A reaction is second order with respect to a reactant. How is the rate of reaction affected if the concentration of the reactant is
(i) doubled
(ii) reduced to half?
A reaction is second order with respect to a reactant. How is the rate of reaction affected if the concentration of the reactant is
(i) doubled
(ii) reduced to half?
Solution and Explanation
Letthe concentration of the reactant be [A] = a
Rate of reaction, R = k[A]2
= ka2
(i) If the concentration of the reactant is doubled, i.e. [A] = 2a, then the rate of the reaction would be
R = k(2a)2
= 4ka2
= 4 R
Therefore, the rate of the reaction would increase by 4 times
(ii) If the concentration of the reactant is reduced to half, i.e.
\([A] = \frac{1}{2} a\)
then the rate of the reaction would be
\(R = k(\frac{1}{2}a)^2\)
=\(\frac{1}{4}ka^2\)
\(= \frac{1}{4} R\)
Therefore, the rate of the reaction would be reduced to \( \frac{1}{4}^{ th}\)
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Concepts Used:
Rate of a Chemical Reaction
The rate of a chemical reaction is defined as the change in concentration of any one of the reactants or products per unit time.
Consider the reaction A → B,
Rate of the reaction is given by,
Rate = −d[A]/ dt=+d[B]/ dt
Where, [A] → concentration of reactant A
[B] → concentration of product B
(-) A negative sign indicates a decrease in the concentration of A with time.
(+) A positive sign indicates an increase in the concentration of B with time.
Factors Determining the Rate of a Reaction:
There are certain factors that determine the rate of a reaction:
- Temperature
- Catalyst
- Reactant Concentration
- Chemical nature of Reactant
- Reactant Subdivision rate