Question

A reaction \( \mathrm{A \rightarrow B} \) follows second order kinetics. How will the rate of reaction be affected if the concentration of (i) A is increased by three times, and (ii) A is reduced to half ?

Hint:

Use rate = k[A]^2. Tripling the concentration gives 3 squared = 9 times, and halving gives one fourth.

Answer 1

Step 1: Write the rate law:
For a second order reaction of A, the rate depends on the square of the concentration of A. \[ \text{Rate} = k[A]^2 \]
Here k is the rate constant. It stays the same because the temperature is not changed.

Step 2: Triple the concentration (part i):
The new concentration is 3[A].
\[ \text{Rate}_1 = k(3[A])^2 = 9\,k[A]^2 \]
So the rate becomes 9 times the original rate.

Step 3: Halve the concentration (part ii):
The new concentration is \( \frac{[A]}{2} \).
\[ \text{Rate}_2 = k\left(\frac{[A]}{2}\right)^2 = \frac{1}{4}\,k[A]^2 \]
So the rate becomes one fourth of the original rate.

Final Answer:
(i) The rate increases 9 times. (ii) The rate falls to one fourth. \[ \boxed{\text{(i) 9 times;\ (ii) } \tfrac{1}{4}\text{ of the original}} \]