Question

Time required for 90% completion of a first order reaction is $t$. What is the time required for completion of 99% reaction?

• A. $t$
• B. $2t$
• C. $t_{1/2}$
• D. $3t$

Hint:

For first-order kinetics, log base 10 values make relationships predictable. Going from 90% complete (10% left, or $10^{-1}$) to 99% complete (1% left, or $10\^{-2}$) means the power of 10 doubles. Therefore, the time required for 99% completion is always exactly twice the time required for 90% completion ($t_{99\%} = 2 \times t_{90\%}$).

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The question asks for the relationship between the time required for a first-order reaction to reach 99% completion and the time $t$ required to reach 90% completion.

Step 2: Key Formula or Approach:
The integrated rate equation for a first-order chemical reaction is given by: $$k = \frac{2.303}{t} \log_{10}\left(\frac{[A]_0}{[A]_t}\right) \implies t = \frac{2.303}{k} \log_{10}\left(\frac{[A]_0}{[A]_t}\right)$$ where $[A]_0$ is the initial concentration and $[A]_t$ is the remaining concentration of the reactant at time $t$.

Step 3: Detailed Explanation:
Let the initial concentration of the reactant be $[A]_0 = 100$. 1. For 90% completion of the reaction: Amount reacted = 90 Remaining concentration $[A]_t = 100 - 90 = 10$ Given time = $t$ Substituting these values into the rate equation: $$t = \frac{2.303}{k} \log_{10}\left(\frac{100}{10}\right) = \frac{2.303}{k} \log_{10}(10)$$ Since $\log_{10}(10) = 1$: $$t = \frac{2.303}{k} \quad \text{--- (Equation 1)}$$ 2. For 99% completion of the reaction: Amount reacted = 99 Remaining concentration $[A]_{t'} = 100 - 99 = 1$ Let the required time be $t_{99\%}$ Substituting these values into the rate equation: $$t_{99\%} = \frac{2.303}{k} \log_{10}\left(\frac{100}{1}\right) = \frac{2.303}{k} \log_{10}(10^2)$$ Using logarithmic properties ($\log x^2 = 2 \log x$): $$t_{99\%} = \frac{2.303}{k} \times 2 \log_{10}(10) = 2 \times \left(\frac{2.303}{k}\right) \quad \text{--- (Equation 2)}$$ Comparing Equation 1 and Equation 2, we can substitute $t$ into the expression for $t_{99\%}$: $$t_{99\%} = 2t$$

Step 4: Final Answer:
The time required for 99% completion of the reaction is $2t$, matching option (B).