Question

For a chemical reaction: $n_{1}A\to n_{2}B;log[-\frac{d\left[A\right]}{dt}]=log\left[\frac{d\left[B\right]}{dt}\right]+0.3$
then the ratio of $n_1$ and $n_2$ is approximately :

• A. $3$
• B. $0.3$
• C. $2$
• D. $0.5$

Answer 1

The correct option is C $2$

$-\frac{1}{n_1}\frac{d\left[A\right]}{dt}=\frac{1}{n_2}\frac{d\left[B\right]}{dt}$

or $-\frac{d\left[A\right]}{dt}=\frac{n_1}{n_2}\frac{d\left[B\right]}{dt}$

or $log[-\frac{d\left[A\right]}{dt}]=log\frac{n_1}{n_2}+log\frac{d\left[B\right]}{dt}$

$\therefore log\frac{n_1}{n_2}=0.3$

$\therefore \frac{n_1}{n_2}=2$