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

Step 1: Write the rate law.
The reaction \(2A + B \rightarrow 2C\) happens in one step, so it is elementary. For an elementary reaction the powers in the rate law equal the number of each molecule in the equation. So rate \(= k[A]^2[B]\).

Step 2: Note the original rate.
Let the starting concentrations be \([A]\) and \([B]\). The original rate is \(r_1 = k[A]^2[B]\).

Step 3: Reduce the volume to one third.
Amount of substance stays the same, but volume becomes one third. Concentration is amount divided by volume, so each concentration becomes 3 times larger: new \([A] = 3[A]\) and new \([B] = 3[B]\).

Step 4: Find the new rate.
\(r_2 = k(3[A])^2(3[B]) = k \cdot 9[A]^2 \cdot 3[B] = 27\,k[A]^2[B] = 27\,r_1\).

Step 5: Compare.
The new rate is 27 times the original rate.

Step 6: Order of reaction.
The order is the sum of the powers, \(2 + 1 = 3\). Changing the volume only changes concentrations, not the rate law, so the order stays the same.

The rate becomes 27 times faster and the order remains third order. \[ \boxed{r_2 = 27\,r_1,\ \text{order} = 3\ \text{(unchanged)}} \]