Question

The time taken for 10% completion of a first order reaction is 20 minutes. The time taken for 19% completion of the same reaction will be:

• A. 20 min
• B. 10 min
• C. 30 min
• D. 40 min

Hint:

For first-order reactions, notice that 100 → 90 → 81 follows a geometric progression. Equal fractional decay implies equal time intervals.

Answer 1

The Correct Option is D

Concept: This question is based on the kinetics of a first-order chemical reaction. Key definitions and mathematical expressions for first-order kinetics include:
• Integrated Rate Law equation: The rate constant $k$ for a first-order reaction can be expressed as: \[ k = \frac{2.303}{t} \log_{10} \left(\frac{[A]_0}{[A]_t}\right) \] where:
• $k$ is the specific reaction rate or rate constant (independent of time and concentration).
• $t$ is the time elapsed for the reaction.
• $[A]_0$ is the initial concentration of the reactant at time $t=0$.
• $[A]_t$ is the remaining concentration of the reactant at time $t$.
• Concentration after decomposition: If a reaction is $x\%$ complete, the amount of reactant consumed is $x\%$ of $[A]_0$, meaning the remaining concentration is: \[ [A]_t = [A]_0 \left(1 - \frac{x}{100}\right) \] By setting up this expression for the two distinct conditions given in the problem, we can eliminate the rate constant $k$ and systematically solve for the unknown time.

Step 1: Analyzing Case 1: 10% completion of the reaction.
Let the initial concentration of the reactant be $[A]_0 = 100$. For a 10% completion of the reaction:
• Amount of reactant reacted = 10% of 100 = 10
• Remaining concentration of the reactant ($[A]_{t1}$) = 100 − 10 = 90
• Time taken ($t_1$) = 20 minutes Substituting these parameters into the integrated rate law formula: \[ k = \frac{2.303}{t_1} \log_{10}\left(\frac{[A]_{t1}}{[A]_0}\right) \] \[ k = \frac{2.303}{20} \log_{10}\left(\frac{90}{100}\right) \] Simplifying: \[ k = \frac{2.303}{20} \log_{10}\left(\frac{9}{10}\right) \cdots (1) \]

Step 2: Analyzing Case 2: 19% completion of the reaction.
Using the same initial concentration, $[A]_0 = 100$. For a 19% completion of the reaction:
• Amount of reactant reacted = 19% of 100 = 19
• Remaining concentration of the reactant ($[A]_{t2}$) = 100 − 19 = 81
• Let the time taken for this process be $t_2$. Substituting into the integrated rate law: \[ k = \frac{2.303}{t_2} \log_{10}\left(\frac{81}{100}\right) \]

Step 3: Equating the expressions to solve for $t_2$.
Since $k$ is constant: \[ \frac{2.303}{20} \log_{10}\left(\frac{9}{10}\right) = \frac{2.303}{t_2} \log_{10}\left(\frac{81}{100}\right) \] Cancel $2.303$: \[ \frac{1}{20} \log_{10}\left(\frac{9}{10}\right) = \frac{1}{t_2} \log_{10}\left(\frac{81}{100}\right) \] \[ t_2 = 20 \times \frac{\log_{10}(9/10)}{\log_{10}(81/100)} \]

Step 4: Simplification using logarithm properties.
\[ \frac{81}{100} = \left(\frac{9}{10}\right)^2 \] So, \[ \log_{10}\left(\frac{81}{100}\right) = 2 \log_{10}\left(\frac{9}{10}\right) \] \[ t_2 = 20 \times \frac{\log_{10}(9/10)}{2\log_{10}(9/10)} \] \[ t_2 = 20 \times \frac{1}{2} = 40 \text{ min} \] Hence, the time required is 40 minutes.