Question

For an isothermal second order aqueous phase reaction A \(\to\) B, the ratio of the time required for 90% conversion to the time required for 45% conversion is _______

• A. 2
• B. 4
• C. 11
• D. 22

Hint:

For any second-order reaction, the time required is proportional to the ratio \(\frac{X_A}{1-X_A}\).
Thus, the ratio of times for two different conversions \(X_1\) and \(X_2\) is simply:
\[ \frac{t_1}{t_2} = \frac{X_1 (1 - X_2)}{X_2 (1 - X_1)} \] Substituting \(X_1 = 0.90\) and \(X_2 = 0.45\) gives:
\[ \frac{t_{90}}{t_{45}} = \frac{0.90 \times 0.55}{0.45 \times 0.10} = 2 \times 5.5 = 11 \] This bypasses writing the constants, saving valuable exam time.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The question asks for the ratio of the time required to achieve \(90\%\) conversion (\(t_{90}\)) to the time required to achieve \(45\%\) conversion (\(t_{45}\)) for a second-order reaction carried out under isothermal, constant-volume (aqueous phase) batch conditions.

Step 2: Key Formula or Approach:
For a second-order reaction of the form \(\text{A} \to \text{B}\) in a constant-volume system, the rate equation is:
\[ -r_A = -\frac{dC_A}{dt} = k C_A^2 \] Integrating this differential equation from \(t = 0\) (where \(C_A = C_{A0}\)) to time \(t\) (where \(C_A = C_A\)):
\[ \int_{C_{A0}}^{C_A} -\frac{dC_A}{C_A^2} = k \int_{0}^{t} dt \] \[ \frac{1}{C_A} - \frac{1}{C_{A0}} = k t \] We express concentration in terms of fractional conversion \(X_A\) using \(C_A = C_{A0}(1 - X_A)\):
\[ t = \frac{1}{k C_{A0}} \left( \frac{1}{1 - X_A} - 1 \right) = \frac{1}{k C_{A0}} \left( \frac{X_A}{1 - X_A} \right) \]

Step 3: Detailed Explanation:
Let us calculate the times for both conversions using the derived formula:
1. For \(90\%\) conversion (\(X_A = 0.90\)):
\[ t_{90} = \frac{1}{k C_{A0}} \left( \frac{0.90}{1 - 0.90} \right) = \frac{1}{k C_{A0}} \left( \frac{0.90}{0.10} \right) = \frac{9}{k C_{A0}} \] 2. For \(45\%\) conversion (\(X_A = 0.45\)):
\[ t_{45} = \frac{1}{k C_{A0}} \left( \frac{0.45}{1 - 0.45} \right) = \frac{1}{k C_{A0}} \left( \frac{0.45}{0.55} \right) = \frac{1}{k C_{A0}} \left( \frac{9}{11} \right) \] 3. Now, find the ratio of \(t_{90}\) to \(t_{45}\):
\[ \frac{t_{90}}{t_{45}} = \frac{\frac{9}{k C_{A0}}}{\frac{9}{11 \cdot k C_{A0}}} \] Canceling the common term \(\frac{9}{k C_{A0}}\) from both the numerator and the denominator:
\[ \frac{t_{90}}{t_{45}} = 11 \] The ratio of the times is exactly 11, which matches Option (C).

Step 4: Final Answer
The ratio of the time required is 11, which corresponds to option (C).