Question

The gas-phase decomposition reaction of ethane is given below: C2H6------ >C2H2 + 3H2 This reaction is investigated at constant temperature and volume. The table below shows the partial pressure (PC2H6) observed for the above reaction at different times. 3 55 Competency Focused Practice Questions | Chemistry | Grade 12 # Time (in mins) PC2H6 (in mm of Hg) 1 0 800 2 100 400 3 200 200 Based on the above data, (i) What is the order of the reaction? Justify with reason. (ii) Calculate the rate constant of the reaction. (iii) What is the ratio of the time required for the completion of 50% of the reaction to that of 75% of the reaction?

Answer 1

(i) The order of the reaction is First Order.
Reason: The half-life of the reactant is constant (100 minutes), which is a characteristic of first-order reactions.

(ii) The rate constant of the reaction is $\mathbf{0.00693 \text{ min}^{-1}}$.

(iii) The ratio of the time required for the completion of 50% of the reaction to that of 75% of the reaction is $\mathbf{1:2}$ or $\mathbf{0.5}$.

Answer 2

The decomposition reaction of ethane is given as: $\text{C}_2\text{H}_6 \longrightarrow \text{C}_2\text{H}_2 + 3\text{H}_2$

The reaction is investigated at constant temperature and volume. The partial pressure of $\text{C}_2\text{H}_6$ at different times is given:

Time (in mins)	$\text{P}_{\text{C}_2\text{H}_6}$ (in mm of Hg)
0	800
100	400
200	200

(i) What is the order of the reaction? Justify with reason.

Observation: From the given data:

• At time $t=0$ min, $\text{P}_{\text{C}_2\text{H}_6} = 800$ mm of Hg.

• At time $t=100$ min, $\text{P}_{\text{C}_2\text{H}_6} = 400$ mm of Hg.
  The pressure has halved (from 800 to 400) in 100 minutes. So, the first half-life ($t_{1/2}^{(1)}$) is 100 minutes.

• At time $t=100$ min, $\text{P}_{\text{C}_2\text{H}_6} = 400$ mm of Hg.

• At time $t=200$ min, $\text{P}_{\text{C}_2\text{H}_6} = 200$ mm of Hg.
  The pressure has halved again (from 400 to 200) in $200 - 100 = 100$ minutes. So, the second half-life ($t_{1/2}^{(2)}$) is 100 minutes.

Justification: Since the half-life ($t_{1/2}$) of the reactant ($\text{C}_2\text{H}_6$) is constant and independent of its initial partial pressure (concentration), the reaction is a first-order reaction. This is a characteristic property of first-order reactions.

(ii) Calculate the rate constant of the reaction.

For a first-order reaction, the integrated rate law is: $k = \frac{1}{t} \ln\left(\frac{P_0}{P_t}\right)$ where $P_0$ is the initial partial pressure and $P_t$ is the partial pressure at time $t$.

Using the data at $t=100$ min: $P_0 = 800$ mm of Hg $P_t = 400$ mm of Hg $t = 100$ min

$k = \frac{1}{100 \text{ min}} \ln\left(\frac{800 \text{ mm of Hg}}{400 \text{ mm of Hg}}\right)$ $k = \frac{1}{100} \ln(2)$ Given $\ln(2) \approx 0.693$ $k = \frac{0.693}{100} \text{ min}^{-1}$ $k = 0.00693 \text{ min}^{-1}$

(iii) What is the ratio of the time required for the completion of 50% of the reaction to that of 75% of the reaction?

For a first-order reaction: Time for 50% completion ($t_{50\%}$ or $t_{1/2}$): $t_{50\%} = \frac{\ln(2)}{k}$

Time for 75% completion ($t_{75\%}$): For 75% completion, the remaining reactant is $100\% - 75\% = 25\%$ of the initial amount. So, $P_t = 0.25 P_0 = \frac{P_0}{4}$. Using the integrated rate law: $t_{75\%} = \frac{1}{k} \ln\left(\frac{P_0}{P_0/4}\right)$ $t_{75\%} = \frac{1}{k} \ln(4)$ Since $\ln(4) = \ln(2^2) = 2 \ln(2)$: $t_{75\%} = \frac{2 \ln(2)}{k}$

Ratio of $t_{50\%}$ to $t_{75\%}$: Ratio $= \frac{t_{50\%}}{t_{75\%}} = \frac{\frac{\ln(2)}{k}}{\frac{2 \ln(2)}{k}}$ Ratio $= \frac{1}{2}$

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Explanation of the solution:

(i) The reaction is first order because the half-life (time taken for the partial pressure of $\text{C}_2\text{H}_6$ to halve) is constant (100 minutes) regardless of the initial pressure. (ii) Using the first-order integrated rate law, $k = \frac{1}{t} \ln\left(\frac{P_0}{P_t}\right)$, and substituting $P_0=800$, $P_t=400$, $t=100$, we get $k = \frac{1}{100} \ln(2) = 0.00693 \text{ min}^{-1}$. (iii) For a first-order reaction, $t_{50\%} = \frac{\ln(2)}{k}$. For 75% completion, 25% reactant remains, so $t_{75\%} = \frac{1}{k} \ln\left(\frac{P_0}{0.25 P_0}\right) = \frac{1}{k} \ln(4) = \frac{2 \ln(2)}{k}$. The ratio $\frac{t_{50\%}}{t_{75\%}} = \frac{\ln(2)/k}{2\ln(2)/k} = \frac{1}{2}$.