The rate of the reaction `A to` Products is given by `(dx)/(dt)=k[A]^(1//2)`. The units of the rate constants are:

Options

A. `lit mol^-1 sec^-1`
B. `lit^(1//2) mol^(-1//2) sec^-1`
C. `lit^(-1//2) mol^(1//2) sec^-1`
D. `lit^(3//2) mol^(-1//2) sec^-1`

Answer

To find the units of the rate constant \( k \) for the reaction given by the rate equation:

\[ \frac{dx}{dt} = k[A]^{1/2} \]

we will follow these steps:

### Step 1: Understand the Rate Equation
The rate of a reaction is defined as the change in concentration of a reactant or product over time. In this case, the rate is given as \( \frac{dx}{dt} \), which has units of concentration per time.

### Step 2: Determine the Units of Rate
The concentration \( [A] \) is typically expressed in moles per liter (mol/L). Therefore, the units of the rate \( \frac{dx}{dt} \) can be expressed as:

\[ \text{Units of rate} = \frac{\text{moles}}{\text{liters} \cdot \text{seconds}} = \text{mol L}^{-1} \text{s}^{-1} \]

### Step 3: Substitute into the Rate Equation
From the rate equation, we have:

\[ \frac{dx}{dt} = k[A]^{1/2} \]

Substituting the units into the equation gives:

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

### Step 4: Determine the Units of Concentration
Since \( [A] \) has units of concentration (mol/L), we can express \( [A]^{1/2} \) as:

\[ [A]^{1/2} = ( \text{mol L}^{-1} )^{1/2} = \text{mol}^{1/2} \text{L}^{-1/2} \]

### Step 5: Rearranging to Solve for \( k \)
Now, we can rearrange the equation to solve for \( k \):

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

### Step 6: Simplifying the Units
Now we simplify the units of \( k \):

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

### Final Result
Thus, the units of the rate constant \( k \) are:

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