For the reaction : `aA to bB`, it is given that
`log[(-dA)/(dt)] = log [(dB)/(dt)] + 0.6020`. What is `a : b` is ?

Options

A. 4
B. 2
C. 1.5
D. 0

Answer

To solve the problem, we need to analyze the given equation and relate it to the stoichiometry of the reaction. Let's break it down step by step.

### Step 1: Understand the given equation
The reaction is given as:
\[ aA \rightarrow bB \]
We are provided with the equation:
\[ \log\left(-\frac{dA}{dt}\right) = \log\left(\frac{dB}{dt}\right) + 0.6020 \]

### Step 2: Relate the rates of change
From the stoichiometry of the reaction, we know that:
\[ -\frac{dA}{dt} = \frac{a}{b} \frac{dB}{dt} \]
This means that the rate of consumption of A is related to the rate of formation of B by their stoichiometric coefficients.

### Step 3: Take logarithms
Taking logarithms on both sides of the rate equation:
\[ \log\left(-\frac{dA}{dt}\right) = \log\left(\frac{a}{b}\right) + \log\left(\frac{dB}{dt}\right) \]

### Step 4: Compare with the given equation
Now, we compare this with the given equation:
\[ \log\left(-\frac{dA}{dt}\right) = \log\left(\frac{dB}{dt}\right) + 0.6020 \]

From this, we can equate the two expressions:
\[ \log\left(\frac{a}{b}\right) + \log\left(\frac{dB}{dt}\right) = \log\left(\frac{dB}{dt}\right) + 0.6020 \]

### Step 5: Simplify the equation
By simplifying, we can eliminate \(\log\left(\frac{dB}{dt}\right)\) from both sides:
\[ \log\left(\frac{a}{b}\right) = 0.6020 \]

### Step 6: Exponentiate to solve for \(\frac{a}{b}\)
To find \(\frac{a}{b}\), we exponentiate both sides:
\[ \frac{a}{b} = 10^{0.6020} \]

### Step 7: Calculate the value
Using a calculator:
\[ 10^{0.6020} \approx 4 \]

### Conclusion
Thus, the ratio \(a : b\) is:
\[ a : b = 4 : 1 \]