# IB DP Chemistry Reactivity 2.2 How fast? The rate of chemical change IB Style Question Bank HL Paper 1

### Question

What happens to the average kinetic energy, KE, of the particles in a gas when the absolute temperature is doubled?

$$KE = \frac{1}{2} mv^2$$

A. increases by a factor of 2
B. Decreases by a factor of 2
C. Increases by a factor of 4
D. Decreases by a factor of 4

Markscheme: A

The average kinetic energy ($$KE$$) of particles in a gas is directly proportional to the absolute temperature ($$T$$). According to the kinetic theory of gases, the relationship is given by:

$KE = \frac{3}{2} kT$

where:
$$KE$$ is the average kinetic energy,
$$k$$ is the Boltzmann constant,
$$T$$ is the absolute temperature.

If the absolute temperature is doubled ($$2T$$), the kinetic energy will also double. Therefore, the correct answer is:

A. Increases by a factor of 2

### Question

Which calculation determines the initial rate of this reaction?

A. $$\frac{0.40 – 0.19}{20 – 0}$$
B. $$\frac{0.40 – 0.10}{40 – 0}$$
C. $$\frac{0.40 – 0}{140 – 0}$$
D. $$\frac{0.40 – 0.20}{10 – 0}$$

Markscheme: D

\begin{aligned} &\text { rate }=-\frac{\text { [reactant at } \left.\mathrm{t}_2\right]-\left[\text { reactant at } \mathrm{t}_1\right]}{\mathrm{t}_2-\mathrm{t}_1}\\ &\text { rate }=-\frac{\Delta[\text { reactant }]}{\Delta \mathrm{t}} \quad \text { or } \quad \text { rate }=\frac{\Delta[\text { product }]}{\Delta \mathrm{t}} \end{aligned}

To calculate initial rate we will draw tangent at t=0.

From this Initial rate= $$\frac{0.40 – 0.20}{10 – 0}$$ 0r $$\frac{0.40 – 0.00}{20 – 0}$$

### Question

What is the order of reaction with respect to A, given the following reaction mechanism?
$\begin{array}{rlr} \mathrm{A}+\mathrm{B} & \rightarrow \mathrm{AB} & \text { fast } \\ \mathrm{AB}+\mathrm{A} & \rightarrow \mathrm{A}_2 \mathrm{~B} & \text { slow } \\ \mathrm{A}_2 \mathrm{~B} & \rightarrow \mathrm{A}_2+\mathrm{B} & \text { fast } \end{array}$

A. 0
B. 1
C. 2
D. 3

Markscheme: C

To determine the order of the overall reaction with respect to A, we can look at the rate-determining step, which is the slow step in the reaction mechanism. In this case, the slow step is the second step:

$\mathrm{AB} + \mathrm{A} \rightarrow \mathrm{A}_2 \mathrm{~B}$

The rate expression for this step can be written as:

$\text{Rate} = k[\mathrm{AB}][\mathrm{A}]$

This indicates that the rate of the reaction is proportional to the concentration of both $$\mathrm{AB}$$ and $$\mathrm{A}$$. Therefore, the overall reaction is second order with respect to A.

So, the correct answer is:

C. 2

### Question

Which pair of graphs indicate the same order of reaction?

Markscheme: D

### Question

Curve X on the following graph shows the volume of oxygen formed during the catalytic decomposition of a 1.0mol $$dm^{-3}$$ solution of hydrogen peroxide.

Which change would produce the curve Y?
B. Adding some 0.1mol $$dm^{-3}$$ hydrogen peroxide solution.
C. Adding some 2.0mol $$dm^{-3}$$ hydrogen peroxide solution.