Question

Sucrose hydrolyses in acidic medium into glucose and fructose by first order rate law with $t_{1/2} = 3$ hour. The percentage of sucrose remaining after 6 hours is ______. (Nearest integer)
(Given : $\log 2 = 0.3010$ and $\log 3 = 0.4771$)

Hint:

Calculate the number of half-lives that have passed in 6 hours. Since one half-life is 3 hours, two half-lives have passed. Each half-life reduces the concentration by half.

Answer 1

Correct Answer: 25

This question is based on the kinetics of a first-order chemical reaction. For a first-order reaction, the time taken for the concentration of a reactant to reduce to half of its initial value is called the half-life ($t_{1/2}$), and it is independent of the initial concentration.

The rate constant ($k$) for a first-order reaction is related to the half-life by the formula:
$$k = \frac{0.693}{t_{1/2}}$$ or more precisely, $$k = \frac{\ln 2}{t_{1/2}}$$

The integrated rate law for a first-order reaction is:
$$\ln \left( \frac{[A]_0}{[A]_t} \right) = kt$$
Where:
$[A]_0$ is the initial concentration of sucrose.
$[A]_t$ is the concentration of sucrose remaining after time $t$.
$t$ is the time elapsed.

Step 1: Identify the given values.
Given half-life, $t_{1/2} = 3$ hours.
Time elapsed, $t = 6$ hours.

Step 2: Express the relationship between initial and final concentration using the half-life relationship.
Alternatively, we can use the formula for the amount remaining after $n$ half-lives:
$$[A]_t = [A]_0 \left( \frac{1}{2} \right)^n$$
Where $n = \frac{t}{t_{1/2}}$.

Step 3: Calculate the number of half-lives ($n$).
$$n = \frac{6 \text{ hours}}{3 \text{ hours}} = 2$$

Step 4: Calculate the fraction of sucrose remaining.
$$[A]_t = [A]_0 \left( \frac{1}{2} \right)^2 = [A]_0 \times \frac{1}{4}$$

Step 5: Convert the fraction to a percentage.
Percentage remaining = $\left( \frac{[A]_t}{[A]_0} \right) \times 100$
Percentage remaining = $\frac{1}{4} \times 100 = 25\%$

The percentage of sucrose remaining after 6 hours is 25.