Question

Consider the reaction, $N_2(g)+3H_2(g) \to 2NH_3(g)$ The equality relationship between $\frac{d[NH_3]}{dt}$ and $-\frac{d [H_2]}{dt}$ is :-

• A. $\frac{d [NH_3]}{dt}=-\frac{1}{3} \frac{d [H_2]}{dt}$
• B. $+\frac{d [NH_3]}{dt}=-\frac{2}{3} \frac{d [H_2]}{dt}$
• C. $+\frac{d [NH_3]}{dt}=-\frac{3}{2} \frac{d [H_2]}{dt}$
• D. $\frac{d [NH_3]}{dt}=-\frac{d [H_2]}{dt}$

Answer 1

The Correct Option is B

The given reaction is: 

\(N_2(g) + 3H_2(g) \to 2NH_3(g)\)

To find the equality relationship between the rate of production of \(NH_3\) and the rate of consumption of \(H_2\), we utilize the concept of stoichiometry in reaction rates. For a general reaction:

\(aA + bB \to cC + dD\)

The rate relationships are given by:

 

• \(\frac{-d[A]}{a\ dt} = \frac{-d[B]}{b\ dt} = \frac{d[C]}{c\ dt} = \frac{d[D]}{d\ dt}\)

 

Applying this to the given equation:

 

• \(\frac{-d[N_2]}{dt} = \frac{-1}{3} \frac{d[H_2]}{dt} = \frac{1}{2} \frac{d[NH_3]}{dt}\)

 

Thus, the specific relationship between \(\frac{d[NH_3]}{dt}\) and \(-\frac{d[H_2]}{dt}\) is:

\(\frac{d[NH_3]}{dt} = -\frac{2}{3} \frac{d[H_2]}{dt}\)

This matches the correct answer option:

\(+\frac{d[NH_3]}{dt} = -\frac{2}{3} \frac{d[H_2]}{dt}\)