DSE Chemistry Diagnostic: Chemical Kinetics

Unit Test 1: Rate Equation Determination

Question

The reaction $A + 2B \rightarrow C$ was studied at constant temperature. The following initial rate data were obtained:

Experiment	$[A]$ (mol/dm$^{3}$)	$[B]$ (mol/dm$^{3}$)	Initial Rate (mol dm$^{-3}$ s$^{-1}$)
1	0.10	0.10	$1.2 \times 10^{-4}$
2	0.20	0.10	$2.4 \times 10^{-4}$
3	0.10	0.20	$4.8 \times 10^{-4}$
4	0.30	0.30	$2.16 \times 10^{-3}$

(a) Determine the order of reaction with respect to $A$ and with respect to $B$. [3 marks]

(b) Write the rate equation and calculate the rate constant $k$, including its units. [3 marks]

(c) Using the rate equation, predict the initial rate when $[A] = 0.15$ and $[B] = 0.25$ mol/dm$^{3}$. [2 marks]

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Worked Solution

(a) Order with respect to $A$: Compare experiments 1 and 2 (where $[B]$ is constant).

$[A]$ doubles (0.10 to 0.20), rate doubles ($1.2 \times 10^{-4}$ to $2.4 \times 10^{-4}$).

Rate $\propto [A]^{1}$, so order with respect to $A$ = 1.

Order with respect to $B$: Compare experiments 1 and 3 (where $[A]$ is constant).

$[B]$ doubles (0.10 to 0.20), rate quadruples ($1.2 \times 10^{-4}$ to $4.8 \times 10^{-4}$).

Rate $\propto [B]^{2}$, so order with respect to $B$ = 2.

Verification with experiment 4: With orders $A = 1$, $B = 2$: predicted rate $= 0.12 \times 0.30 \times 0.30^{2} = 3.24 \times 10^{-3}$. The given value of $2.16 \times 10^{-3}$ does not match. Experiment 4 may contain a typo, or is designed to test whether students verify all data points. Proceeding with orders $A = 1$, $B = 2$ as determined from experiments 1--3.

(b) Rate equation: $\text{Rate} = k[A][B]^{2}$

From experiment 1:

$k = \frac{1.2 \times 10^{-4}}{0.10 \times (0.10)^{2}} = \frac{1.2 \times 10^{-4}}{0.001} = 0.12 \text{ dm}^{6} \text{ mol}^{-2} \text{ s}^{-1}$

Units: $\frac{\text{mol dm}^{-3} \text{ s}^{-1}}{(\text{mol dm}^{-3})^{3}} = \text{dm}^{6} \text{ mol}^{-2} \text{ s}^{-1}$

(c) Rate = $0.12 \times 0.15 \times (0.25)^{2} = 0.12 \times 0.15 \times 0.0625 = 1.125 \times 10^{-3} \text{ mol dm}^{-3} \text{ s}^{-1}$

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Unit Test 2: Maxwell-Boltzmann Distribution

Question

(a) Sketch a Maxwell-Boltzmann distribution curve for a gas at temperature $T_{1}$, and on the same axes, sketch the curve for the same gas at a higher temperature $T_{2}$ ($T_{2} \gt T_{1}$). Indicate the activation energy $E_{a}$ on both curves. [3 marks]

(b) Explain why increasing the temperature increases the rate of a reaction more significantly than increasing the concentration of reactants. [3 marks]

(c) A student claims that "adding a catalyst shifts the Maxwell-Boltzmann distribution to the left, so more molecules have energy above $E_{a}$." Evaluate this claim. [2 marks]

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Worked Solution

(a) Key features of the sketch:

• Both curves start at the origin, peak, and tail off towards higher energies.
• The $T_{2}$ curve has a lower peak that is shifted to the right (higher average energy) compared to $T_{1}$.
• Both curves have the same total area (same number of molecules).
• A horizontal line at $E_{a}$ intersects both curves. The area under the curve to the right of $E_{a}$ is larger for $T_{2}$.

(b) Increasing concentration increases the rate because there are more collisions per unit time, but the proportion of successful collisions (those with $E \geq E_{a}$) remains the same.

Increasing temperature increases the rate because:

1. More molecules have energy $\geq E_{a}$ (larger area under the curve beyond $E_{a}$), so a greater proportion of collisions are successful.
2. The average kinetic energy increases, so molecules move faster and collide more frequently. The combined effect (higher proportion AND higher frequency) makes temperature much more impactful than concentration alone.

(c) The claim is incorrect. A catalyst lowers the activation energy ($E_{a}$) rather than shifting the distribution. The Maxwell-Boltzmann distribution itself does not change. By lowering $E_{a}$, a larger area under the same distribution curve now exceeds the threshold, meaning more molecules have sufficient energy to react.

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Unit Test 3: Catalyst and Activation Energy

Question

The decomposition of hydrogen peroxide is catalysed by manganese(IV) oxide:

$2H_{2}O_{2}(aq) \rightarrow 2H_{2}O(l) + O_{2}(g)$

(a) Define the term activation energy. [1 mark]

(b) Explain how $MnO_{2}$ increases the rate of decomposition of $H_{2}O_{2}$, using the term "alternative reaction pathway." [3 marks]

(c) Does $MnO_{2}$ change the enthalpy change ($\Delta H$) of the reaction? Explain. [2 marks]

(d) The reaction is zero order with respect to $H_{2}O_{2}$ when $MnO_{2}$ is present in large excess. Explain what "zero order" means and why this occurs. [2 marks]

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Worked Solution

(a) Activation energy is the minimum energy that colliding reactant particles must possess for a reaction to occur.

(b) $MnO_{2}$ provides an alternative reaction pathway with a lower activation energy. The reactant molecules adsorb onto the surface of the solid catalyst, where bonds are weakened, lowering the energy barrier for the reaction. More molecules now possess energy greater than or equal to this lower activation energy, so the rate of successful collisions increases.

(c) No, $MnO_{2}$ does not change the enthalpy change of the reaction. A catalyst affects the rate of the reaction (kinetics) by providing a lower-energy pathway, but it does not affect the thermodynamics ($\Delta H$). The enthalpy of the reactants and products remains the same regardless of whether a catalyst is used.

(d) "Zero order" means the rate of reaction is independent of the concentration of $H_{2}O_{2}$.

This occurs because when $MnO_{2}$ is in large excess, the surface of the catalyst is fully covered (saturated) with $H_{2}O_{2}$ molecules. Adding more $H_{2}O_{2}$ cannot increase the number of molecules adsorbed on the surface at any given time, so the rate depends only on the available catalyst surface area, not on $[H_{2}O_{2}]$.

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Integration Test 1: Rate Equation + Mechanism Prediction

Question

The reaction $2NO_{2}(g) + F_{2}(g) \rightarrow 2NO_{2}F(g)$ has the experimentally determined rate equation:

$\text{Rate} = k[NO_{2}][F_{2}]$

(a) What is the overall order of the reaction? [1 mark]

(b) A student proposes the following one-step mechanism:

$2NO_{2} + F_{2} \rightarrow 2NO_{2}F$

Explain why this proposed mechanism is inconsistent with the rate equation. [2 marks]

(c) Propose a two-step mechanism that is consistent with the rate equation, identifying the rate-determining step. [3 marks]

(d) The reaction is first order with respect to $NO_{2}$. If the concentration of $NO_{2}$ is tripled while $[F_{2}]$ is kept constant, by what factor does the rate change? [1 mark]

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Worked Solution

(a) Overall order = $1 + 1 = 2$ (first order in $NO_{2}$, first order in $F_{2}$).

(b) If the reaction occurred in a single step involving the collision of $2NO_{2}$ and $F_{2}$ simultaneously (a termolecular collision), the rate equation would be:

$\text{Rate} = k[NO_{2}]^{2}[F_{2}]$

This predicts second order with respect to $NO_{2}$, which contradicts the experimentally determined rate equation (first order in $NO_{2}$). Therefore, the one-step mechanism is inconsistent.

(c) A consistent two-step mechanism:

Step 1 (slow, rate-determining): $NO_{2} + F_{2} \rightarrow NO_{2}F + F$

Step 2 (fast): $NO_{2} + F \rightarrow NO_{2}F$

Overall: $2NO_{2} + F_{2} \rightarrow 2NO_{2}F$

The rate equation is determined by the rate-determining step:

$\text{Rate} = k[NO_{2}][F_{2}]$

This matches the experimental rate equation.

(d) Since the reaction is first order in $NO_{2}$, tripling $[NO_{2}]$ will triple the rate (factor of 3).

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Integration Test 2: Initial Rates + Concentration-Time Graph

Question

For the reaction $A \rightarrow B + C$, the following data were collected:

Time (s)	$[A]$ (mol/dm$^{3}$)
0	0.80
100	0.60
200	0.45
300	0.34
400	0.25

(a) Plot a concentration-time graph and determine the order of reaction with respect to $A$. [3 marks]

(b) Calculate the rate of reaction at $t = 200$ s by drawing a tangent. [2 marks]

(c) The half-life of this reaction between $t = 0$ and $t = 200$ s is approximately 175 s. The half-life between $t = 200$ and $t = 400$ s is approximately 250 s. What does this tell you about the order of the reaction? [2 marks]

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Worked Solution

(a) From the data, as time increases by equal intervals (100 s), the concentration of $A$ decreases by a decreasing amount:

• 0 to 100 s: $\Delta[A] = -0.20$
• 100 to 200 s: $\Delta[A] = -0.15$
• 200 to 300 s: $\Delta[A] = -0.11$
• 300 to 400 s: $\Delta[A] = -0.09$

The rate of decrease is proportional to the current concentration (as $[A]$ decreases, the rate of decrease also decreases proportionally). This is characteristic of a first-order reaction.

For a first-order reaction, a plot of $\ln[A]$ vs time should be a straight line:

Time (s)	$[A]$	$\ln[A]$
0	0.80	$-0.223$
100	0.60	$-0.511$
200	0.45	$-0.799$
300	0.34	$-1.079$
400	0.25	$-1.386$

The $\ln[A]$ values decrease approximately linearly (drops by about 0.29 per 100 s interval), confirming first order.

(b) At $t = 200$ s, $[A] = 0.45$ mol/dm$^{3}$.

For a first-order reaction: Rate $= k[A]$.

From the gradient of $\ln[A]$ vs $t$:

$k = \frac{(-1.386) - (-0.223)}{400 - 0} = \frac{-1.163}{400} = 2.91 \times 10^{-3} \text{ s}^{-1}$

Rate at $t = 200$ s: $k[A] = 2.91 \times 10^{-3} \times 0.45 = 1.31 \times 10^{-3} \text{ mol dm}^{-3} \text{ s}^{-1}$

By tangent method: The gradient of the $[A]$ vs $t$ curve at $t = 200$ s gives $-d[A]/dt \approx 1.3 \times 10^{-3}$ mol dm$^{-3}$ s$^{-1}$.

(c) The half-life is increasing with time. This is characteristic of a second-order reaction (where $t_{1/2} \propto 1/[A]_{0}$, so as $[A]$ decreases, $t_{1/2}$ increases). However, from part (a) the reaction appears first-order.

The resolution: For a first-order reaction, the half-life should be constant. The apparent increase in half-life (175 s vs 250 s) suggests the data may not perfectly follow first-order kinetics, or the half-life estimates from the limited data are imprecise. The more reliable analysis from part (a) (linearity of $\ln[A]$ vs $t$) suggests the reaction is approximately first-order, with the half-life variation due to experimental data limitations.

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Integration Test 3: Temperature Effect and Arrhenius Concept

Question

The rate constant for a reaction at 300 K is $2.5 \times 10^{-3}$ s$^{-1}$ and at 320 K is $1.0 \times 10^{-2}$ s$^{-1}$.

(a) Calculate the activation energy for this reaction using the Arrhenius equation:

$\ln\left(\frac{k_{2}}{k_{1}}\right) = \frac{E_{a}}{R}\left(\frac{1}{T_{1}} - \frac{1}{T_{2}}\right)$

($R = 8.314$ J mol$^{-1}$ K$^{-1}$) [4 marks]

(b) Calculate the rate constant at 340 K. [2 marks]

(c) A student states that "at very high temperatures, all reactions will be instantaneous because all molecules will have energy above $E_{a}$." Evaluate this statement. [2 marks]

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Worked Solution

(a) $\ln\left(\frac{1.0 \times 10^{-2}}{2.5 \times 10^{-3}}\right) = \frac{E_{a}}{8.314}\left(\frac{1}{300} - \frac{1}{320}\right)$

$\ln(4) = \frac{E_{a}}{8.314}\left(\frac{320 - 300}{300 \times 320}\right)$

$1.386 = \frac{E_{a}}{8.314} \times \frac{20}{96000}$

$1.386 = \frac{E_{a}}{8.314} \times 2.083 \times 10^{-4}$

$E_{a} = \frac{1.386 \times 8.314}{2.083 \times 10^{-4}}$

$E_{a} = \frac{11.523}{2.083 \times 10^{-4}} = 55320 \text{ J/mol} = 55.3 \text{ kJ/mol}$

(b) Using the Arrhenius equation with $T_{1} = 300$ K, $k_{1} = 2.5 \times 10^{-3}$ s$^{-1}$, $T_{2} = 340$ K:

$\ln\left(\frac{k_{2}}{2.5 \times 10^{-3}}\right) = \frac{55320}{8.314}\left(\frac{1}{300} - \frac{1}{340}\right)$

$= 6654.7 \times \frac{40}{102000} = 6654.7 \times 3.922 \times 10^{-4} = 2.610$

$\frac{k_{2}}{2.5 \times 10^{-3}} = e^{2.610} = 13.60$

$k_{2} = 13.60 \times 2.5 \times 10^{-3} = 3.40 \times 10^{-2} \text{ s}^{-1}$

(c) The statement is an overgeneralisation. While it is true that at higher temperatures, a larger proportion of molecules have energy above $E_{a}$, the Maxwell-Boltzmann distribution always has a tail extending to very low energies. At any finite temperature, some molecules will always have energy below $E_{a}$. Furthermore, even if all molecules exceed $E_{a}$, the rate still depends on the collision frequency and the correct orientation of collisions (the steric/entropy factor). The rate would increase dramatically but would not be truly "instantaneous." In practice, at very high temperatures, other factors such as reactant decomposition or competing reactions may become significant.