Consider following two reactions
`A to "Product", =(d[A])/(dt) =-k-(1)[A]^(0)`
`B to "Product", -(d[B])/(dt) = k_(2)[B]`
`k_(1)` and `k_(2)` are expressed in terms of molarity `("mol" L^(-1)` and time `s^(-1)`) as

Options

A. `s^(-1),Ms^(-1)L^(-1)`
B. `Ms^(-1),Ms^(-1)`
C. `s^(-1),M^(-1)s^(-1)`
D. `Ms^(-1),s^(-1)`

Answer

To solve the problem, we need to analyze the two given reactions and determine the units of the rate constants \( k_1 \) and \( k_2 \).

### Step-by-Step Solution:

1. **Identify the Rate Expression for Reaction A:**
The rate of the reaction for A is given by:
\[ -\frac{d[A]}{dt} = k_1 [A]^0 \]
Since \([A]^0 = 1\), we can simplify this to:
\[ -\frac{d[A]}{dt} = k_1 \]

2. **Determine the Units for \( k_1 \):**
The rate of reaction (left side) has units of concentration change over time. Therefore, the units for \(-\frac{d[A]}{dt}\) are:
\[ \text{mol L}^{-1} \text{s}^{-1} \]
Since \( k_1 \) is equal to the rate, it must also have the same units:
\[ [k_1] = \text{mol L}^{-1} \text{s}^{-1} \]

3. **Identify the Rate Expression for Reaction B:**
The rate of the reaction for B is given by:
\[ -\frac{d[B]}{dt} = k_2 [B] \]

4. **Determine the Units for \( k_2 \):**
The left side \(-\frac{d[B]}{dt}\) again has units of concentration change over time:
\[ \text{mol L}^{-1} \text{s}^{-1} \]
The concentration \([B]\) has units of:
\[ \text{mol L}^{-1} \]
Therefore, we can express the units for \( k_2 \) as follows:
\[ \text{mol L}^{-1} \text{s}^{-1} = k_2 \cdot \text{mol L}^{-1} \]
Rearranging gives:
\[ k_2 = \frac{\text{mol L}^{-1} \text{s}^{-1}}{\text{mol L}^{-1}} = \text{s}^{-1} \]

### Final Units:
- The unit of \( k_1 \) is:
\[ [k_1] = \text{mol L}^{-1} \text{s}^{-1} \]
- The unit of \( k_2 \) is:
\[ [k_2] = \text{s}^{-1} \]