Sucrose hydrolyses in acid solution into glucose and fructose following first order rate law with a half- life of 3.33 h at $25^{\circ} \mathrm{C}$. After 9 h , the fraction of sucrose remaining is f . The value of $\log _{10}\left(\frac{1}{\mathrm{f}}\right)$ is $\_\_\_\_$ $\times 10^{-2}$. (Rounded off to the nearest integer) [Assume: $\ln 10=2.303, \ln 2=0.693$ ]

JEE Main · 2021 · 24 Feb 2021 · Shift 2 · Chemical Kinetics

Sucrose hydrolyses in acid solution into glucose and fructose following first order rate law with a half- life of 3.33 h at

2 5^{\circ} C

. After 9 h , the fraction of sucrose remaining is f . The value of

lo g_{10} (\frac{1}{f})

____

\times 1 0^{- 2}

.
(Rounded off to the nearest integer)
[Assume:

ln 10 = 2.303, ln 2 = 0.693

]

Answer: 81

For a first-order reaction the rate constant follows from the half-life:

k = \frac{ln 2}{t _{1/2}} = \frac{0.693}{3.33} = 0.2081 h^{- 1}

The integrated first-order law relates the fraction

f

still unreacted to the elapsed time:

ln \frac{1}{f} = k t = 0.2081 \times 9 = 1.873

lo g_{10} \frac{1}{f} = \frac{1.873}{2.303} = 0.813

Expressed in the units asked for,

0.813 = 81.3 \times 1 0^{- 2}

, which to the nearest integer makes the required value

81

JEE Main · 2021 · 24 Feb 2021 · Shift 2 · Integer answer