JEE MAIN - Chemistry (2021 - 24th February Evening Shift - No. 20)
Sucrose hydrolyses in acid solution into glucose and fructose following first order rate law with a half-life of 3.33 h at 25$$^\circ$$C. After 9 h, the fraction of sucrose remaining is f. The value of $${\log _{10}}\left( {{1 \over f}} \right)$$ is ________ $$\times$$ 10$$-$$2. (Rounded off to the nearest integer)
[Assume : ln 10 = 2.303, ln 2 = 0.693]
[Assume : ln 10 = 2.303, ln 2 = 0.693]
Answer
81
Explanation
Given, $$\mathop {{C_{12}}{H_{22}}{O_{11}}}\limits_{Sucrose} + {H_2}O\buildrel {1st\,order} \over
\longrightarrow \mathop {{C_6}{H_{12}}{O_6}}\limits_{Glu\cos e} + \mathop {{C_6}{H_{12}}{O_6}}\limits_{Fructose} $$
$${t_{1/2}} = {{10} \over 3}h$$
At t = 0, a = [A]0 (initial conc.)
t = 9h, a $$-$$ x = [A]t [conc. at time t]
For using 1st order equation,
$$K = {{2.303} \over t}\log {{{{[A]}_0}} \over {{{[A]}_t}}} $$
$$\Rightarrow {{K \times t} \over {2.303}} = \log {{{{[A]}_0}} \over {{{[A]}_t}}}$$
$${{\ln 2 \times 9} \over {10/3 \times 2.303}} = \log \left( {{1 \over F}} \right) \Rightarrow \log \left( {{1 \over F}} \right) = 0.8124$$ ($$\because$$ $$k = {{\ln 2} \over {{t_{1/2}}}}$$)
$$\log \left( {{1 \over F}} \right) = 81.24 \times {10^{ - 2}}$$
x = 81.24 or x $$\approx$$ 81
$${t_{1/2}} = {{10} \over 3}h$$
At t = 0, a = [A]0 (initial conc.)
t = 9h, a $$-$$ x = [A]t [conc. at time t]
For using 1st order equation,
$$K = {{2.303} \over t}\log {{{{[A]}_0}} \over {{{[A]}_t}}} $$
$$\Rightarrow {{K \times t} \over {2.303}} = \log {{{{[A]}_0}} \over {{{[A]}_t}}}$$
$${{\ln 2 \times 9} \over {10/3 \times 2.303}} = \log \left( {{1 \over F}} \right) \Rightarrow \log \left( {{1 \over F}} \right) = 0.8124$$ ($$\because$$ $$k = {{\ln 2} \over {{t_{1/2}}}}$$)
$$\log \left( {{1 \over F}} \right) = 81.24 \times {10^{ - 2}}$$
x = 81.24 or x $$\approx$$ 81
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