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 `mole^(-2//3) lit^(2//3) time^-1`
C. 2,`mole lit^-1 time^-1`
D. None of these

Answer

To solve the question regarding the reaction \( A + 2B \rightarrow D \) with the rate law given as \( \text{Rate} = k[A][B]^{\frac{2}{3}} \), we need to determine the total order of the reaction and the units of the rate constant \( k \).

### Step-by-Step Solution:

1. **Identify the Rate Law**:
The rate law is given as:
\[ \text{Rate} = k[A][B]^{\frac{2}{3}} \]

2. **Determine the Order with Respect to Each Reactant**:
- The order with respect to \( A \) is \( 1 \) (since the exponent of \( [A] \) is \( 1 \)).
- The order with respect to \( B \) is \( \frac{2}{3} \) (since the exponent of \( [B] \) is \( \frac{2}{3} \)).

3. **Calculate the Total Order of the Reaction**:
The total order of the reaction is the sum of the orders with respect to each reactant:
\[ \text{Total Order} = 1 + \frac{2}{3} = \frac{3}{3} + \frac{2}{3} = \frac{5}{3} \]
Thus, the total order of the reaction is \( \frac{5}{3} \) or approximately \( 1.67 \).

4. **Determine the Units of the Rate Constant \( k \)**:
The general formula for the units of the rate constant \( k \) for an \( n \)-th order reaction is:
\[ [k] = \text{mol}^{1-n} \cdot \text{L}^{n-1} \cdot \text{s}^{-1} \]
Here, \( n \) is the total order of the reaction, which we found to be \( \frac{5}{3} \).

5. **Substituting the Total Order into the Units Formula**:
Substitute \( n = \frac{5}{3} \) into the units formula:
\[ [k] = \text{mol}^{1 - \frac{5}{3}} \cdot \text{L}^{\frac{5}{3} - 1} \cdot \text{s}^{-1} \]
Simplifying this gives:
\[ [k] = \text{mol}^{-\frac{2}{3}} \cdot \text{L}^{\frac{2}{3}} \cdot \text{s}^{-1} \]

6. **Final Answer**:
- Total Order: \( \frac{5}{3} \) or approximately \( 1.67 \)
- Units of \( k \): \( \text{mol}^{-\frac{2}{3}} \cdot \text{L}^{\frac{2}{3}} \cdot \text{s}^{-1} \)