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Step-by-Step Solution
Step 1: Identify the Given Data
• The hydrolysis of sucrose follows a first-order rate law.
• The half-life of sucrose ( t_{1/2} ) at 25°C is 3.33 h.
• We want to find \log_{10}\!\bigl(\tfrac{1}{f}\bigr) , where f is the fraction of sucrose remaining after 9 hours.
Step 2: Recall the First-Order Kinetics Formula
For a first-order reaction, the rate constant k is related to the half-life by:
\[
k = \frac{\ln(2)}{t_{1/2}}.
\]
Also, the integrated first-order rate equation (in base 10 logarithm) can be expressed as:
\[
\log_{10} \left(\frac{[A]_0}{[A]_t}\right) = \frac{k\,t}{2.303},
\]
where [A]_0 is the initial concentration of sucrose and [A]_t is the concentration at time t .
Step 3: Determine the Rate Constant k
Given t_{1/2} = 3.33\text{ h} ,
\[
k = \frac{\ln(2)}{t_{1/2}}
= \frac{0.693}{3.33}\,\text{h}^{-1}.
\]
(We will substitute exact values in a later step.)
Step 4: Apply the Formula for the Fraction Remaining
After t = 9\text{ h} , let the fraction of sucrose that remains be f . Then:
\[
\log_{10}\!\bigl(\tfrac{1}{f}\bigr)
= \log_{10}\!\left(\frac{[A]_0}{[A]_t}\right)
= \frac{k \times 9}{2.303}.
\]
Substitute k = \tfrac{\ln(2)}{3.33} into the above:
\[
\log_{10}\!\bigl(\tfrac{1}{f}\bigr)
= \frac{\bigl(\frac{\ln(2)}{3.33}\bigr)\times 9}{2.303}.
\]
Step 5: Numerical Evaluation
1. Compute k :
\[
k \approx \frac{0.693}{3.33} \approx 0.208\,\text{h}^{-1}.
\]
2. Multiply by 9\text{ h} :
\[
0.208 \times 9 = 1.872.
\]
3. Divide by 2.303 :
\[
\frac{1.872}{2.303} \approx 0.812.
\]
Therefore,
\[
\log_{10}\!\bigl(\tfrac{1}{f}\bigr) \approx 0.812.
\]
Step 6: Express the Value in Terms of 10^{-2}
To write 0.812 as some number \times\,10^{-2} , observe that
\[
0.812 = 81.2 \times 10^{-2}.
\]
When rounded to the nearest integer for the coefficient, it becomes 81 \times 10^{-2} .
Final Answer
\[
\log_{10}\!\bigl(\tfrac{1}{f}\bigr) \approx 81 \times 10^{-2}.
\]
