Consider the reaction N2(g)+3H2(g)→2NH3(g)The equality relationship between dNH3dt and −dH2dt is

A
$+ \frac{d [{NH}_{3}]}{dt} = - \frac{2}{3} \frac{d [H_{2}]}{dt}$
B
$+ \frac{d [{NH}_{3}]}{dt} = - \frac{3}{2} \frac{d [H_{2}]}{dt}$
C
$\frac{d [{NH}_{3}]}{dt} = - \frac{d [H_{2}]}{dt}$
D
$\frac{d [{NH}_{3}]}{dt} = - \frac{1}{3} \frac{d [H_{2}]}{dt}$

Solution:

If we write rate of reaction in terms of concentration of

${NH}_{3} and H_{2}, then$
Rate of reaction $= \frac{1}{2} \frac{d [{NH}_{3}]}{dt} = - \frac{1}{3} \frac{d [H_{2}]}{dt}$
So $\frac{d [{NH}_{3}]}{dt} = - \frac{2}{3} \frac{d [H_{2}]}{dt}$

Consider the reaction $N_{2} (g) + 3 H_{2} (g) \to 2 {NH}_{3} (g)$
The equality relationship between $\frac{d [{NH}_{3}]}{dt} and - \frac{d [H_{2}]}{dt} is$

Solution:

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