Question

Half life for a first order reaction is 6.93 hour. What is the time required for 80% completion of the reaction?

• A. 12 hours
• B. 18 hours
• C. 6 hours
• D. 16 hours

Hint:

For a quick mental check: 50% completion takes 1 half-life (6.93 hours). 75% completion takes 2 half-lives (13.86 hours). Since 80% completion requires slightly more than 2 half-lives, the answer must be just above 13.86 hours. This easily rules out 6, 12, and 18 hours, leaving 16 hours as the only logical option!

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
We are given the half-life ($t_{1/2}$) of a first-order chemical reaction. We need to determine the total time period ($t$) necessary for the reaction to reach exactly 80% completion.

Step 2: Key Formula or Approach:
For a first-order reaction, the rate constant ($k$) is calculated from half-life using: $$ k = \frac{0.693}{t_{1/2}} $$ The integrated first-order rate expression 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 at time $t$.

Step 3: Detailed Explanation:
First, let's find the rate constant ($k$) using the provided half-life value ($t_{1/2} = 6.93\ \text{hours}$): $$ k = \frac{0.693}{6.93\ \text{hour}} = 0.1\ \text{hour}^{-1} $$ Next, set up the concentration parameters for 80% completion:

• Let the initial reactant concentration be $[A]_0 = 100$.

• Since 80% of the reactant has been consumed, the remaining concentration is: $$ [A]_t = 100 - 80 = 20 $$
Now, substitute $k$ and the concentrations into the integrated rate law to find time ($t$): $$ t = \frac{2.303}{0.1} \log_{10}\left(\frac{100}{20}\right) $$ $$ t = 23.03 \times \log_{10}(5) $$ Given that $\log_{10}(5) \approx 0.6990$: $$ t = 23.03 \times 0.6990 \approx 16.10\ \text{hours} $$ Rounding to the nearest whole integer matching the options gives 16 hours.

Step 4: Final Answer: The total time required for 80% completion is 16 hours, which corresponds to option (D).