Question

For the reaction:

\[\ce{2A + B -> A2B}\] 

the rate = k[A][B]² with k = 2.0 × 10⁻⁶ mol⁻² L² s⁻¹. Calculate the initial rate of the reaction when [A] = 0.1 mol L⁻¹, [B] = 0.2 mol L⁻¹. Calculate the rate of reaction after [A] is reduced to 0.06 mol L⁻¹.

Answer 1

Given: [A] = 0.1 mol L⁻¹

[B] = 0.2 mol L⁻¹

k = 2.0 × 10⁻⁶

The initial rate of the reaction is

Rate = k[A][B]²

= 2.0 × 10⁻⁶ × 0.1 × (0.2)²

= 8 × 10⁻⁹ mol L⁻¹ s⁻¹

When [A] reduces to 0.06 mol L⁻¹, i.e. 0.04 mol L⁻¹ of A has reacted, then the reactant B

= `1/2 xx 0.04`

= 0.02 mol L⁻¹

Hence, the new [B] = 0.2 − 0.02 = 0.18 mol L⁻¹

Thus, the new concentrations of A and B are

[A] = 0.06 mol L⁻¹,

[B] = 0.18 mol L⁻¹,

Now rate = 2.0 × 10⁻⁶ × (0.06) × (0.18)²

= 3.89 × 10⁻⁹ mol L⁻¹ s⁻¹