In the following reaction:

\(ax \rightarrow by\) \(log_{10}\left(\frac{dx}{dt}\right) = log_{10}\left(\frac{dy}{dt}\right) + 0.4771\)

\(X\) and \(Y\) respectively can be :

[Given: log 3 = 0.4771]

\(C_2H_2\) and \(C_6H_6\)

The correct answer is option 1. \(C_2H_2\) and \(C_6H_6\).

Given,

\(log_{10}\left(\frac{dx}{dt}\right) = log_{10}\left(\frac{dy}{dt}\right) + 0.4771\)

or, \(log_{10}\left(\frac{dx}{dt}\right) = log_{10}\left(\frac{dy}{dt}\right) + log_{10}(3)\)

or, \(log_{10}\left(\frac{dx}{dt}\right) = log_{10}\left(\frac{3 dy}{dt}\right)\)

or, \(\frac{dx}{dt} = \frac{3dy}{dt} ---------(i)\)

For the given reaction

\(ax \rightarrow by\)

Overall rate, \('r'\) can be expressed as:

\(r = -\frac{1}{a}\frac{dx}{dt} = \frac{1}{b}\frac{dy}{dt}\)

Comparing equation (i) with the above equation we get

\(x:y = 1:3\)

The only option that follows the ratio is option (1) \(C_2H_2\) and \(C_6H_6\).