Question

For a 1st order change R \(\rightarrow\) P, the concentration of Reactant R changes from 0.1 M to 0.025 M in 40 minutes. The rate of reaction when the concentration of R is 0.01 M is________.

• A. 1.73 \(\times\) 10\(^{-5}\) M min\(^{-1}\)
• B. 3.47 \(\times\) 10\(^{-4}\) M min\(^{-1}\)
• C. 3.47 \(\times\) 10\(^{-5}\) M min\(^{-1}\)
• D. 1.73 \(\times\) 10\(^{-4}\) M min\(^{-1}\)

Hint:

Always calculate the rate constant \(k\) first using the given concentration change over time. Then use this \(k\) with the specific concentration for which the rate is requested. Pay attention to units (M, min\(^{-1}\)).

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The question asks to calculate the instantaneous rate of a first-order reaction at a specific reactant concentration, given the initial and final concentrations over a time interval.
Step 2: Key Formula or Approach:
1. Integrated Rate Law for First-Order Reaction:
\[ k = \frac{2.303}{t} \log_{10}\left(\frac{[R]_0}{[R]_t}\right) \]
Where:
- \( k \) is the rate constant.
- \( t \) is the time.
- \([R]_0\) is the initial concentration of reactant.
- \([R]_t\) is the concentration of reactant at time \(t\).
2. Rate Law for First-Order Reaction:
\[ \text{Rate} = k[R] \]
Step 3: Detailed Explanation:
Part 1: Calculate the rate constant (\(k\))
Given:
Initial concentration, \([R]_0 = 0.1 \text{ M}\)
Concentration at time \(t\), \([R]_t = 0.025 \text{ M}\)
Time, \(t = 40 \text{ minutes}\)
Substitute these values into the integrated rate law:
\[ k = \frac{2.303}{40 \text{ min}} \log_{10}\left(\frac{0.1 \text{ M}}{0.025 \text{ M}}\right) \]
\[ k = \frac{2.303}{40} \log_{10}(4) \]
Using \(\log_{10}(4) \approx 0.6021\):
\[ k = \frac{2.303}{40} \times 0.6021 \]
\[ k \approx 0.057575 \times 0.6021 \]
\[ k \approx 0.03467 \text{ min}^{-1} \]
Rounding to a reasonable number of significant figures, \(k \approx 0.0347 \text{ min}^{-1}\).
Part 2: Calculate the rate of reaction when \([R] = 0.01 \text{ M}\)
Using the rate law for a first-order reaction:
\[ \text{Rate} = k[R] \]
Substitute the calculated \(k\) and the given concentration:
\[ \text{Rate} = (0.0347 \text{ min}^{-1}) \times (0.01 \text{ M}) \]
\[ \text{Rate} = 0.000347 \text{ M min}^{-1} \]
Expressing in scientific notation:
\[ \text{Rate} = 3.47 \times 10^{-4} \text{ M min}^{-1} \]
Step 4: Final Answer:
The rate of reaction when the concentration of R is 0.01 M is $3.47 \times 10^{-4}$ M min$^{-1}$.