For the reaction: `aA + bB rarr cC+dD`
Rate `= (dx)/(dt) = (-1)/(a)(d[A])/(dt) = (-1)/(b)(d[B])/(dt) = (1)/( c)(d[C])/(dt) = (1)/(d)(d[D])/(dt)`
In the following reaction,
`xA rarr yB`
`log.[-(d[A])/(dt)] = log.[(d[B])/(dt)] + 0.3`
where negative isgn indicates rate of disappearance of the reactant. Thus, `x:y` is:

Options

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

Explanation

To solve the problem, we need to analyze the given reaction and the logarithmic relationship provided. Let's break it down step by step.

### Step 1: Understand the Reaction
We have a reaction represented as:
\[ xA \rightarrow yB \]
This means that \( x \) moles of reactant \( A \) produce \( y \) moles of product \( B \).

### Step 2: Write the Rate Expressions
From the reaction, we can express the rates of change of concentrations:
- The rate of disappearance of \( A \) is given by:
\[ -\frac{1}{x} \frac{d[A]}{dt} \]
- The rate of appearance of \( B \) is given by:
\[ \frac{1}{y} \frac{d[B]}{dt} \]

### Step 3: Set Up the Logarithmic Equation
According to the problem, we have:
\[ \log\left(-\frac{d[A]}{dt}\right) = \log\left(\frac{d[B]}{dt}\right) + 0.3 \]

### Step 4: Simplify the Logarithmic Equation
Using properties of logarithms, we can rewrite the equation:
\[ \log\left(-\frac{d[A]}{dt}\right) - \log\left(\frac{d[B]}{dt}\right) = 0.3 \]
This can be simplified to:
\[ \log\left(\frac{-\frac{d[A]}{dt}}{\frac{d[B]}{dt}}\right) = 0.3 \]

### Step 5: Convert Logarithm to Exponential Form
From the logarithmic equation, we can convert it to its exponential form:
\[ \frac{-\frac{d[A]}{dt}}{\frac{d[B]}{dt}} = 10^{0.3} \]
Calculating \( 10^{0.3} \):
\[ 10^{0.3} \approx 2 \]

### Step 6: Relate the Rates to Stoichiometric Coefficients
From the rate expressions we set up earlier:
\[ \frac{-\frac{d[A]}{dt}}{\frac{d[B]}{dt}} = \frac{x}{y} \]
Thus, we have:
\[ \frac{x}{y} = 2 \]

### Step 7: Determine the Ratio \( x:y \)
From the equation \( \frac{x}{y} = 2 \), we can express this as:
\[ x:y = 2:1 \]

### Final Answer
Thus, the ratio \( x:y \) is \( 2:1 \).

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