Question

Sucrose hydrolyses in acid solution into glucose and fructose following first order rate law with a half-life of 3.33 h at 25°C. After 9hm the fraction of sucrose remaining is f. The value of $\log_{10}(\frac{1}{f})$ is _______ $\times 10^{-2}$. (Rounded off to the nearest integer)

Answer 1

81

Answer 2

The hydrolysis of sucrose in acid solution follows first-order kinetics. The integrated rate law for a first-order reaction is given by:

$\ln(\frac{[A]_t}{[A]_0}) = -kt$

where $[A]_t$ is the concentration of sucrose at time $t$, $[A]_0$ is the initial concentration, and $k$ is the rate constant.

The fraction of sucrose remaining after time $t$ is $f = \frac{[A]_t}{[A]_0}$. So, $\ln(f) = -kt$.

The half-life ($t_{1/2}$) for a first-order reaction is related to the rate constant by:

$k = \frac{\ln(2)}{t_{1/2}}$

Given $t_{1/2} = 3.33 \text{ h}$ and $t = 9 \text{ h}$ (assuming "9hm" is a typo for 9 hours).

$k = \frac{\ln(2)}{3.33}$

Substitute the value of $k$ into the integrated rate law:

$\ln(f) = - \frac{\ln(2)}{3.33} \times 9$

We are asked to find the value of $\log_{10}(\frac{1}{f})$.

$\log_{10}(\frac{1}{f}) = \log_{10}(f^{-1}) = - \log_{10}(f)$

We know that $\ln(x) = 2.303 \log_{10}(x)$, so $\log_{10}(x) = \frac{\ln(x)}{2.303}$.

$\log_{10}(f) = \frac{\ln(f)}{2.303}$

Substitute the expression for $\ln(f)$:

$\log_{10}(f) = \frac{- \frac{\ln(2)}{3.33} \times 9}{2.303} = - \frac{\ln(2)}{2.303} \times \frac{9}{3.33}$

We know that $\frac{\ln(2)}{2.303} = \log_{10}(2)$.

$\log_{10}(f) = - \log_{10}(2) \times \frac{9}{3.33}$

Now, calculate $\log_{10}(\frac{1}{f})$:

$\log_{10}(\frac{1}{f}) = - \log_{10}(f) = \log_{10}(2) \times \frac{9}{3.33}$

Using the value $\log_{10}(2) \approx 0.30103$:

$\log_{10}(\frac{1}{f}) \approx 0.30103 \times \frac{9}{3.33}$

Calculate the value:

$\frac{9}{3.33} = \frac{900}{333} = \frac{100}{37}$

$\log_{10}(\frac{1}{f}) \approx 0.30103 \times \frac{100}{37}$

$0.30103 \times \frac{100}{37} \approx 0.30103 \times 2.7027027... \approx 0.81355$

The value of $\log_{10}(\frac{1}{f})$ is approximately 0.81355.

We need to express this in the form $\_\_\_\_\_\_ \times 10^{-2}$ and round off the blank to the nearest integer.

$0.81355 = X \times 10^{-2}$

$X = 0.81355 \times 100 = 81.355$

Rounding off 81.355 to the nearest integer gives 81.