Question

The reaction \[\ce{2A + B2 -> 2AB}\] is an elementary reaction. For a certain quantity of reactants, if the volume of the reaction vessel is reduced by a factor of 3, the rate of the reaction increases by a factor of ______ (Round off to the nearest integer).

Answer 1

The reaction \[\ce{2A + B2 -> 2AB}\] is an elementary reaction. For a certain quantity of reactants, if the volume of the reaction vessel is reduced by a factor of 3, the rate of the reaction increases by a factor of 27.

Explanation:

The given reaction is 

\[\ce{2A + B2 -> 2AB}\]

Since the reaction is elementary, the rate law can be directly written based on the stoichiometry of the reaction:

Rate = k(A)²(B)

Where

k is the rate constant.

(A) is the concentration of reactant A,

(B) is the concentration of reactant B.

Let, n_A be the number of moles of A,

n_B be the number of moles of B,

The initial concentrations are

\[\ce{(A) = \frac{n_A}{V}}\] and

\[\ce{(B) = \frac{n_B}{V}}\]

When the volume is reduced by a factor of 3, the new volume V' is

\[\ce{V' = \frac{V}{3}}\]

Thus, the new concentrations become:

\[\ce{(A') = \frac{n_A}{V'} = \frac{n_A}{\frac{V}{3}} = 3 * \frac{n_A}{V} = 3(A)}\]

\[\ce{(B') = \frac{n_B}{V'} = \frac{n_B}{\frac{V}{3}} = 3 * \frac{n_B}{V} = 3(B)}\]

Now we substitute the new concentrations into the rate law:

\[\ce{Rate' = k(A')^2(B') = k(3(A))^2(3(B))}\]

Calculating this gives

\[\ce{Rate' = k * 9(A)^3(B) = 27 k(A)^2(B)}\]

Since the original rate is 

Rate = k(A)²(B)

We can see that

Rate' = 27 × Rate

Thus, the rate of the reaction increases by a factor of 27 when the volume of the reaction vessel is reduced by a factor of 3.