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Hydrolysis of cane sugar in presence ...

Hydrolysis of cane sugar in presence of acid is a first order reaction with half life time of 4 hours what fraction of sucrose undergoes hydrolysis in 11 hrs ?

A

`0.9345`

B

`0.6122`

C

`0.8511`

D

`0.412`

Text Solution

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The correct Answer is:
To solve the problem of determining the fraction of sucrose that undergoes hydrolysis in 11 hours, we will follow these steps: ### Step 1: Determine the rate constant (k) For a first-order reaction, the half-life (\(t_{1/2}\)) is related to the rate constant (k) by the formula: \[ k = \frac{0.693}{t_{1/2}} \] Given that the half-life is 4 hours, we can substitute this value into the formula: \[ k = \frac{0.693}{4} = 0.17325 \, \text{hours}^{-1} \] ### Step 2: Use the first-order kinetics equation The first-order kinetics equation is given by: \[ \ln\left(\frac{A_0}{A}\right) = kt \] Where: - \(A_0\) is the initial concentration of sucrose, - \(A\) is the concentration of sucrose remaining after time \(t\), - \(k\) is the rate constant, - \(t\) is the time in hours. ### Step 3: Set up the equation for 11 hours Assuming the initial concentration \(A_0 = 1\) (for simplicity), we can express the remaining concentration \(A\) as \(1 - X\), where \(X\) is the fraction of sucrose that has undergone hydrolysis. Substituting the values into the equation: \[ \ln\left(\frac{1}{1 - X}\right) = k \cdot t \] For \(t = 11\) hours, we have: \[ \ln\left(\frac{1}{1 - X}\right) = 0.17325 \cdot 11 \] Calculating the right-hand side: \[ 0.17325 \cdot 11 = 1.90575 \] Thus, we have: \[ \ln\left(\frac{1}{1 - X}\right) = 1.90575 \] ### Step 4: Solve for \(1 - X\) To eliminate the natural logarithm, we exponentiate both sides: \[ \frac{1}{1 - X} = e^{1.90575} \] Calculating \(e^{1.90575}\): \[ e^{1.90575} \approx 6.707 \] Thus, we have: \[ 1 - X = \frac{1}{6.707} \] Calculating this gives: \[ 1 - X \approx 0.149 \] ### Step 5: Solve for \(X\) Now, we can find \(X\): \[ X = 1 - 0.149 \approx 0.851 \] ### Conclusion The fraction of sucrose that undergoes hydrolysis in 11 hours is approximately 0.851. Therefore, the correct answer is option C. ---
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