Question

Sucrose Decomposes In Acid Solution Into Glucose And Fructose According To The First Order Rate Law, With T1/2 = 3.00 Hours. What Fraction Of Sample Of Sucrose Remains After 8 Hours?

Answer 1

Step 1: Finding the value of k

According to first order rate law equation,

$$\begin{gathered} k=\frac{0.063}{t_{1/2}} \\ =\frac{0.063}{3.0} \\ =0.231hr \end{gathered}$$

Step 2: Finding the value of the sample

Let initial concentration of sucrose be 1 M

After 8 hours, the concentration will be $1-\mathrm{x}$. $\mathrm{x}$ represents the amount of sucrose decomposed. By the following steps $\mathrm{x}$ is being calculated

$$\begin{gathered} k=\frac{2.303}{t}\log \frac{{\left[A\right]}_0}{\left[A\right]} \\ 0.231=\frac{2.303}{8}\log \frac{1}{1-x} \\ l\mathrm{og}\frac{1}{1-x}=0.8024 \\ \frac{1}{1-x}=6.345 \\ x=0.842 \end{gathered}$$

$\therefore$The concentration of a sample of sucrose after 8 hours will be

$1-0.842=0.158M$