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:

For gas-phase reactions, remember: Decrease in volume \(\rightarrow\) Increase in Pressure \(\rightarrow\) Increase in Concentration \(\rightarrow\) Increase in Rate.

Answer 1

The rate of the reaction is determined by the concentrations of the reactants and the rate law expression. The rate law can be written as: \[ \text{Rate} = k[A]^m[B]^n \] where \(k\) is the rate constant, \(m\) and \(n\) are the orders of the reaction with respect to reactants A and B, respectively.

When the volume of the reaction vessel is decreased to one third of its original volume, the concentrations of the reactants will increase because concentration is inversely proportional to volume. Specifically, the concentrations of \(A\) and \(B\) will be multiplied by 3 (since concentration = number of moles / volume). Therefore, the new concentrations will be: \[ [A]' = 3[A], \quad [B]' = 3[B] \]

If the volume is reduced to one third, the rate will increase by a factor of \( (3)^m \times (3)^n = 3^{m+n} \), since the rate law is dependent on the concentrations raised to their respective powers.

Will there be any change in the order of the reaction?
No, the order of the reaction does not change with volume. The order of the reaction is determined by the molecularity and the mechanism of the reaction, which remain unaffected by the change in volume. The reaction will still have the same order with respect to \(A\) and \(B\), even though their concentrations increase with reduced volume.