Question

A first order reaction is 50% complete in 20 minutes. Calculate the time for 75% completion of the reaction. $[\log~4=0\cdot6]$

Hint:

For first-order reactions, 75% completion is always exactly $2 \times t_{1/2}$. 87.5% completion is $3 \times t_{1/2}$.

Answer 1

Step 1: Concept
The half-life ($t_{1/2}$) and integrated rate equation for a first-order reaction.

Step 2: Meaning
The time required for 50% completion is the half-life ($t_{1/2} = 20\text{ min}$).

Step 3: Analysis
For a first-order reaction, $k = \frac{0.693}{t_{1/2}} = \frac{0.693}{20}$. Using the first-order rate equation for 75% completion: $t = \frac{2.303}{k} \log\left(\frac{100}{100 - 75}\right) = \frac{2.303}{k} \log 4$ Substitute $k = \frac{2.303}{t_{1/2}} \log 2$: $t_{75\%} = \frac{t_{1/2}}{\log 2} \log 4 = \frac{20}{0.3} \times 0.6 = 40\text{ minutes}$.

Final Answer: 40 minutes. (Since 75% completion exactly represents two half-lives: $2 \times 20 = 40\text{ min}$).