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Consider the reaction
$N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)$
The equality relationship between $\dfrac{d[NH_3]}{dt}$ and $-\dfrac{d[H_2]}{dt}$ is :-
$N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)$
The equality relationship between $\dfrac{d[NH_3]}{dt}$ and $-\dfrac{d[H_2]}{dt}$ is :-
A
$\dfrac{d[NH_3]}{dt} = -\dfrac{1}{3}\dfrac{d[H_2]}{dt}$
B
$+\dfrac{d[NH_3]}{dt} = -\dfrac{2}{3}\dfrac{d[H_2]}{dt}$
C
$+\dfrac{d[NH_3]}{dt} = -\dfrac{3}{2}\dfrac{d[H_2]}{dt}$
D
$\dfrac{d[NH_3]}{dt} = -\dfrac{d[H_2]}{dt}$
Detailed Solution
For the reaction $N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)$, the rate is:
Rate $= -\dfrac{d[N_2]}{dt} = -\dfrac{1}{3}\dfrac{d[H_2]}{dt} = +\dfrac{1}{2}\dfrac{d[NH_3]}{dt}$
Taking the last two terms: $\dfrac{1}{2}\dfrac{d[NH_3]}{dt} = -\dfrac{1}{3}\dfrac{d[H_2]}{dt}$
Multiplying both sides by 2:
$\dfrac{d[NH_3]}{dt} = -\dfrac{2}{3}\dfrac{d[H_2]}{dt}$
This agrees with the stoichiometry: 3 moles of $H_2$ consumed give 2 moles of $NH_3$.
Rate $= -\dfrac{d[N_2]}{dt} = -\dfrac{1}{3}\dfrac{d[H_2]}{dt} = +\dfrac{1}{2}\dfrac{d[NH_3]}{dt}$
Taking the last two terms: $\dfrac{1}{2}\dfrac{d[NH_3]}{dt} = -\dfrac{1}{3}\dfrac{d[H_2]}{dt}$
Multiplying both sides by 2:
$\dfrac{d[NH_3]}{dt} = -\dfrac{2}{3}\dfrac{d[H_2]}{dt}$
This agrees with the stoichiometry: 3 moles of $H_2$ consumed give 2 moles of $NH_3$.