Question

Rate law for the reaction $A + 2B \to C$ is found to be, $Rate=k\times[A]\times[B]$
The concentration of reactant ‘B’ is doubled, keeping the concentration of ‘A’ constant, the value of the rate constant will be?
A. The same
B. Doubled
C. Quadrupled
D. Halved

Answer 1

Hint: We will write the reaction given in the question. Then we will understand how the concentration of the reactants has an effect on the rate of the reaction. Then we will write the answer and choose the correct option.

Complete answer:
Step1. We are given a reaction to the question. It is $A + 2B \to C$. Here the one mole of the ‘A’ reacts to two moles of ‘B’ and produces one mole of the ‘C’.
Step2. Now the concentration of ‘B’ is doubled but the concentration of A is kept constant. Now it may seem that it has an effect on the rate of the reaction but it is not true. The rate of reaction is completely not dependent on the concentration of the reactants or the product. Even if we add something in them or we remove some moles. It does not affect the rate of the reaction. The rate of the reaction depends on other circumstances like the temperature. Often when the temperature is increased it is faster than the rate of the reaction.
Step3. Since we know that the rate of reaction does not depend on the concentration hence the change will not occur.
So, ‘A. The same’ is the correct answer.

Note: The rate of reaction depends on how fast the reaction is going on or how fast the concentration of the two reactants decrease. All the variables are not needed to measure the rate it can be measured in terms of reactant or product.