In the following reaction, `xA rarryB`
`log_(10)[-(d[A])/(dt)]=log_(10)[(d([B]))/(dt)]+0.3010` ‘A’ and ‘B’ respectively can be:

Options

A. `C_(2)H_(2) and C_(6)H_(6)`
B. `C_(2)H_(4) and C_(4)H_(8)`
C. `N_(2)O_(4) and NO_(2)`
D. n - Butane and iso - butane

Explanation

To solve the given problem, we need to analyze the relationship between the rates of change of concentrations of reactants and products in a chemical reaction, as described by the logarithmic equation provided.

### Step-by-Step Solution:

1. **Understanding the Given Equation**:
The equation provided is:
\[ \log_{10}\left(-\frac{d[A]}{dt}\right) = \log_{10}\left(\frac{d[B]}{dt}\right) + 0.3010 \]
Here, \(-\frac{d[A]}{dt}\) represents the rate of decrease of concentration of A, and \(\frac{d[B]}{dt}\) represents the rate of increase of concentration of B.

2. **Rewriting the Constant**:
The constant \(0.3010\) can be rewritten as:
\[ 0.3010 = \log_{10}(2) \]
This means we can express the equation as:
\[ \log_{10}\left(-\frac{d[A]}{dt}\right) = \log_{10}\left(\frac{d[B]}{dt} \cdot 2\right) \]

3. **Using Logarithmic Properties**:
By applying the property of logarithms that states \(\log_{10}(m) = \log_{10}(n)\) implies \(m = n\), we can equate the arguments:
\[ -\frac{d[A]}{dt} = 2 \cdot \frac{d[B]}{dt} \]

4. **Rearranging the Equation**:
Rearranging this gives us:
\[ \frac{d[A]}{dt} = -2 \cdot \frac{d[B]}{dt} \]
This indicates that for every 2 moles of B produced, 1 mole of A is consumed.

5. **Identifying the Stoichiometry**:
From the equation, we can deduce the stoichiometry of the reaction:
\[ xA \rightarrow yB \quad \text{where } x = 2 \text{ and } y = 1 \]
Therefore, the reaction can be represented as:
\[ 2A \rightarrow B \]

6. **Evaluating the Options**:
Now we need to check the options provided to see which reaction fits the stoichiometry \(2A \rightarrow B\).

- **Option 1**: \(C_2H_2 \rightarrow C_6H_6\) (not correct)
- **Option 2**: \(2C_2H_4 \rightarrow C_4H_8\) (correct)
- **Option 3**: \(N_2O_4 \rightarrow NO_2\) (not correct)
- **Option 4**: \(n\)-butane and iso-butane (not correct)

The only option that fits the stoichiometry \(2A \rightarrow B\) is **Option 2**.

### Final Answer:
The correct answer is **Option 2: \(2C_2H_4 \rightarrow C_4H_8\)**.