Reaction:`A+2B to D` follows the rate law: Rate =`k[A][B]^(2//3)` . The total order and unit of k will be :

Options

A. 1.30 `time^-1`
B. 1.67 `mol^(-2//3) lit^(2//3) time^-1`
C. 2,`mol lit^-1 time^-1`
D. None of these

Answer

To solve the question regarding the reaction \( A + 2B \to D \) that follows the rate law \( \text{Rate} = k[A][B]^{2/3} \), we need to determine the total order of the reaction and the unit of the rate constant \( k \).

### Step 1: Determine the Total Order of the Reaction
The total order of a reaction is the sum of the powers of the concentration terms in the rate law expression.

Given the rate law:
\[ \text{Rate} = k[A][B]^{2/3} \]

- The order with respect to \( A \) is 1 (since the concentration of \( A \) is raised to the power of 1).
- The order with respect to \( B \) is \( \frac{2}{3} \).

Now, we can calculate the total order:
\[ \text{Total Order} = 1 + \frac{2}{3} = \frac{3}{3} + \frac{2}{3} = \frac{5}{3} \]

### Step 2: Determine the Unit of the Rate Constant \( k \)
The unit of the rate constant \( k \) depends on the order of the reaction. The general formula for the unit of \( k \) for a reaction of order \( n \) is:
\[ \text{Unit of } k = \text{(concentration)}^{1-n} \cdot \text{(time)}^{-1} \]

In our case, the order \( n \) is \( \frac{5}{3} \). Therefore, we can substitute \( n \) into the formula:
\[ \text{Unit of } k = \text{(mole/L)}^{1 - \frac{5}{3}} \cdot \text{(time)}^{-1} \]
\[ = \text{(mole/L)}^{-\frac{2}{3}} \cdot \text{(time)}^{-1} \]

This can be rewritten as:
\[ = \frac{1}{\text{(mole)}^{\frac{2}{3}} \cdot \text{(L)}^{\frac{2}{3}}} \cdot \text{(time)}^{-1} \]

Thus, the unit of \( k \) is:
\[ \text{Unit of } k = \text{mole}^{-\frac{2}{3}} \cdot \text{L}^{\frac{2}{3}} \cdot \text{time}^{-1} \]

### Final Answer
- Total Order: \( \frac{5}{3} \) or approximately \( 1.67 \)
- Unit of \( k \): \( \text{mole}^{-\frac{2}{3}} \cdot \text{L}^{\frac{2}{3}} \cdot \text{time}^{-1} \)