JEE PYQ: Chemical Kinetics Question 7
Question 7 - 2021 (24 Feb Shift 2)
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 $\log_{10}\left(\frac{1}{f}\right)$ is ____ $\times 10^{-2}$. (Rounded off to the nearest integer) [Assume: $\ln 10 = 2.303$, $\ln 2 = 0.693$]
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Answer: 81
Solution
Sucrose $\xrightarrow{\text{Hydrolysis}}$ Glucose + Fructose. $t_{1/2} = 3.33,\text{h} = \frac{10}{3},\text{h}$. $C_t = \frac{C_0}{2^{t/t_{1/2}}}$. Fraction remaining $= f = \frac{C_t}{C_0} = \frac{1}{2^{9/t_{1/2}}}$. $\log(1/f) = \log(2^{9/t_{1/2}}) = \frac{9}{t_{1/2}}\log(2) = \frac{9}{\frac{10}{3}} \times 0.3 = \frac{8.1}{10} = 0.81 = 81 \times 10^{-2}$.
BETA
