Question

If the time required for 40% completion of a first order reaction is 50 minutes, what is the time required for completion of 80% of the same reaction?
(\( \log 2 = 0.3010, \log 5 = 0.6990, \log 6 = 0.7782 \))

• A. 157.8 min
• B. 147.8 min
• C. 78.8 min
• D. 73.8 min

Hint:

For a first-order reaction, use \( k = \frac{1}{t} \ln \left( \frac{[\text{A}]_0}{[\text{A}]} \right) \) to find \( k \), then calculate the time for a different percentage using the same \( k \). Convert given \( \log \) values to \( \ln \) using \( \ln x = \log x \times 2.3026 \).

Answer 1

The Correct Option is A

For a first-order reaction, the rate constant \( k \) can be found using the formula: \[ k = \frac{1}{t} \ln \left( \frac{[\text{A}]_0}{[\text{A}]} \right) \] where \( [\text{A}]_0 \) is the initial concentration, and \( [\text{A}] \) is the concentration at time \( t \). For 40% completion, 60% of the reactant remains: \[ \frac{[\text{A}]}{[\text{A}]_0} = 0.6, \quad \ln \left( \frac{1}{0.6} \right) = \ln \left( \frac{10}{6} \right) = \ln 10 - \ln 6 \] Given \( \log 6 = 0.7782 \), \( \ln 6 = \log 6 \times \ln 10 \approx 0.7782 \times 2.3026 \approx 1.792 \), and \( \ln 10 \approx 2.3026 \), so: \[ \ln \left( \frac{10}{6} \right) \approx 2.3026 - 1.792 = 0.5106 \] For 40% completion, \( t = 50 \, \text{min} \): \[ k = \frac{0.5106}{50} \approx 0.010212 \, \text{min}^{-1} \] Now, for 80% completion, 20% of the reactant remains: \[ \frac{[\text{A}]}{[\text{A}]_0} = 0.2, \quad \ln \left( \frac{1}{0.2} \right) = \ln 5 \] Given \( \log 5 = 0.6990 \), \( \ln 5 = \log 5 \times \ln 10 \approx 0.6990 \times 2.3026 \approx 1.6094 \). The time \( t \) for 80% completion is: \[ t = \frac{\ln 5}{k} = \frac{1.6094}{0.010212} \approx 157.6 \, \text{min} \] This is closest to 157.8 min.
So, the time required for 80% completion is 157.8 min.

Answer 2

If the time required for 40% completion of a first order reaction is 50 minutes, what is the time required for completion of 80% of the same reaction?

Given:
- Time for 40% completion = 50 min
- For a first-order reaction: \( k = \frac{2.303}{t} \log\left(\frac{[A]_0}{[A]}\right) \)

Step 1: Calculate rate constant (k) using 40% completion:
If 40% is completed, then 60% of the reactant remains, so:
\[ \frac{[A]_0}{[A]} = \frac{100}{60} = \frac{5}{3} \]
Now apply the first-order formula:
\[ k = \frac{2.303}{50} \log\left(\frac{5}{3}\right) \]
\[ \log\left(\frac{5}{3}\right) = \log 5 - \log 3 = 0.6990 - 0.4771 = 0.2219 \]
\[ k = \frac{2.303}{50} \times 0.2219 = 0.01023\ \text{min}^{-1} \]

Step 2: Use k to find time for 80% completion:
If 80% is completed, then 20% remains, so:
\[ \frac{[A]_0}{[A]} = \frac{100}{20} = 5 \]
Use the formula again:
\[ t = \frac{2.303}{k} \log(5) \]
\[ t = \frac{2.303}{0.01023} \times 0.6990 = 225.1 \times 0.6990 = 157.38\ \text{min} \]

Final Answer:
\[ \boxed{157.8\ \text{min}} \]