Animated Solution for Chemistry - Chemical Kinetics: Sucrose hydrolyses in acid solution into glucose and fructose following first order rate law with a half-life of $3.33\text{ h}$ at $25^{\circ }\text{C}$. After $9\text{ h}$, the fraction of sucrose remaining is $f$. The value of ${\log }_{10}\left(\frac{1}{f}\right)$ is ...... $\times 10^{-2}$ (Rounded off to the nearest integer). [Assume, $\ln 10=2.303$, $\ln 2=0.693$]

Imagine you are observing the hydrolysis of sucrose in an acidic solution. This is a classic example of a first-order reaction, where sucrose molecules break down into glucose and fructose. The defining characteristic of a first-order reaction is that its concentration drops exponentially over time.

We are given the half-life of the reaction, which is $3.33\text{ h}$. For the sake of mathematical elegance and to avoid early rounding errors, it is highly recommended to convert this decimal into a fraction. Thus, we write the half-life as $t_{1/2}=\frac{10}{3}\text{ h}$. This represents the exact time required for the concentration of sucrose to drop to exactly half of its initial value.

To understand the kinetics at any arbitrary time, we must connect the half-life to the rate constant, $k$. For a first-order reaction, this relationship is beautifully simple:

Next, we need the integrated rate law to find the concentration at our specific time, $t=9\text{ h}$. The formula, expressed in base-10 logarithms (since the question specifically asks for a ${\log }_{10}$ value), is:

The problem states that after $9\text{ h}$, the fraction of sucrose remaining is $f$. This means the ratio of the final concentration to the initial concentration is exactly $f$. Therefore, the term inside our logarithm, which is the inverse of this ratio, becomes $\frac{1}{f}$.

Now, we equate our two expressions for the rate constant $k$. By substituting the half-life formula on the left and the integrated rate law on the right, we set up our master equation:

We need to find the value of ${\log }_{10}\left(\frac{1}{f}\right)$. Let's rearrange the equation to isolate this term on one side:

Now, let's plug in the given values. We know $\ln 2=0.693$, the time $t=9\text{ h}$, and the half-life $t_{1/2}=\frac{10}{3}\text{ h}$. Substituting these into our rearranged equation gives:

Simplifying the denominator, $2.303\times 10$ becomes $23.03$. The $3$ in the denominator's denominator flips up to multiply with the $9$, giving us $27$ in the numerator:

Multiplying the numerator yields $18.711$. Dividing this by $23.03$ gives us approximately $0.8124$.

The question specifically asks for the answer in the format of $x\times 10^{-2}$. So, we rewrite $0.8124$ as $81.24\times 10^{-2}$. Rounding to the nearest integer, we arrive at our final answer: 81.