The unit of rate constant for the reaction,
`2H_(2) + 2NO to 2H_(2)O + N_(2)`
Which has rate `= k [H_(2))[NO]^(2)`, is

Options

A. mol `L^(-1) s^(-1)`,
B. `S^(-1)`
C. `mol^(-2) L^(2) s^(-1)`
D. mol `L^(-1)`

Explanation

To determine the unit of the rate constant (k) for the reaction:

\[ 2H_2 + 2NO \rightarrow 2H_2O + N_2 \]

with the rate law given as:

\[ \text{Rate} = k [H_2][NO]^2 \]

we will follow these steps:

### Step 1: Identify the overall order of the reaction
The overall order of a reaction is determined by summing the powers of the concentration terms in the rate law expression.

From the rate law:
- The concentration of \( H_2 \) is raised to the power of 1.
- The concentration of \( NO \) is raised to the power of 2.

Thus, the overall order \( n \) is calculated as follows:

\[ n = 1 + 2 = 3 \]

### Step 2: Write the expression for the unit of the rate constant
The unit of the rate constant \( k \) can be expressed in terms of concentration and time. The general formula for the unit of the rate constant is:

\[ \text{Unit of } k = \text{(Unit of Rate)} \div \text{(Unit of Concentration)}^n \]

Where:
- The unit of rate is typically expressed in terms of molarity per second (mol L\(^{-1}\) s\(^{-1}\)).
- The unit of concentration is molarity (mol L\(^{-1}\)).

### Step 3: Substitute the values into the expression
Now, substituting the known values into the expression for the unit of \( k \):

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

Since \( n = 3 \):

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

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

\[ \text{Unit of } k = \frac{\text{mol L}^{-1} \text{s}^{-1}}{\text{mol}^3 \text{L}^{-3}} = \text{mol}^{-2} \text{L}^{2} \text{s}^{-1} \]

### Conclusion
Thus, the unit of the rate constant \( k \) for the given reaction is:

\[ \text{Unit of } k = \text{mol}^{-2} \text{L}^{2} \text{s}^{-1} \]