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Q3.6 (ii):
A reaction is second order with respect to a reactant. How is the rate of reaction affected if the concentration of the reactant is (ii) reduced to half ?

Solution :

Step 1: Defining the Rate Law

For a reaction that is second order with respect to a reactant $A$, the rate of the reaction ($r_1$) is expressed by the rate law equation:

$r_1 = k[A]^2$

[Per the definition of Rate Law, where $k$ is the rate constant and $[A]$ is the molar concentration of the reactant].

Step 2: Defining the New Concentration

The problem states that the concentration of the reactant is reduced to half of its initial value. Let the initial concentration be $[A]_1$ and the new concentration be $[A]_2$.

$[A]_2 = \frac{1}{2}[A]_1$

Step 3: Calculating the New Rate of Reaction

The new rate of reaction ($r_2$) can be calculated by substituting the new concentration into the rate law equation:

$r_2 = k[A]_2^2$

$r_2 = k \left( \frac{1}{2}[A]_1 \right)^2$

[By the Principle of Substitution in algebraic expressions].

Step 4: Algebraic Simplification

Expanding the squared term:

$r_2 = k \left( \frac{1}{4} \right) [A]_1^2$

$r_2 = \frac{1}{4} (k[A]_1^2)$

Since $r_1 = k[A]_1^2$, we can substitute $r_1$ into the equation:

$r_2 = \frac{1}{4} r_1$

Visual Representation of Kinetic Dependency

Concentration [A] Rate (r) Initial Half

Analytical Conclusion

By comparing the initial rate ($r_1$) and the new rate ($r_2$), it is evident that the rate of the reaction decreases by a factor of 4.

Final Solution: The rate of reaction becomes one-fourth ($\frac{1}{4}$) of its original value when the concentration of the reactant is reduced to half.


More Questions from Class 12 Chemistry Chemical Kinetics Exercises


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