Question

A sample of glucose was decomposed at 140°C in a solution containing 0.030 M HCl. The velocity constant k, was found to be 0.0080 hr^–1. If the spontaneous rate constant, is 0.0010 hr^–1 and the catalysis due to hydroxyl ions in this acidic solution is considered

• A. 0.22 per mole per hour
• B. 0.233 per mole per hour
• C. 0.27 per mole per hour
• D. 0.29 per mole per hour

Answer 1

The Correct Option is B

To determine the catalysis rate constant due to hydroxyl ions, we need to use the given data and apply the concept of catalysis in chemical kinetics. The given data includes: 

• Total rate constant, \( k = 0.0080 \, \text{hr}^{-1} \)
• Spontaneous rate constant, \( k_0 = 0.0010 \, \text{hr}^{-1} \)
• Concentration of HCl which is a measure of H⁺ ions, \( [\text{H}^+] = 0.030 \, \text{M} \)

The reaction in the presence of acid involves three components:

1. The spontaneous decomposing component, characterized by \( k_0 \).
2. The catalysis by hydrogen ions \( (\text{H}^+) \), which is not considered here explicitly since we focus on hydroxyl ions.
3. The catalysis by hydroxyl ions \( (\text{OH}^-) \).

Since the reaction takes place in an acidic solution, \( [\text{OH}^-] \) is negligible compared to \( [\text{H}^+] \). However, in constructing the relationship, we consider the effective rate constant formula:

\(k = k_0 + k_{\text{catalysis}} [\text{H}^+]\)

We rearrange to find the catalysis rate constant \( k_{\text{catalysis}} \):

\(k_{\text{catalysis}} = \frac{k - k_0}{[\text{H}^+]}\)

Substitute the given values into the equation:

\(k_{\text{catalysis}} = \frac{0.0080 \, \text{hr}^{-1} - 0.0010 \, \text{hr}^{-1}}{0.030 \, \text{M}}\)

Perform the calculation:

\(k_{\text{catalysis}} = \frac{0.0070 \, \text{hr}^{-1}}{0.030 \, \text{M}} = 0.233 \, \text{per mole per hour}\)

Therefore, the catalysis rate constant due to hydroxyl ions in this acidic solution is 0.233 per mole per hour.