Question

In the following reaction: $xA \rightarrow yB$ ,
$\log\left[- \frac{d \left[\right. A \left]\right.}{dt}\right]=\log\left[\frac{d \left[\right. B \left]\right.}{dt}\right]+0.3$ ,
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
$xA \rightarrow yB$ ,
rate $=$ $-\frac{1}{x}\frac{d \left[\right. A \left]\right.}{dt}=\frac{1}{y}\frac{d \left[\right. B \left]\right.}{dt}$
$-\frac{d \left[\right. A \left]\right.}{dt}=\frac{x}{y}\frac{d \left[B\right]}{dt}$ .
Taking the log on both the sides:
$\log\left(\frac{- d \left[\right. A \left]\right.}{dt}\right)=\log\left(\frac{x}{y}\right)\left(\frac{+ d \left[\right. B \left]\right.}{dt}\right)$
$\log\left[\frac{- d \left[\right. A \left]\right.}{dt}\right]=\log\left[\frac{d \left[\right. B \left]\right.}{dt}\right]+\log\left(\frac{x}{y}\right)$ .....(1)
From the given equation,
$\log\left[- \frac{d \left[\right. A \left]\right.}{dt}\right]=\log\left[\frac{d \left[\right. B \left]\right.}{dt}\right]+0.3$ .......(2)
By comparison of equation (1) and equation (2),
$\log\left(\frac{x}{y}\right)=0.3$ . Since $\log2=0.3$ ,
$\Rightarrow \frac{x}{y}=\frac{2}{1}$
$\Rightarrow x:y=2:1$ .