Question

The rate of a first order reaction, $A\to B$ is $5.4\times {{10}^{-6}}M{{s}^{-1}}$ when [A] is 0.3M. Calculate the rate constant of the reaction.

Answer 1

Hint: Think about the concept of first order kinetics. Write the rate law for the given reaction. Just substitute the given values and get the answer.


Complete step by step solution:
- The rate of the reaction depends on the concentration of the reactants at the given temperature.
- The rate law is an experimentally determined equation that gives the rate of a chemical reaction in terms of molar concentration of the reactants.
\[\text{Rate }\propto \text{ Concentration of reactants}\]
- The rate constant, k is the rate of reaction if all the concentration terms are set equal to unity.
- Order of the reaction is the sum of exponential terms to which the concentration terms in the rate law are raised.
- According to the question, the given reaction follows first order kinetics. Therefore, the concentration term will have exponential value as 1.
- We have been given rate of the reaction is $5.4\times {{10}^{-6}}M{{s}^{-1}}$ and [A] is 0.3M and we need to calculate the rate constant, k.
\[A\to B\]
- Therefore, the rate law for this reaction is given as,
Rate of the reaction, v = k [A]
\[5.4\times {{10}^{-6}}=k\times 0.3\]
\[k=\dfrac{5.4\times {{10}^{-6}}}{0.3}=1.8\times {{10}^{-5}}{{s}^{-1}}\]

- Therefore, the rate constant of the reaction, $k=1.8\times {{10}^{-5}}{{s}^{-1}}$



Note: Remember rate of the reaction is directly proportional to the molar concentration of the products. Order of the reaction is experimentally determined. It is the sum of powers of concentration terms in the rate law.