Reaction Kinetics

Statement 1 follows a reactant being used up per unit time — the definition of rate. Correct.

Statement 2 follows a product being formed — the same definition, from the product side. Correct.

Statement 3 follows the gas volume produced per unit time at constant pressure — the gas-volume method in the table above. Correct.

State the relationship: $\text{average rate} = \frac{\text{change in volume of CO}_2}{\text{time taken}}$

Change in volume: $48 - 12 = 36\ \text{cm}^3$.

Time taken: $80 - 20 = 60\ \text{s}$.

Divide: $\text{rate} = \frac{36}{60} = 0.60$

Units: volume unit over time unit, $\text{cm}^3\text{ s}^{-1}$. So the average rate is $0.60\ \text{cm}^3\text{ s}^{-1}$.

State the relationship: $\text{average rate} = \frac{\text{change in }[\text{Br}_2]}{\text{time taken}}$

Read the two values off Fig. 4.1: at $t=0$, $[\text{Br}_2] = 100\times10^{-5}\ \text{mol dm}^{-3}$; at $t=600\ \text{s}$, $[\text{Br}_2] = 12\times10^{-5}\ \text{mol dm}^{-3}$.

Change in concentration: $(100\times10^{-5}) - (12\times10^{-5}) = 88\times10^{-5}\ \text{mol dm}^{-3}$

Divide by the time taken, $600\ \text{s}$: $\text{rate} = \frac{88\times10^{-5}}{600} = 1.47\times10^{-6}$

Units: concentration ($\text{mol dm}^{-3}$) over time ($\text{s}$), giving $\text{mol dm}^{-3}\text{ s}^{-1}$.

Convert the gas volume to moles of $\text{O}_2$, using the molar gas volume at room conditions, $24\,000\ \text{cm}^3\text{ mol}^{-1}$: $n(\text{O}_2) = \frac{180}{24\,000} = 0.00750\ \text{mol}$

Use the equation to find moles of $\text{H}_2\text{O}_2$ used. The ratio is $2\ \text{H}_2\text{O}_2 : 1\ \text{O}_2$: $n(\text{H}_2\text{O}_2) = 2 \times 0.00750 = 0.0150\ \text{mol}$

Convert the time to minutes, because the answer is wanted in $\text{mol min}^{-1}$: $90\ \text{s} = \frac{90}{60} = 1.50\ \text{min}$

Divide: $\text{rate} = \frac{0.0150}{1.50} = 0.0100\ \text{mol min}^{-1}$

Find $P$, the rate over the first $30\ \text{s}$: the volume changes from $0$ to $64\ \text{cm}^3$, so $P = \frac{64 - 0}{30} = 2.133\ \text{cm}^3\text{ s}^{-1}$

Find $Q$, the rate over the last $30\ \text{s}$ (from $150\ \text{s}$ to $180\ \text{s}$): the volume changes from $161$ to $167\ \text{cm}^3$, so $Q = \frac{167 - 161}{30} = 0.200\ \text{cm}^3\text{ s}^{-1}$

Subtract: $P - Q = 2.133 - 0.200 = 1.93\ \text{cm}^3\text{ s}^{-1}$

State the relationship: $\text{rate} = \left|\,\text{gradient of the tangent}\,\right| = \frac{\text{change in concentration}}{\text{change in time}}$

Initial rate — change in concentration along the tangent at $t=0$: $0.100 - 0 = 0.100\ \text{mol dm}^{-3}$.

Change in time along the same tangent: $50 - 0 = 50\ \text{s}$.

Divide: $\text{initial rate} = \frac{0.100}{50} = 2.0\times10^{-3}\ \text{mol dm}^{-3}\text{ s}^{-1}$

Rate at 50 s — change in concentration along the second tangent: $0.074 - 0 = 0.074\ \text{mol dm}^{-3}$.

Change in time along the same tangent: $100 - 0 = 100\ \text{s}$.

Divide: $\text{rate at 50 s} = \frac{0.074}{100} = 7.4\times10^{-4}\ \text{mol dm}^{-3}\text{ s}^{-1}$

Draw a tangent to the absorbance–time curve at $20\ \text{s}$ (where the absorbance is $0.30$).

Find its gradient from two points far apart. The tangent falls by about $0.45$ absorbance units over about $50\ \text{s}$: $\text{gradient} \approx \frac{0.45}{50} \approx 0.009\ \text{absorbance units per second}$

Read the conversion factor off the calibration curve. Absorbance $0.50$ matches $0.25\times10^{-3}\ \text{mol dm}^{-3}$, and the line passes through the origin (absorbance $0.10$ matches $0.05\times10^{-3}$), so $1\ \text{absorbance unit} \equiv 0.50\times10^{-3}\ \text{mol dm}^{-3}$

Convert the gradient into a rate: $\text{rate} \approx 0.009 \times 0.50\times10^{-3} \approx 4.5\times10^{-6} \approx 5\times10^{-6}\ \text{mol dm}^{-3}\text{ s}^{-1}$

State the relationship: the same amount of sulfur hides the cross in each run, so $\text{rate} \propto \dfrac{1}{t}$.

At $20\,°\text{C}$: $\dfrac{1}{t} = \dfrac{1}{80} = 0.0125\ \text{s}^{-1}$.

At $40\,°\text{C}$: $\dfrac{1}{t} = \dfrac{1}{20} = 0.0500\ \text{s}^{-1}$.

Compare: $\frac{0.0500}{0.0125} = 4$ The reaction is 4 times faster at $40\,°\text{C}$.

Link pressure to particles: at higher pressure the same number of gas molecules is in a smaller volume, so there are more molecules per unit volume.

Link that to collisions: the molecules are closer together, so $\text{H}_2$ and $\text{I}_2$ molecules collide more frequently.

Link that to effective collisions: the temperature is unchanged, so the same fraction of collisions have enough energy. More collisions per second therefore means more effective collisions per second.

Conclude: a higher frequency of effective collisions means a higher rate.

A is wrong. The temperature is constant, so the particles' energies are unchanged. "More particles with energy above the activation energy" is how temperature works.

B is wrong. Higher pressure means the particles are squeezed into less space, not more.

C is wrong. Particle speed depends on temperature, which has not changed.

D is right. More particles per unit volume collide more often, so effective collisions are more frequent too.

Spot the one thing that changes. Same volume of acid, same mass of the same $\text{CuCO}_3$ powder (so the same surface area), and no change of temperature is mentioned — only the acid concentration differs: $0.1\ \text{mol dm}^{-3}$ in reaction 1 against $0.5\ \text{mol dm}^{-3}$ in reaction 2.

Apply collision theory. Reaction 2 has five times as many $\text{H}^+$ ions in each $\text{dm}^3$, so acid particles strike the surface of the powder more often: the frequency of collisions, and with it the frequency of effective collisions, is higher in reaction 2. So the ratio (reaction 1 ÷ reaction 2) of effective collisions per second is less than 1.

Link collisions to rate. Rate depends on the frequency of effective collisions, so reaction 2 is also faster — the ratio of rates (reaction 1 ÷ reaction 2) is less than 1 as well.