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 9 h, the fraction of sucrose remaining is f. The value of log10 1 f 1f is ......x _10^-2. (Rounded off to the nearest integer) [Assume : ln 10 = 2.303, ln 2 = 0.693]

Answer 1

Step 1 :- Find the value of k, if the half life period is given by using formula

$$\begin{gathered} \mathrm{k}=\frac{0.693}{{\mathrm{t}}_{1/2}} \\ \mathrm{k}=\frac{0.693}{3.33} \\ \mathrm{k}=0.207/\mathrm{hr} \end{gathered}$$

Step 2 :- To find the value of k, after the 9hrs of the reaction.

${\mathrm{C}}_{11}{\mathrm{H}}_{22}{\mathrm{O}}_{12}\rightleftharpoons {\mathrm{C}}_6{\mathrm{H}}_{12}{\mathrm{O}}_6+{\mathrm{C}}_6{\mathrm{H}}_{10}{\mathrm{O}}_5$

Sucrose Glucose Fructose

At t=0 1 0 0

After 9hrs, 1 - x x x

Let the initial concentration (A₀) be 1M.

After 9hrs, the sucrose remaining (A_t) be( 1-x )M

$$\begin{gathered} \mathrm{K}=\frac{2.303}{{\mathrm{t}}_{1/2}}\log \frac{\left[{\mathrm{A}}_0\right]}{\left[{\mathrm{A}}_{\mathrm{t}}\right]} \\ 0.207=\frac{2.303}{9}\log \frac{\left[{\mathrm{A}}_0\right]}{\left[{\mathrm{A}}_{\mathrm{t}}\right]} \\ \frac{0.207*9}{2.303}=\log \frac{\left[{\mathrm{A}}_0\right]}{\left[{\mathrm{A}}_{\mathrm{t}}\right]} \\ 0.8124=\log \frac{1}{\mathrm{f}} \\ \log \frac{1}{\mathrm{f}}=81*{10}^{-2} \end{gathered}$$

Final answer :- The value of Log₁₀1/f is 81 x 10^-2.