The rate of reaction `A+Bto` Products is given by Rate `=k[A]^(1//2)[B]^(2)`. What are the units of the rate constant ?

Explanation

To determine the units of the rate constant \( k \) for the reaction \( A + B \to \text{Products} \) with the rate expression given by:

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

we will follow these steps:

### Step 1: Write down the rate expression
The rate of the reaction is given by the equation:

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

### Step 2: Identify the units of the rate
The rate of a reaction is typically expressed in terms of concentration change over time. Therefore, the units of the rate are:

\[ \text{Rate} = \frac{\text{moles}}{\text{liter} \cdot \text{second}} = \text{mol L}^{-1} \text{s}^{-1} \]

### Step 3: Identify the units of concentration
The concentration of reactants \( [A] \) and \( [B] \) is expressed in moles per liter:

\[ [A] = \text{mol L}^{-1} \]
\[ [B] = \text{mol L}^{-1} \]

### Step 4: Substitute the units into the rate expression
Substituting the units of \( [A] \) and \( [B] \) into the rate expression:

\[ \text{Rate} = k \cdot ([A]^{1/2}) \cdot ([B]^{2}) \]

Substituting the units:

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

### Step 5: Simplify the right side
Calculating the right side:

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

Now, multiplying these together:

\[ \text{mol}^{1/2} \cdot \text{L}^{-1/2} \cdot \text{mol}^{2} \cdot \text{L}^{-2} = \text{mol}^{2 + 1/2} \cdot \text{L}^{-1/2 - 2} = \text{mol}^{5/2} \cdot \text{L}^{-5/2} \]

### Step 6: Set up the equation for \( k \)
Now we have:

\[ \text{mol L}^{-1} \text{s}^{-1} = k \cdot \text{mol}^{5/2} \cdot \text{L}^{-5/2} \]

### Step 7: Solve for \( k \)
Rearranging to solve for \( k \):

\[ k = \frac{\text{mol L}^{-1} \text{s}^{-1}}{\text{mol}^{5/2} \cdot \text{L}^{-5/2}} = \text{mol}^{-5/2} \cdot \text{L}^{5/2} \cdot \text{mol L}^{-1} \cdot \text{s}^{-1} \]

This simplifies to:

\[ k = \text{mol}^{-3/2} \cdot \text{L}^{3/2} \cdot \text{s}^{-1} \]

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

\[ k = \text{mol}^{-3/2} \cdot \text{L}^{3/2} \cdot \text{s}^{-1} \]

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