Question

The decomposition of a hydrocarbon has value of rate constant as 2.5x${{10}^{4}}{{s}^{-1}}$ at ${{27}^{\circ }}C$. At what temperature would the rate constant be 7.5 x ${{10}^{4}}{{s}^{-1}}$. If the energy of activation is 19.147x ${{10}^{3}}Jmo{{l}^{-1}}$?

Answer 1

Hint: We need to find the order of the reaction first to be able to solve the question which can be obtained by knowing the unit of rate constant also. The rate constant increases with the increase in the temperature generally. It can be studied by the Arrhenius equation.

Complete step by step solution:
-For us to be able to solve the question, we first need to know what is the order of a reaction. To know that, we should know the rate law. It links the rate of a reaction to the concentration of the reactants and some constant parameters.
- The rate law can be shown as
\[rate=k{{\left[ A \right]}^{x}}{{\left[ B \right]}^{y}}\]
where A and B are reactants of the equation and k is a constant parameter. The sum of x and y gives the overall order of the reaction.
-If the rate of the equation does not depend on the reactant concentration; i.e; it is constant irrespective of the reactant concentration, then the equation is zero order.
-If the rate depends on one reactant, it is a first order reaction and if it depends on two reactants or square of 1 reactant, then it is a second order reaction.
-The unit for the first order reaction is ${{s}^{-1}}$ which is the same as that given in the question. So the order of the reaction is 1 and the equation for first order reaction is $\ln \left[ A \right]=-kt+\ln \left[ {{A}_{0}} \right]$
-The rate constant and the temperature are directly proportional to each other. Generally, the rate of reaction doubles itself for every ${{10}^{\circ }}$ rise in the temperature. For the first order reaction, their relation is shown by Arrhenius equation which is
\[\log \dfrac{{{k}_{2}}}{{{k}_{1}}}=\dfrac{{{E}_{a}}}{2.303R}\left( \dfrac{{{T}_{2}}-{{T}_{1}}}{{{T}_{1}}{{T}_{2}}} \right)\]
Where T denotes temperature, k denotes the rate constant and E is the activation energy.
Putting the values in the equation we get
\[\begin{align}
& \log \dfrac{7.5x{{10}^{4}}}{2.5x{{10}^{4}}}=\dfrac{19.147x1000}{2.303x8.314}\left( \dfrac{{{T}_{2}}-300}{{{T}_{2}}x300} \right) \\
& \Rightarrow \text{log}\left( 3 \right)=\left( \dfrac{{{T}_{2}}-300}{{{T}_{2}}x300} \right)x999.99 \\
 & \Rightarrow 0.477=\left( \dfrac{{{T}_{2}}-300}{{{T}_{2}}x300} \right)x1000 \\
 & \Rightarrow \dfrac{0.477x3}{10}=\dfrac{{{T}_{2}}-300}{{{T}_{2}}} \\
 & \Rightarrow {{T}_{2}}=350K \\
\end{align}\]

Note: Always use the SI unit before putting the values of the quantities in the equation. The SI unit for temperature is K and so we converted the temperature in Kelvin before using it by the formula K=C+273.