Q.74. Sucrose hydrolyses in acidic medium into glucose and fructose by first order rate law with t =

Q: What is the correct answer for JEE Main Chemistry Q74?

The correct answer is 25. The hydrolysis of sucrose follows first-order kinetics. Given half-life (t$_{1/2}$) = 3 hours. For a first-order reaction, the amount remaining after 'n' half-lives is given by $N = N_0 \left( \frac{1}{2} \right)^n$. Total time = 6 hours. Number of half-lives (n) = $\frac{\text{Total time}}{\text{Half-life}} = \frac{6 \text{ hours}}{3 \text{ hours}} = 2$ Percentage of sucrose remaining = $\left

Q: Q.74. Sucrose hydrolyses in acidic medium into glucose and fructose by first order rate law with t = 3 hour. The percentage of sucrose remaining after 6 hours is ______ (Nearest integer) (Given: log…

The correct answer is 25. The hydrolysis of sucrose follows first-order kinetics. Given half-life (t$_{1/2}$) = 3 hours. For a first-order reaction, the amount remaining after 'n' half-lives is given by $N = N_0 \left( \frac{1}{2} \right)^n$. Total time = 6 hours. Number of half-lives (n) = $\frac{\text{Total time}}{\text{Half-life}} = \frac{6 \text{ hours}}{3 \text{ hours}} = 2$ Percentage of sucrose remaining = $\left

Question

Q.74. Sucrose hydrolyses in acidic medium into glucose and fructose by first order rate law with t = 3 hour. The percentage of sucrose remaining after 6 hours is ______ (Nearest integer) (Given: log 2 = 0.3010 and log 3 = 0.4771)

FirstInTest · Accepted Answer

The hydrolysis of sucrose follows first-order kinetics.
Given half-life (t$_{1/2}$) = 3 hours.

For a first-order reaction, the amount remaining after 'n' half-lives is given by $N = N_0 \left( \frac{1}{2} ight)^n$.

Total time = 6 hours.
Number of half-lives (n) = $\frac{	ext{Total time}}{	ext{Half-life}} = \frac{6 	ext{ hours}}{3 	ext{ hours}} = 2$

Percentage of sucrose remaining = $\left( \frac{1}{2} ight)^2 	imes 100\% = \frac{1}{4} 	imes 100\% = 25\%$

Alternatively, using the first-order rate constant (k):
$k = \frac{0.693}{t_{1/2}} = \frac{0.693}{3 	ext{ hr}} = 0.231 	ext{ hr}^{-1}$

For a first-order reaction, $\ln\left( \frac{N_0}{N} ight) = kt$
$\ln\left( \frac{N_0}{N} ight) = (0.231 	ext{ hr}^{-1}) 	imes (6 	ext{ hr}) = 1.386$
$\frac{N_0}{N} = e^{1.386}$
Using $\ln 2 = 0.693$, $e^{0.693} = 2$. So $e^{1.386} = e^{2 	imes 0.693} = (e^{0.693})^2 = 2^2 = 4$.
$\frac{N_0}{N} = 4 \Rightarrow \frac{N}{N_0} = \frac{1}{4}$

Percentage remaining = $\frac{N}{N_0} 	imes 100\% = \frac{1}{4} 	imes 100\% = 25\%$

Rounding to the nearest integer, the percentage of sucrose remaining is 25%.

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