Conisder the following elementary reaction,
`2A + B + C rarr` Products. All reactants are present in the gaseous state and reactant `C` is taken in excess.
What is the unit of rate constant of the reaction?

Options

A. `mol^(-1) time^(-1)`
B. `time^(-1)`
C. `mol^(-1) L^(2) time^(-1)`
D. `mol^(-2) L^(2) time^(-1)`

Explanation

To determine the unit of the rate constant (k) for the given reaction \(2A + B + C \rightarrow \text{Products}\), we can follow these steps:

### Step 1: Write the Rate Law Expression
For the elementary reaction, the rate law can be expressed based on the stoichiometry of the reactants. Since reactant C is in excess, it does not affect the rate of the reaction. Therefore, the rate law is given by:

\[ \text{Rate} = k [A]^2 [B]^1 \]

### Step 2: Determine the Overall Order of the Reaction
The overall order of the reaction is the sum of the powers of the concentration terms in the rate law. Here, the order with respect to A is 2 and with respect to B is 1. Therefore, the total order (n) of the reaction is:

\[ n = 2 + 1 = 3 \]

### Step 3: Use the Formula for the Unit of the Rate Constant
The unit of the rate constant (k) can be derived from the general formula:

\[ \text{Unit of } k = \text{(concentration)}^{1-n} \cdot \text{(time)}^{-1} \]

In this case, concentration is typically expressed in moles per liter (mol/L), and time is in seconds (s). Therefore, we substitute n = 3 into the formula:

\[ \text{Unit of } k = \text{(mol/L)}^{1-3} \cdot \text{s}^{-1} \]

### Step 4: Simplify the Expression
Now we simplify the expression:

\[ \text{Unit of } k = \text{(mol/L)}^{-2} \cdot \text{s}^{-1} = \frac{1}{\text{mol}^2/\text{L}^2} \cdot \frac{1}{\text{s}} = \frac{\text{L}^2}{\text{mol}^2 \cdot \text{s}} \]

### Final Answer
Thus, the unit of the rate constant (k) for the reaction \(2A + B + C \rightarrow \text{Products}\) is:

\[ \text{L}^2 \cdot \text{mol}^{-2} \cdot \text{s}^{-1} \]