the factor $\dfrac{{ - {E_a}}}{{RT}}$ , this factor represents the fraction of molecules.
Now let’s start solving equation for ${T_2}$
We have Arrhenius equation as-$k = A\,\exp (\dfrac{{ - {E_a}}}{{RT}})$
If we take logarithm on both side, we get this equation, $\ln \,k = \,\ln A - \,\dfrac{{{E_a}}}{{RT}}$
Now as we have to find value for ${T_2}$ it means there must be ${T_1}$ also,
Let’s take rate constants be ${k_1}\,$ and ${k_2}\,$ for temperature ${T_1}$ and ${T_2}$ , then the above equations for two rate constants become as $\ln \,{k_1} = \,\ln A - \,\dfrac{{{E_a}}}{{R{T_1}}}$ and $\ln \,{k_2} = \,\ln A - \,\dfrac{{{E_a}}}{{R{T_2}}}$
Subtracting these equations we get, $\ln \,{k_2} - \ln {k_1} = \,\,\dfrac{{{E_a}}}{{R{T_1}}} - \dfrac{{{E_a}}}{{R{T_2}}}$
In the next step if we take $\dfrac{{{E_a}}}{R}$ common it will changes like this,
$\ln \,{k_2} - \ln {k_1} = \,\,\dfrac{{{E_a}}}{R}(\dfrac{1}{{{T_1}}} - \dfrac{1}{{{T_2}}})$
$\ln \,\left( {\dfrac{{{k_2}}}{{{k_1}}}} \right) = \,\,\dfrac{{{E_a}}}{R}\left[ {\dfrac{1}{{{T_1}}} - \dfrac{1}{{{T_2}}}} \right]$
On multiplying in the brackets we get, $\ln \,\left( {\dfrac{{{k_2}}}{{{k_1}}}} \right) = \,\,\dfrac{{{E_a}}}{R} \times \dfrac{1}{{{T_1}}} - \dfrac{{{E_a}}}{R} \times \dfrac{1}{{{T_2}}}$

This is the value of ${T_2}$ . If we put the value of all terms in the right hand side, we can easily calculate the value of ${T_2}$ .

Note: Arrhenius equation gives relation between temperature of a reaction and rate constant, so take temperature and rate constant for initial and final stages of a reaction, final stage may be any temperature you want. While solving the equation for two conditions, the open bracket wisely multiplies the coefficient with inner terms. In most cases mistakes happen in the calculation part.