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. `2C_(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 problem, we start with the given equation:

\[ \log_{10}\left[-\frac{d[A]}{dt}\right] = \log_{10}\left[\frac{d[B]}{dt}\right] + 0.3010 \]

### Step 1: Understand the logarithmic relationship
The equation can be interpreted using properties of logarithms. The term \(0.3010\) is equivalent to \(\log_{10}(2)\), since \(10^{0.3010} \approx 2\). Therefore, we can rewrite the equation as:

\[ \log_{10}\left[-\frac{d[A]}{dt}\right] = \log_{10}\left[\frac{d[B]}{dt}\right] + \log_{10}(2) \]

### Step 2: Use properties of logarithms
Using the property of logarithms that states \(\log(a) + \log(b) = \log(ab)\), we can combine the right-hand side:

\[ \log_{10}\left[-\frac{d[A]}{dt}\right] = \log_{10}\left[2 \cdot \frac{d[B]}{dt}\right] \]

### Step 3: Remove the logarithm
By taking the antilogarithm of both sides, we can eliminate the logarithm:

\[ -\frac{d[A]}{dt} = 2 \cdot \frac{d[B]}{dt} \]

### Step 4: Rearranging the equation
Rearranging this equation gives us:

\[ \frac{d[A]}{dt} = -2 \cdot \frac{d[B]}{dt} \]

### Step 5: Interpret the stoichiometry
This relationship indicates that for every 2 moles of \(B\) produced, 1 mole of \(A\) is consumed. This suggests a stoichiometric ratio of:

\[ A \rightarrow 2B \]

### Conclusion
Thus, the chemical reaction can be represented as:

\[ 2A \rightarrow B \]

From the options given, we can conclude that the correct answer is that \(A\) and \(B\) are related by a stoichiometric coefficient of 2 for \(A\) and 1 for \(B\).