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

Answer: (A)

L mole-1 s-1

Answer 2

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.