Question

Reaction: $2A \rightarrow B + C$

Given:

1. The reaction follows a first-order rate law with respect to $[A]$.
2. Rate constant ($k$) = $0.025 \, \text{s}^{-1}$ at $300 \, \text{K}$.
3. Initial concentration of $A$, $[A]_0 = 0.6 \, \text{M}$.

Questions:

1. Calculate the time (in seconds) for $[A]$ to decrease to $0.15 \, \text{M}$.
2. Determine the instantaneous rate of disappearance of $A$ when $[A] = 0.3 \, \text{M}$.
3. Find the half-life of the reaction.
4. If the reaction were zero-order instead, how would the half-life change? (Calculate for $[A]_0 = 0.6 \, \text{M}$).

Answer 1

1. The time for $[A]$ to decrease to $0.15 \, \text{M}$ is $55.45 \, \text{s}$.
2. The instantaneous rate of disappearance of $A$ when $[A] = 0.3 \, \text{M}$ is $0.0075 \, \text{M s}^{-1}$.
3. The half-life of the reaction is $27.72 \, \text{s}$.
4. If the reaction were zero-order, the half-life would be $12 \, \text{s}$.

Answer 2

The problem involves applying concepts of chemical kinetics, specifically integrated rate laws, differential rate laws, and half-life calculations for first-order and zero-order reactions.

1. Calculate the time (in seconds) for $[A]$ to decrease to $0.15 \, \text{M}$.

The reaction is first-order with respect to $[A]$. The integrated rate law for a first-order reaction is: $\ln\left(\frac{[A]_t}{[A]_0}\right) = -kt$ Rearranging to solve for time $t$: $t = \frac{1}{k} \ln\left(\frac{[A]_0}{[A]_t}\right)$ Given:

• Initial concentration, $[A]_0 = 0.6 \, \text{M}$
• Final concentration, $[A]_t = 0.15 \, \text{M}$
• Rate constant, $k = 0.025 \, \text{s}^{-1}$

Substitute the values: $t = \frac{1}{0.025 \, \text{s}^{-1}} \ln\left(\frac{0.6 \, \text{M}}{0.15 \, \text{M}}\right)$ $t = 40 \, \text{s} \times \ln(4)$ $t = 40 \, \text{s} \times 1.38629$ $t = 55.45 \, \text{s}$

2. Determine the instantaneous rate of disappearance of $A$ when $[A] = 0.3 \, \text{M}$.

For a first-order reaction, the instantaneous rate of disappearance of $A$ is given by the differential rate law: $\text{Rate} = -\frac{d[A]}{dt} = k[A]$ Given:

• Rate constant, $k = 0.025 \, \text{s}^{-1}$
• Concentration of $A$, $[A] = 0.3 \, \text{M}$

Substitute the values: $\text{Rate} = (0.025 \, \text{s}^{-1}) \times (0.3 \, \text{M})$ $\text{Rate} = 0.0075 \, \text{M s}^{-1}$

3. Find the half-life of the reaction.

For a first-order reaction, the half-life ($t_{1/2}$) is independent of the initial concentration and is given by: $t_{1/2} = \frac{\ln(2)}{k}$ Given:

• Rate constant, $k = 0.025 \, \text{s}^{-1}$

Substitute the value: $t_{1/2} = \frac{0.6931}{0.025 \, \text{s}^{-1}}$ $t_{1/2} = 27.72 \, \text{s}$

4. If the reaction were zero-order instead, how would the half-life change? (Calculate for $[A]_0 = 0.6 \, \text{M}$).

If the reaction were zero-order, we assume the numerical value of the rate constant remains $0.025$, but its units would be $\text{M s}^{-1}$ for a zero-order reaction (i.e., $k_{zero-order} = 0.025 \, \text{M s}^{-1}$). For a zero-order reaction, the half-life ($t_{1/2}$) depends on the initial concentration and is given by: $t_{1/2} = \frac{[A]_0}{2k}$ Given:

• Initial concentration, $[A]_0 = 0.6 \, \text{M}$
• Hypothetical zero-order rate constant, $k = 0.025 \, \text{M s}^{-1}$

Substitute the values: $t_{1/2} = \frac{0.6 \, \text{M}}{2 \times 0.025 \, \text{M s}^{-1}}$ $t_{1/2} = \frac{0.6 \, \text{M}}{0.05 \, \text{M s}^{-1}}$ $t_{1/2} = 12 \, \text{s}$ Compared to the first-order half-life (27.72 s), the half-life would be significantly shorter (12 s) if the reaction were zero-order with the same numerical rate constant.

---

Explanation of the solution:

1. Time for $[A]$ to decrease: Use the integrated rate law for a first-order reaction, $t = \frac{1}{k} \ln\left(\frac{[A]_0}{[A]_t}\right)$, substituting the given initial and final concentrations and the rate constant.
2. Instantaneous rate: Apply the differential rate law for a first-order reaction, $\text{Rate} = k[A]$, using the given rate constant and concentration of $A$.
3. Half-life (first-order): Calculate the half-life using the formula for a first-order reaction, $t_{1/2} = \frac{\ln(2)}{k}$, which is independent of concentration.
4. Half-life (zero-order): For a hypothetical zero-order reaction, assume the numerical value of $k$ is the same but its units are $\text{M s}^{-1}$. Use the half-life formula for a zero-order reaction, $t_{1/2} = \frac{[A]_0}{2k}$, and the given initial concentration.