The rate equation for the reaction between compounds A and B is rate = k[A]^x[B]^y - AQA - A-Level Chemistry - Question 2 - 2017 - Paper 2

Calculate the gradient of the tangent. Based on the graph, the concentration of A decreases from 0.50 mol dm⁻³ to approximately 0.45 mol dm⁻³ over a small time interval of about 0.5 seconds. Thus, the initial rate can be determined using:

$ext{Rate} = -\frac{\Delta[A]}{\Delta t} = -\frac{0.50 - 0.45}{0.5} = -\frac{0.05}{0.5} = 0.10 \, \text{mol dm}^{-3} \, ext{s}^{-1}$

Let the initial concentration of A in the first experiment be [A]₁, and in the second experiment [A]₂. The relationship between the two rates is given by:

$\text{Rate}_{\text{new}} = 1.7 \times \text{Rate}_{\text{original}}$

Using the original initial rate as 0.10 mol dm⁻³ s⁻¹:

$\text{Rate}_{\text{new}} = 1.7 \times 0.10 \, \text{mol dm}^{-3} \, \text{s}^{-1} = 0.17 \, \text{mol dm}^{-3} \, \text{s}^{-1}$

From the original rate equation, we know:

$\text{Rate} = k[A]^x[B]^y$

Assuming the concentration of B is constant, the new initial concentration can be calculated as follows: Let [A]₁ = 0.50 mol dm⁻³ (initial),

Using the relationship: [ [A]₂ = [A]₁ \times \textfactor\n] [ \text{Rate}_{\text{new}} = k [A]₂^x[B]^y\n] [ 0.17 = k[A]^x[B]^y\n] Solving gives the new concentration as:

$[A]₂ = \frac{0.17}{0.10} [A]₁ = 0.50 \times 1.7 = 0.85 \, \text{mol dm}^{-3}$