If the rate law of a reaction `n(A) to B ` is expressed as
`underset("mol/L/s")(Rate)=-1/n (d[A])/(dt) =+(d[B])/(dt)=k[A]^(x)`
The unit of the rate constant will be

Options

A. `mol^(x)//L^(x)//s`
B. `L^(x)//mol^(x)//s`
C. `mol^((1-x)).L^((x-1)).s^(-1)`
D. `mol^((x-1)).L^((1-x)).s^(-1)`

Answer

To determine the unit of the rate constant \( k \) for the reaction \( n(A) \rightarrow B \) with the rate law expressed as:

\[ \text{Rate} = -\frac{1}{n} \frac{d[A]}{dt} = \frac{d[B]}{dt} = k[A]^x \]

we can follow these steps:

### Step 1: Understand the Rate Expression
The rate of the reaction is given in the form of:

\[ \text{Rate} = k[A]^x \]

where:
- \(\text{Rate}\) has units of \(\text{mol/L/s}\)
- \([A]\) is the concentration of reactant A with units of \(\text{mol/L}\)
- \(k\) is the rate constant we need to find
- \(x\) is the order of the reaction with respect to A

### Step 2: Rearranging the Rate Law
We can rearrange the rate law to isolate \( k \):

\[ k = \frac{\text{Rate}}{[A]^x} \]

### Step 3: Substitute the Units
Now, substituting the units into the equation:

\[ \text{Rate} = \text{mol/L/s} \]
\[ [A] = \text{mol/L} \]

Substituting these into the expression for \( k \):

\[ k = \frac{\text{mol/L/s}}{(\text{mol/L})^x} \]

### Step 4: Simplifying the Units
Now we simplify the units for \( k \):

\[ k = \frac{\text{mol/L/s}}{(\text{mol}^x/\text{L}^x)} = \frac{\text{mol} \cdot \text{L}^x}{\text{L} \cdot \text{s} \cdot \text{mol}^x} \]

This can be rewritten as:

\[ k = \frac{\text{mol}^{1-x}}{\text{L}^{1-x} \cdot \text{s}} \]

### Step 5: Final Unit of the Rate Constant
Thus, the unit of the rate constant \( k \) can be expressed as:

\[ k = \text{mol}^{1-x} \cdot \text{L}^{x-1} \cdot \text{s}^{-1} \]

### Conclusion
The unit of the rate constant \( k \) depends on the order of the reaction \( x \). For a reaction of order \( x \), the unit of \( k \) is:

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