*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

**▶️Answer/Explanation**

**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. \(\frac{0.40 – 0.20}{10 – 0}\)

**Answer/Explanation**

**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

**▶️Answer/Explanation**

**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?

**Answer/Explanation**

**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?

A. Adding water.

B. Adding some 0.1mol \(dm^{-3}\) hydrogen peroxide solution.

C. Adding some 2.0mol \(dm^{-3}\) hydrogen peroxide solution.

D. Repeating the experiment without a catalyst.

**▶️Answer/Explanation**

**Markscheme: B**

The curve Y is described as shallow in shape with a greater volume than curve X, which is steeper. This implies that the reaction represented by curve Y is slower than the one represented by curve X.