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. Figure 2 Draw a tangent to the curve at t = 0. Use this tangent to deduce the initial rate of the reaction. The experiment was repeated at the same temperature and with the same initial concentration of B but with a different initial concentration of A. The new initial rate was 1.7 times greater than in the original experiment. Calculate the new initial concentration of A.

Photo AI

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

Question 2

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, ... show full transcript

Worked Solution & Example Answer: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

Step 1

Draw a tangent to the curve at t = 0.

96%

114 rated

Answer

At t = 0, draw a straight line that touches the curve at the point where [A] = 0.50 mol dm⁻³. The tangent should intersect the time axis between 0 and 12.5 seconds.

Step 2

Use this tangent to deduce the initial rate of the reaction.

99%

104 rated

Answer

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:

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

Step 3

Calculate the new initial concentration of A.

96%

101 rated

Answer

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:

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

97% of Students

Report Improved Results

98% of Students

Recommend to friends

100,000+

Students Supported

1 Million+

Questions answered

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

Worked Solution & Example Answer: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

Draw a tangent to the curve at t = 0.

Use this tangent to deduce the initial rate of the reaction.

Calculate the new initial concentration of A.

Join the A-Level students using SimpleStudy...

Other A-Level Chemistry topics to explore