Question

The rate constant of the reaction $A \to B$ is $0.6 \times {10^{ - 3}}mol{L^{ - 1}}{s^{ - 1}}$. If the concentration of A is 5M, then concentration of B after 20 minutes is:
(A) 0.36M
(B) 0.72M
(C) 1.08M
(D) 3.60M

Answer 1

Hint: First find the order of the reaction by the unit of the rate constant. Here, in this type of reaction, the rate of the reaction is constant over time as it does not depend upon the concentration of reactants or products.

Complete step by step solution:
First of all, we need to find the order of the reaction in order to calculate the concentration of B. We can find the order of the reaction from the unit of the rate constant.
- We are given that the rate constant has the unit $mol{L^{ - 1}}{s^{ - 1}}$. All the zero order reactions have the rate constant of this unit only. So, we found that this is the zero order reaction.
- For all zero order reaction, we can say that the concentration of the product after time t will be
\[x = kt\]
Where x is the concentration of the product, k is the rate constant of the reaction and t is time in seconds.
We are given that the time is 20 minutes. We know that 1 minute = 60 seconds. So, we can write that 20 minutes = $20 \times 60$=1200 seconds
- We can put the available values in the above equation to get
\[x = 1200 \times 0.6 \times {10^{ - 3}} = 0.72M\]
Thus, we obtained that the concentration of B after 20 minus will be 0.72M.

So, the correct answer is (B).

Note: Note that if the unit of rate constant is $tim{e^{ - 1}}$, then the reaction is of first order. If the unit of rate constant is $mo{l^{ - 1}}L{s^{ - 1}}$ , then the order of the reaction is accepted as second order.