Question

Following reaction takes place in one step:

\(2A + B \rightarrow 2C\)

How will the rate of above reaction change if the volume of the reaction vessel is decreased to one third of its original volume? Will there be any change in the order of reaction with the reduced volume?

Hint:

Because the reaction takes place in one step (it is an elementary reaction), its rate law can be written directly from the equation.

Answer 1

Concept:
Because the reaction takes place in one step (it is an elementary reaction), its rate law can be written directly from the equation. The powers of the concentrations are equal to the number of each reactant molecule in the balanced equation. Also, concentration means amount per unit volume, so if the volume changes, every concentration changes.

Step 1 (write the rate law):
For the one-step reaction \(2A + B \rightarrow 2C\), the rate law is:
rate = \(k[A]^2[B]^1\).
The order of reaction = 2 + 1 = 3.

Step 2 (effect of reducing the volume):
If the volume is reduced to one third (1/3) of the original, the same amount of each substance is now packed into one third the space, so each concentration becomes 3 times larger.
New rate = \(k(3[A])^2(3[B]) = k \times 9[A]^2 \times 3[B] = 27 \times k[A]^2[B]\).

Step 3 (compare):
New rate = 27 times the original rate.

Step 4 (order):
The order depends only on the powers in the rate law, which do not change when we change the volume. So the order remains the same.

Answer: The rate increases to 27 times its original value (because each concentration becomes 3 times larger and the overall order is 3, giving \(3^3 = 27\)). There is no change in the order of the reaction; it stays third order.