Question

For the following reaction:
$xA\to yB$
${\log }_{10}[-\frac{d\left[A\right]}{dt}]={\log }_{10}\left[\frac{d\left[B\right]}{dt}\right]+{\log }_{10}\left(3\right)$
Here $x,y$ are the stoichiometric coefficients of reactant A and product B respectively.
Negative sign indicates rate of disappearance of the reactant.
Calculate the value of $x:y$

• A. $3:1$
• B. $1:3$
• C. $1:1$
• D. $2:1$

Answer 1

The correct option is A $3:1$

${\log }_{10}[-\frac{d\left[A\right]}{dt}]={\log }_{10}\left[\frac{d\left[B\right]}{dt}\right]+{\log }_{10}\left(3\right)$

${\log }_{10}[-\frac{d\left[A\right]}{dt}]={\log }_{10}\left[\frac{3d\left[B\right]}{dt}\right]$

$\Rightarrow -\frac{d\left[A\right]}{dt}=3\times \frac{d\left[B\right]}{dt}$

For a given reaction, if:
$aA+bB\to cC+dD$

Overall rate $r$ can be expressed as:

$r=\frac{-1}{a}\times \frac{d\left[A\right]}{dt}=\frac{-1}{b}\times \frac{d\left[B\right]}{dt}=\frac{+1}{c}\times \frac{d\left[C\right]}{dt}=\frac{+1}{d}\frac{d\left[D\right]}{dt}$
Comparing eqn (i) with above:
$x:y=1:\frac{1}{3}x:y=3:1$