Question

For the reaction; $N_2+3H_2\to 2N{H}_3$, how does the rate expressions of th reaction inter-relate $\frac{d\left[H_2\right]}{dt}$ and $\frac{d\left[N{H}_3\right]}{dt}$?

• A. $-\frac{1}{3}\frac{d\left[H_2\right]}{dt}=+\frac{1}{2}\frac{d\left[N{H}_3\right]}{dt}$
• B. $-\frac{1}{2}\frac{d\left[H_2\right]}{dt}=+\frac{1}{3}\frac{d\left[N{H}_3\right]}{dt}$
• C. $+\frac{1}{2}\frac{d\left[H_2\right]}{dt}=1\frac{1}{3}\frac{d\left[N{H}_3\right]}{dt}$
• D. $+\frac{1}{3}\frac{d\left[H_2\right]}{dt}=-\frac{1}{2}\frac{d\left[N{H}_3\right]}{dt}$

Answer 1

The correct option is A $-\frac{1}{3}\frac{d\left[H_2\right]}{dt}=+\frac{1}{2}\frac{d\left[N{H}_3\right]}{dt}$

Rate Law is an experimentally determined equation of the rate of the reaction is called the rate law of the reaction. In rate law, the rate of the reaction is expressed as a function of the concentration of the reactants and products present in the reaction.

For this reaction, the rate of reaction is written as:-

$-\frac{d\left[N_2\right]}{dt}=-\frac{1}{3}\frac{d\left[H_2\right]}{dt}=+\frac{1}{2}\frac{d\left[N{H}_3\right]}{dt}$

Hence, the correct answer is option $\text{A}$.