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Q3.25:
Sucrose decomposes in acid solution into glucose and fructose according to the first order rate law, with \(t_{1/2} = 3.00\) hours. What fraction of sample of sucrose remains after 8 hours ?

Solution :

Step 1: Establishing the Kinetic Parameters

The decomposition of sucrose follows a first-order rate law. The rate constant $k$ for a first-order reaction is related to the half-life $t_{1/2}$ by the following expression:

$k = \frac{\ln(2)}{t_{1/2}}$ [Derived from the integration of the first-order rate equation, $\ln[R]_t = -kt + \ln[R]_0$]

Given $t_{1/2} = 3.00 \text{ h}$:

$k = \frac{0.693}{3.00 \text{ h}} = 0.231 \text{ h}^{-1}$

Step 2: Formulating the Concentration-Time Relationship

For a first-order reaction, the amount of substance remaining after time $t$ is given by the integrated rate law:

$\ln\left(\frac{[R]_t}{[R]_0}\right) = -kt$

To find the fraction remaining, we calculate the ratio $\frac{[R]_t}{[R]_0}$:

$\frac{[R]_t}{[R]_0} = e^{-kt}$

Step 3: Quantitative Calculation

Given $t = 8.00 \text{ h}$ and $k = 0.231 \text{ h}^{-1}$, substitute these values into the exponential function:

$\frac{[R]_t}{[R]_0} = e^{-(0.231 \text{ h}^{-1} \times 8.00 \text{ h})}$

$\frac{[R]_t}{[R]_0} = e^{-1.848}$

Using the exponential constant $e \approx 2.718$:

$\frac{[R]_t}{[R]_0} \approx 0.1575$

Step 4: Graphical Representation of First-Order Decay

Time (h) [Sucrose] 8h

Step 5: Analysis of the Result

The calculation $e^{-1.848}$ represents the fraction of the initial concentration that remains undecayed after 8 hours of reaction time. The value $0.1575$ indicates that approximately $15.75\%$ of the initial sucrose concentration exists in the system after the elapsed period.

Final Solution: The fraction of the sample of sucrose remaining after 8 hours is approximately 0.1575.


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