Question

The conversion \[A \to B\] follows second order kinetics. Doubling the concentration of A will increase the rate of formation of B by the factor of:
A.2
B.4
C.0.5
D.0.25

Answer 1

Hint: The rate law in chemical kinetics can be defined as a mathematical relationship obtained by experimenting and comparing the rate of the reaction to the concentration of the reactant. The order of the reaction is basically the sum of the powers to which the concentration term of the reactants is raised to.
Formula Used:
  Rate = \[k{[X]^m}{[Y]^n}\]

Complete step by step answer:
Suppose two reactants X and Y, with chemical concentrations ‘x’ and ‘y’ respectively, are reacted to form Z with chemical concentration ‘z’. We can represent this equation as: \[xX + yY \to zZ\]
Hence, rate law for this reaction can be expressed as:
Rate = \[k{[X]^m}{[Y]^n}\], where k is the rate constant.
Now, depending on the order of the reaction, the value for the sum of the exponents, i.e. (m + n) will vary.
For a first order reaction, m + n = 1
For a second order reaction, m + n = 2, and so on.
Now, coming back to the problem at hand,
The given reaction is: \[A \to B\] and it is of the second order.
Hence, the rate law for the given reaction can be written as:
Rate = \[k{[A]^2}\]
Now, if the concentration of the reactant A is doubled, then we must substitute the value of ‘A’ as ‘2A’
Hence, Rate = \[k{[2A]^2}\]
             Rate = \[4k{[A]^2}\]
Which is 4 times the original rate of the reaction.
Therefore, we can conclude that the rate of the reaction increases by 4 times.

Hence, Option B is the correct option.

Note:
The unit of k that is the rate constant is for different ordered reactions is different. This is because the powers to which the concentrations of the reactants will be raised to in the rate law, will be dependent on the order of the reaction.