Question

Rate constant \( k = 4.5 \times 10^{-2}\ \mathrm{L^2\ mol^{-2}\ s^{-1}} \), then what is the order of reaction?

• A. 0
• B. 3
• C. 2
• D. 4

Hint:

Memorize these common units of rate constant: zero order \(=\mathrm{mol\ L^{-1}\ s^{-1}}\), first order \(=\mathrm{s^{-1}}\), second order \(=\mathrm{L\ mol^{-1}\ s^{-1}}\), and third order \(=\mathrm{L^2\ mol^{-2}\ s^{-1}}\).

Answer 1

The Correct Option is B

Step 1: Recall the unit of rate constant for \(n\)-th order reaction.
For an \(n\)-th order reaction, the unit of rate constant is given by
\[ \mathrm{(concentration)^{1-n}\ time^{-1}} \] If concentration is expressed in \( \mathrm{mol\ L^{-1}} \), then the unit becomes
\[ \mathrm{L^{n-1}\ mol^{1-n}\ s^{-1}} \] Step 2: Compare with the given unit.
Given unit of rate constant is
\[ \mathrm{L^2\ mol^{-2}\ s^{-1}} \] Now compare this with the general form
\[ \mathrm{L^{n-1}\ mol^{1-n}\ s^{-1}} \] So,
\[ n - 1 = 2 \] \[ n = 3 \] Also,
\[ 1 - n = -2 \] which again gives
\[ n = 3 \] Step 3: Conclusion.
Therefore, the reaction is of third order.
Final Answer:3.