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]
Answer & explanation
Correct answer: option 1
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 \(C_6H_6\).