Question

The rate of reaction:
\(\text{CH}_3\text{COOC}_2\text{H}_5 + \text{NaOH} \rightarrow \text{CH}_3\text{COONa} + \text{C}_2\text{H}_5\text{OH}\)
is given by the equation. Rate \(K[\text{CH}_2\text{COOC}_2\text{H}_5] [\text{NaOH}]\). If concentration is expressed in \(\text{mol L}^{-1}\). The unit of K is

• A. L mole^-1 s^-1
• B. S^-1
• C. mol^-2 L² s^-1
• D. mol L^-1 s^-1

Answer 1

The Correct Option is A

In the given rate equation:
\(\text{Rate} = k[\text{CH}_3\text{COOC}_2\text{H}_5][\text{NaOH}]\)

The rate constant (K) is determined by the units of the rate equation. Let's analyze the units:
\(\text{Rate} = k [\text{CH}_3\text{COOC}_2\text{H}_5] [\text{NaOH}]\)

Rate has units of \(\text{mol L}^{-1} \text{ s}^{-1}\) (since it is expressed as concentration change per unit time).
\([\text{CH}_3\text{COOC}_2\text{H}_5] \text{ has units of mol L}^{-1}\)
\([\text{NaOH}] \text{ also has units of mol L}^{-1}\)

By substituting the units into the rate equation, we have:
\(\text{mol L}^{-1} \text{ s}^{-1} = k \times (\text{mol L}^{-1}) \times (\text{mol L}^{-1})\)

To balance the units on both sides of the equation, K must have units of \(\text{L mol}^{-1} \text{ s}^{-1}\).
 Therefore, option (A) \(\text{L mole}^{-1} \text{ s}^{-1}\) is the correct unit for K.

Answer 2

The given rate law is:
\( \text{Rate} = K[\text{CH}_3\text{COOC}_2\text{H}_5][\text{NaOH}] \)
Since the rate law is of the form: Rate = K [A][B], this is a second-order reaction.

Units of Rate: mol L^-1 s^-1
Units of [A] and [B]: mol L^-1

So, plug into the rate equation:
\( \text{mol L}^{-1} \text{ s}^{-1} = K (\text{mol L}^{-1}) (\text{mol L}^{-1}) \)
\( K = \frac{\text{mol L}^{-1} \text{ s}^{-1}}{(\text{mol L}^{-1})^2} = \text{L mol}^{-1} \text{ s}^{-1} \)

Correct Answer: L mol^-1 s^-1