Sucrose is converted to a mixture of glucose and fructose in a pseudo first order process under alkaline conditions. The reaction has half life of $28.4 min$ . The time period (in min) upto one decimal, required for the reduction of a $8.0 mM$ sample of sucrose to $1.0 mM$ is
(Given: $l o g 8 = 0.903$ )

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Solution

Given: $t_{1 / 2} = 28.4 min$
$\Rightarrow t_{1 / 2} = \frac{0.693}{k}$
$\Rightarrow k = \frac{0.693}{28.4} = 0.0244 {min}^{- 1}$
We know,
$t = \frac{2.303}{k} l o g \frac{A_{o}}{A_{t}}$
$\Rightarrow t = \frac{2.303}{0.0244} l o g \frac{8}{1}$
$\Rightarrow t = 85.2 min$

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Sucrose is converted to a mixture of glucose and fructose in a pseudo first order process under alkaline conditions. The reaction has half life of 28.4 min. The time period (in min) upto one decimal, required for the reduction of a 8.0 mM sample of sucrose to 1.0 mM is (Given: log8=0.903)

Given: t1/2=28.4 min ⇒t1/2=0.693k ⇒k=0.69328.4=0.0244 min−1 We know, t=2.303klogAoAt ⇒t=2.3030.0244log81 ⇒t=85.2 min

Given: $t_{1 / 2} = 28.4 min$
$\Rightarrow t_{1 / 2} = \frac{0.693}{k}$
$\Rightarrow k = \frac{0.693}{28.4} = 0.0244 {min}^{- 1}$
We know,
$t = \frac{2.303}{k} l o g \frac{A_{o}}{A_{t}}$
$\Rightarrow t = \frac{2.303}{0.0244} l o g \frac{8}{1}$
$\Rightarrow t = 85.2 min$