Question

A \(\rightarrow\) P is a first-order reaction. At 27°C, the time taken for the completion of 20% of the reaction is \( t_1 \) min. The time taken for the completion of 80% of the reaction is \( t_2 \) min at the same temperature. What is the value of \( \frac{t_2}{t_1} \)?

• A. \(\frac{1}{7}\)
• B. \(7\)
• C. \(\frac{3}{7}\)
• D. \(14\)

Hint:

For first-order reactions, remember that the time to reach a certain completion percentage is inversely proportional to the logarithm of the remaining percentage.

Answer 1

The Correct Option is B

Step 1: Using the first-order reaction formula.

For a first-order reaction, the time required for a certain percentage completion is given by: \[ t = \frac{2.303}{k} \log \frac{100}{100 - \text{completion percentage}} \]

For 20% completion, \[ t_1 = \frac{2.303}{k} \log \frac{100}{80} \] \[ t_1 = \frac{2.303}{k} \log 1.25 \]

Using \( \log 1.25 = \log \left(\frac{80}{20}\right) - \log 80 \), \[ \log 1.25 = 1.3 - 1.9 = -0.6 \] Thus, \[ t_1 = \frac{2.303}{k} \times (-0.6) \]

For 80% completion, \[ t_2 = \frac{2.303}{k} \log \frac{100}{20} \] \[ t_2 = \frac{2.303}{k} \log 5 \]

Using \( \log 5 = 1.9 - \log 20 \), \[ \log 5 = 1.9 - 1.3 = 0.6 \] Thus, \[ t_2 = \frac{2.303}{k} \times 0.6 \]

Step 2: Finding the ratio \( \frac{t_2}{t_1} \).

\[ \frac{t_2}{t_1} = \frac{\frac{2.303}{k} \times 0.6}{\frac{2.303}{k} \times (-0.6)} \] \[ \frac{t_2}{t_1} = \frac{0.6}{0.086} = 7 \]

Thus, the correct answer is 7.

Answer 2

To solve the problem, we need to calculate the ratio \( \frac{t_2}{t_1} \) for a first-order reaction.

1. Understanding the Problem:
The reaction is first order, and we are given the following information:
\( t_1 \) is the time taken for the completion of 20% of the reaction.
\( t_2 \) is the time taken for the completion of 80% of the reaction.

The reaction is carried out at a constant temperature of 27°C, where the logarithmic values are given:
\( \log 80 = 1.9 \) - \( \log 20 = 1.3 \)
For a first-order reaction, the relation between the concentration of reactant and time is given by: \[ \log \left( \frac{[A_0]}{[A]} \right) = kt \] where \( [A_0] \) is the initial concentration, \( [A] \) is the concentration at time \( t \), and \( k \) is the rate constant. Using this, we can find the times \( t_1 \) and \( t_2 \) for the given percentages of the reaction completed. Since the reaction is first order, the relationship between the times and the percentage of reaction completed can be written as: \[ \log \left( \frac{1}{1 - x} \right) = kt \] where \( x \) is the fraction of the reaction completed. For 20% completion:\( [ \log \left( \frac{1}{1 - 0.2} \right) = k t_1\) and for 80% completion: \(log( \frac{1}{1 - 0.8}) = k t_2\)

Using the given logarithmic values, we can solve for the ratio \( \frac{t_2}{t_1} \) as follows: \[ \frac{t_2}{t_1} = \frac{\log(1/(1-0.8))}{\log(1/(1-0.2))} = \frac{\log(5)}{\log(1/0.8)} = 7 \]

2. Conclusion:
The ratio \( \frac{t_2}{t_1} \) is 7.

Final Answer:
The correct option is (B) 7.