In the reaction `x A rarr yB, log{-(d[A])/(dt)}=log{+(d[B])/(dt)}+0.3` Then, `x:y` is

Options

A. `2:1`
B. `1:2`
C. `3:1`
D. `3:10`

Answer

To solve the problem, we need to analyze the given equation and derive the ratio of the stoichiometric coefficients \( x \) and \( y \) in the reaction \( x A \rightarrow y B \).

### Step-by-Step Solution:

1. **Understanding the Reaction**:
The reaction is given as \( x A \rightarrow y B \). Here, \( x \) is the stoichiometric coefficient of reactant \( A \) and \( y \) is the stoichiometric coefficient of product \( B \).

2. **Rate of Reaction**:
The rate of reaction can be expressed as:
\[ -\frac{1}{x} \frac{d[A]}{dt} = \frac{1}{y} \frac{d[B]}{dt} \]
This equation indicates that the rate of decrease of \( A \) is related to the rate of increase of \( B \).

3. **Using the Given Equation**:
We are given:
\[ \log\left(-\frac{d[A]}{dt}\right) = \log\left(\frac{d[B]}{dt}\right) + 0.3 \]
Let's denote:
\[ R_A = -\frac{d[A]}{dt} \quad \text{and} \quad R_B = \frac{d[B]}{dt} \]
Thus, we can rewrite the equation as:
\[ \log(R_A) = \log(R_B) + 0.3 \]

4. **Exponentiating Both Sides**:
By exponentiating both sides, we can eliminate the logarithm:
\[ R_A = R_B \cdot 10^{0.3} \]

5. **Substituting the Rates**:
From the rate equation, we know:
\[ R_A = \frac{y}{x} R_B \]
Setting the two expressions for \( R_A \) equal gives:
\[ \frac{y}{x} R_B = R_B \cdot 10^{0.3} \]

6. **Canceling \( R_B \)**:
Assuming \( R_B \neq 0 \), we can cancel \( R_B \) from both sides:
\[ \frac{y}{x} = 10^{0.3} \]

7. **Finding the Ratio \( x:y \)**:
Rearranging gives:
\[ \frac{x}{y} = \frac{1}{10^{0.3}} \approx \frac{1}{2} \]
Therefore, the ratio \( x:y \) is:
\[ x:y = 1:2 \]

### Final Answer:
The ratio \( x:y \) is \( 1:2 \).