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
Show Hint
For first-order reactions, always check if the concentration change is a power of 2 (like \( 1/2, 1/4, 1/8 \)). If it is, you can skip the logarithmic formula and find the half-life directly by dividing the total time by the number of half-lives.
Step 1: Understanding the Concept:
To find the rate at a specific concentration, we first need to determine the first-order rate constant ($k$) from the provided concentration-time data. Step 2: Key Formula or Approach:
1. Recognize half-life intervals to quickly find $t_{1/2}$, or use $k = \frac{2.303}{t} \log\frac{[R]_0}{[R]}$.
2. Then calculate $k = \frac{0.693}{t_{1/2}}$.
3. Finally, calculate the specific rate using the differential rate law: $\text{Rate} = k[R]$. Step 3: Detailed Explanation:
1. Calculate the Rate Constant ($k$):
Initial concentration, $[R]_{0} = 0.1$ M.
Final concentration, $[R]_{t} = 0.025$ M.
Total time, $t = 40$ min.
Notice the simple ratio: $0.1 \xrightarrow{t_{1/2}} 0.05 \xrightarrow{t_{1/2}} 0.025$.
The concentration dropped to one-fourth, meaning exactly two half-lives have passed.
$2 \times t_{1/2} = 40 \text{ min} \implies t_{1/2} = 20 \text{ min}$.
Now, calculate $k$:
\[ k = \frac{0.693}{t_{1/2}} = \frac{0.693}{20} = 0.03465 \text{ min}^{-1}. \]
2. Calculate the Rate at $[R] = 0.01$ M:
\[ \text{Rate} = k \times [R] \]
\[ \text{Rate} = 0.03465 \text{ min}^{-1} \times 0.01 \text{ M} \]
\[ \text{Rate} = 3.465 \times 10^{-4} \text{ M min}^{-1} \approx 3.47 \times 10^{-4} \text{ M min}^{-1}. \] Step 4: Final Answer:
The reaction rate at 0.01 M is $3.47 \times 10^{-4}$ M min$^{-1}$.
