In acidic solution sucrose is converted to a mixture of glucose and fructose in pseudo first order reaction. It has been found that the concentration of sucrose decreased from $20 m m o l L^{- 1}$ to $8 m m o l L^{- 1}$ in 38 minutes. What is the half life of the reaction?

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Solution

For $1^{s t}$ order reaction, $K = \frac{1}{t} l n \frac{a_{o}}{a_{t}}$
Here $a_{o} = 20 m m o l / L, a_{t} = 8 m m o l / L, t = 38 m i n$
$∴ K = \frac{1}{38} l n \frac{20}{8} = 0.02411 m i n^{- 1}$
$∵ t_{1 / 2} = \frac{l n 2}{K}$
$\Rightarrow t_{1 / 2} = \frac{l n 2}{0.02411} = 28.75 m i n s$

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In acidic solution sucrose is converted to a mixture of glucose and fructose in pseudo first order reaction. It has been found that the concentration of sucrose decreased from 20 mmol L−1 to 8 mmol L−1 in 38 minutes. What is the half life of the reaction?

For 1st order reaction, K=1tlnaoatHere ao=20mmol/L,at=8mmol/L,t=38min∴K=138ln208=0.02411min−1∵t1/2=ln2K⇒t1/2=ln20.02411=28.75mins

In acidic solution sucrose is converted to a mixture of glucose and fructose in pseudo first order reaction. It has been found that the concentration of sucrose decreased from $20 m m o l L^{- 1}$ to $8 m m o l L^{- 1}$ in 38 minutes. What is the half life of the reaction?

For $1^{s t}$ order reaction, $K = \frac{1}{t} l n \frac{a_{o}}{a_{t}}$
Here $a_{o} = 20 m m o l / L, a_{t} = 8 m m o l / L, t = 38 m i n$
$∴ K = \frac{1}{38} l n \frac{20}{8} = 0.02411 m i n^{- 1}$
$∵ t_{1 / 2} = \frac{l n 2}{K}$
$\Rightarrow t_{1 / 2} = \frac{l n 2}{0.02411} = 28.75 m i n s$