Sucrose decomposes in acid solution into glucose and fructose according to first order rate law with $t_{1 / 2} = 3$ Hrs. What fraction of the sample of sucrose remains after $8$ hours?

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Solution

First order rate law is:
$ln (\frac{A_{0}}{A_{t}}) = k t a t t = t_{1 / 2} = 3 h, A_{t} = 0.5 A_{0}$
On substitution: $ln (\frac{A_{0}}{{0.5 A}_{0}}) = k 3 k = 0.231$
At $t = 8 h, A_{t} =$ concentration at time $t$
Substitution gives: $ln (\frac{A_{0}}{A_{t}}) = 0.231 \times 8 (\frac{A_{t}}{A_{0}}) = 0.1575$
Therefore $15.75$ % of sucrose remains undecomposed after 8 hours.

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Sucrose decomposes in acid solution into glucose and fructose according to first order rate law with t1/2=3 Hrs. What fraction of the sample of sucrose remains after 8 hours?

First order rate law is:ln(A0At)=ktatt=t1/2=3h,At=0.5A0On substitution: ln(A00.5A0)=k3k=0.231At t=8h,At=concentration at time tSubstitution gives: ln(A0At)=0.231×8(AtA0)=0.1575Therefore 15.75% of sucrose remains undecomposed after 8 hours.

Sucrose decomposes in acid solution into glucose and fructose according to first order rate law with $t_{1 / 2} = 3$ Hrs. What fraction of the sample of sucrose remains after $8$ hours?