Question

The temperature coefficient of a reaction is 2.5. If the rate constant at ${{T}_{1}}\text{ K}$ is $2.5\times {{10}^{-3}}{{s}^{-1}}$, the rate constant at ${{T}_{2}}\text{ K}$is: $\left( {{T}_{2}}>{{T}_{1}} \right)$
[A] $6.25\times {{10}^{-3}}$
[B] $1.0\times {{10}^{-2}}$
[C] $6.25\times {{10}^{-2}}$
[D] $1.0\times {{10}^{-3}}$

Answer 1

Hint: To solve this question, we can use the relation between the rate constant and the temperature coefficient. The relation states that, temperature coefficient is the ratio of the rate constant of a reaction at two different temperatures.

Complete step by step answer:
In the question, the temperature coefficient is given to us and we have to find out the rate constant at a different, higher temperature. Before getting into how to solve this, let us understand what is temperature coefficient and rate constant.
In chemical kinetics, we use the term rate constant to indicate a relationship between the rate of chemical reaction and the molar concentration of the reactant.
We can find out the rate constant by the concentration of reactants and the order of the reaction.
However, here we can use the temperature coefficient given to us to find out the rate constant at the given temperature.
Temperature coefficient of a reaction is the ratio of rate constants at two different temperatures when the temperature increases 10 degree Celsius.
Therefore, using this we can write that the temperature coefficient given to us will be equal to the ratio of the rate constant at the higher temperature to the rate constant at the lower temperature.

It is given to us that ${{T}_{2}}$ is higher than ${{T}_{1}}$. Therefore, we can write that-
     \[\dfrac{{{K}_{{{T}_{2}}}}}{{{K}_{{{T}_{1}}}}}=2.5\]
Where the rate constant at temperature ${{T}_{2}}$ kelvin is denoted by ${{K}_{{{T}_{2}}}}$ and the rate constant at temperature ${{T}_{1}}$ kelvin is denoted by ${{K}_{{{T}_{1}}}}$.
The value of rate constant at ${{T}_{1}}$ kelvin is given to us which is $2.5\times {{10}^{-3}}{{s}^{-1}}$.

Therefore, we can write that-
     \[\begin{align}
  & \dfrac{{{K}_{{{T}_{2}}}}}{2.5\times {{10}^{-3}}{{s}^{-1}}}=2.5 \\
 & or,{{K}_{{{T}_{2}}}}=2.5\times 2.5\times {{10}^{-3}}{{s}^{-1}}=6.25\times {{10}^{-3}}{{s}^{-1}} \\
\end{align}\]
Or, the rate constant at ${{T}_{2}}\text{ K}$ is \[6.25\times {{10}^{-3}}{{s}^{-1}}\].
So, the correct answer is “Option A”.

Note: The rate of a reaction increases upon increasing the temperature according to the equation given by Arrhenius. As we can see the rate constant increased at a higher temperature. However, the rate constant rapidly increases with the increase in temperature but at a very high temperature, the rate of increase falls off eventually.