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

The initial rate of the reaction is

Rate = k [A][B]²

[A] = 0.1 mol L⁻¹, [B] = 0.2 mol L⁻¹, k = 2.0 × 10⁻⁶

= 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

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

= 3.89 × 10⁻⁹ mol L⁻¹ s⁻¹