The rate equation for the reaction between compounds A and B is rate = k[A]^m[B]^n Figure 2 shows how, in an experiment, the concentration of A changes with time, t, in this reaction - AQA - A-Level Chemistry - Question 2 - 2017 - Paper 2

To find the initial rate, calculate the gradient of the tangent line. From the tangent, the reaction appears to decrease from 0.50 mol dm⁻³ to 0.40 mol dm⁻³ over a time interval of 0.5 seconds. Thus, the initial rate can be calculated as:

$ext{Rate} = \frac{[A]_0 - [A]}{\text{Time}} = \frac{0.50 - 0.40}{0.5} = 0.20 \text{ mol dm}^{-3} ext{s}^{-1}$

Given that the new initial rate is 1.7 times the original rate:

$ext{New Rate} = 1.7 \times 0.20 = 0.34 \text{ mol dm}^{-3} ext{s}^{-1}$

Using the rate equation again and assuming the concentration of B remains the same:

Let the new initial concentration of A be [A]. The rate varies with [A] as:

$\text{Rate} \propto [A]^m$

Assuming m = 1 (for simplicity), if the rate increases by 1.7 times, we have:

$\text{New [A]} = \frac{0.34}{0.20} \times [A]_{original}$
Substituting the original concentration of A as 0.50 mol dm⁻³:

$New [A] = 1.7 \times 0.50 = 0.85 \text{ mol dm}^{-3}$