For a chemical reaction `A rarr` Products, the rate of disappearance of `A` is given by:
`-(dC_(A))/(dt)=(K_(1)C_(A))/(1+K_(2)C_(A))` At low `C_(A)`, the reaction is of the …. Order with rate constant …..: `("Assume" K_(1), K_(2) "are lesser than" 1)`

Options

A. `I,(K_(1))/(K_(2))`
B. `I,K_(1)`
C. `II,K_(1)//K_(2)`
D. `II,(K_(1))/(K_(1)+K_(2))`

Explanation

To solve the problem, we need to analyze the given rate equation and determine the order of the reaction and the rate constant at low concentrations of A.

### Step-by-Step Solution:

1. **Understanding the Rate Equation**:
The rate of disappearance of A is given by:
\[ -\frac{dC_A}{dt} = \frac{K_1 C_A}{1 + K_2 C_A} \]
Here, \(C_A\) is the concentration of A, and \(K_1\) and \(K_2\) are constants.

2. **Assuming Low Concentration of A**:
We are told to assume that \(C_A\) is low. Under this condition, \(K_2 C_A\) will also be very small because \(K_2\) is less than 1. Therefore, we can approximate:
\[ 1 + K_2 C_A \approx 1 \]

3. **Simplifying the Rate Equation**:
Substituting this approximation into the rate equation gives:
\[ -\frac{dC_A}{dt} \approx \frac{K_1 C_A}{1} = K_1 C_A \]

4. **Identifying the Order of Reaction**:
The simplified equation can be rewritten as:
\[ -\frac{dC_A}{dt} = K_1 C_A \]
This is in the form of a first-order reaction, which is generally expressed as:
\[ \text{Rate} = k C_A^n \]
where \(n\) is the order of the reaction. Here, we can see that \(n = 1\).

5. **Determining the Rate Constant**:
From the simplified equation, we can identify that the rate constant \(k\) is equal to \(K_1\):
\[ k = K_1 \]

### Final Answer:
At low \(C_A\), the reaction is of **first order** with rate constant **\(K_1\)**.