Question

In the following reaction: $xA\to yB$ ,
$\log \left[-\frac{d[A]}{dt}\right]=\log \left[\frac{d[B]}{dt}\right]+0.3$ ,
where the $-ve$ sign indicates the rate of disappearance of the reactant. Thus, $x:y$ is:

• A. $1:2$
• B. $2:1$
• C. $3:1$
• D. $3:10$

Answer 1

Answer: (B)

Rate of a chemical reaction can be determined by the time derivative of the concentration of any reactant or product.
For a general reaction
$xA\to yB$ ,
rate $=$ $-\frac{1}{x}\frac{d[A]}{dt}=\frac{1}{y}\frac{d[B]}{dt}$
$-\frac{d[A]}{dt}=\frac{x}{y}\frac{d\left[B\right]}{dt}$ .
Taking the log on both the sides:
$\log \left(\frac{-d[A]}{dt}\right)=\log \left(\frac{x}{y}\right)\left(\frac{+d[B]}{dt}\right)$
$\log \left[\frac{-d[A]}{dt}\right]=\log \left[\frac{d[B]}{dt}\right]+\log \left(\frac{x}{y}\right)$ .....(1)
From the given equation,
$\log \left[-\frac{d[A]}{dt}\right]=\log \left[\frac{d[B]}{dt}\right]+0.3$ .......(2)
By comparison of equation (1) and equation (2),
$\log \left(\frac{x}{y}\right)=0.3$ . Since $\log 2=0.3$ ,
$\Rightarrow \frac{x}{y}=\frac{2}{1}$
$\Rightarrow x:y=2:1$ .