Understanding Rate Constant Units and Reaction Order

The rate constant ($k$) is a proportionality constant in the rate law of a chemical reaction. Its units depend on the overall order of the reaction. Understanding these units is key to determining the reaction order.

General Formula for Rate Constant Units

The general formula for the units of a rate constant for a reaction of overall order $n$ is given by:

\(\text{Units of } k = (\text{Concentration})^{1-n} (\text{Time})^{-1}\)

Using the common units for concentration (mol L\({}^{-1}\)) and time (sec\({}^{-1}\)), the formula becomes:

\(\text{Units of } k = (\text{mol L}^{-1})^{1-n} (\text{sec})^{-1}\)

\(\text{Units of } k = \text{mol}^{1-n} \text{L}^{-(1-n)} \text{sec}^{-1}\)

\(\text{Units of } k = \text{mol}^{1-n} \text{L}^{n-1} \text{sec}^{-1}\)

We can use this general formula to find the order ($n$) for each given unit by comparing the exponents of mol, L, and sec.

Determining Reaction Order from Rate Constant Units

Let's analyze each given rate constant unit:

(A) sec\({}^{-1}\)

Comparing with \(\text{mol}^{1-n} \text{L}^{n-1} \text{sec}^{-1}\):

• The exponent of sec is -1, which matches the general formula.
• There are no mol or L terms, which means their exponents must be zero.
• For mol: \(1-n = 0 \implies n = 1\)
• For L: \(n-1 = 0 \implies n = 1\)

Thus, the unit sec\({}^{-1}\) corresponds to a 1st-order reaction.

(B) mol L\({}^{-1}\) sec\({}^{-1}\)

Comparing with \(\text{mol}^{1-n} \text{L}^{n-1} \text{sec}^{-1}\):

• The exponent of sec is -1, which matches.
• For mol: \(1-n = 1 \implies n = 1 - 1 = 0\)
• For L: \(n-1 = -1 \implies n = -1 + 1 = 0\)

Thus, the unit mol L\({}^{-1}\) sec\({}^{-1}\) corresponds to a 0th-order reaction.

(C) mol\({}^{-1}\) L sec\({}^{-1}\)

Comparing with \(\text{mol}^{1-n} \text{L}^{n-1} \text{sec}^{-1}\):

• The exponent of sec is -1, which matches.
• For mol: \(1-n = -1 \implies n = 1 - (-1) = 2\)
• For L: \(n-1 = 1 \implies n = 1 + 1 = 2\)

Thus, the unit mol\({}^{-1}\) L sec\({}^{-1}\) corresponds to a 2nd-order reaction.

(D) mol\({}^{-2}\) L\({}^{2}\) sec\({}^{-1}\)

Comparing with \(\text{mol}^{1-n} \text{L}^{n-1} \text{sec}^{-1}\):

• The exponent of sec is -1, which matches.
• For mol: \(1-n = -2 \implies n = 1 - (-2) = 3\)
• For L: \(n-1 = 2 \implies n = 2 + 1 = 3\)

Thus, the unit mol\({}^{-2}\) L\({}^{2}\) sec\({}^{-1}\) corresponds to a 3rd-order reaction.

Summary of Rate Constants and Orders

Arranging Units by Increasing Reaction Order

We need to arrange the units (A), (B), (C), (D) in increasing order of their reaction order. The orders we found are:

• (B): Order 0
• (A): Order 1
• (C): Order 2
• (D): Order 3

Arranging these orders from smallest to largest: 0 < 1 < 2 < 3.

Corresponding to the units, the increasing order is:

(B) < (A) < (C) < (D)

Revision Table: Chemical Kinetics Rate Constant Units

Additional Information: Chemical Kinetics and Rate Laws

Chemical kinetics is the study of reaction rates. The rate law of a reaction expresses the relationship between the reaction rate and the concentrations of reactants. For a general reaction aA + bB → products, the rate law is often written as:

\(\text{Rate} = k[\text{A}]^x[\text{B}]^y\)

Here:

• Rate is the reaction rate (e.g., change in concentration per unit time, typically mol L\({}^{-1}\) sec\({}^{-1}\)).
• k is the rate constant.
• [A] and [B] are the molar concentrations of reactants A and B.
• x is the order of the reaction with respect to reactant A.
• y is the order of the reaction with respect to reactant B.
• The overall order of the reaction is \(n = x + y\).

The orders x and y are determined experimentally and are not necessarily equal to the stoichiometric coefficients a and b. The units of the rate constant k must be such that the overall rate has units of concentration/time.

By analyzing the units of k, we can directly deduce the overall order of the reaction without knowing the specific rate law expression, as demonstrated in this problem.