Consider the reaction, N2(g)+3H2(g) → 2NH3(g) The equality relationship between (d[NH3]/dt) and -(d [H2]/dt) is :-

Question

Consider the reaction, N2(g)+3H2(g) → 2NH3(g) The equality relationship between (d[NH3]/dt) and -(d [H2]/dt) is :- (A) (d [NH3]/dt)=-(1/3) (d [H2]/dt) (B) +(d [NH3]/dt)=-(2/3) (d [H2]/dt) (C) +(d [NH3]/dt)=-(3/2) (d [H2]/dt) (D) (d [NH3]/dt)=-(d [H2]/dt)

Tardigrade Admin · Accepted Answer

For the reaction N 2( g )+3 H 2( g ) → 2 NH 3( g ) (- d [ N 2]/ dt )=-(1/3) ( d / dt )([ H 2])=(1/2) ( d / dt )([ NH 3]) ∴ ( d / dt )([ NH 3])=-(2/3) ( d / dt )([ H 2])

Q. Consider the reaction,
$N_{2} (g) + 3 H_{2} (g) \to 2 N H_{3} (g)$
The equality relationship between $\frac{d [ N H _{3} ]}{d t}$ and $- \frac{d [ H _{2} ]}{d t}$ is :-

$\frac{d [ N H _{3} ]}{d t} = - \frac{1}{3} \frac{d [ H _{2} ]}{d t}$

$+ \frac{d [ N H _{3} ]}{d t} = - \frac{2}{3} \frac{d [ H _{2} ]}{d t}$

$+ \frac{d [ N H _{3} ]}{d t} = - \frac{3}{2} \frac{d [ H _{2} ]}{d t}$

$\frac{d [ N H _{3} ]}{d t} = - \frac{d [ H _{2} ]}{d t}$

For the reaction $N_{2} (g) + 3 H_{2} (g) \to 2 N H_{3} (g)$

$\frac{- d [ N _{2} ]}{d t} = - \frac{1}{3} \frac{d}{d t} ([H_{2}]) = \frac{1}{2} \frac{d}{d t} ([N H_{3}])$

$∴ \frac{d}{d t} ([N H_{3}]) = - \frac{2}{3} \frac{d}{d t} ([H_{2}])$

Q. Consider the reaction, N2​(g)+3H2​(g)→2NH3​(g) The equality relationship between dtd[NH3​]​ and −dtd[H2​]​ is :-

dtd[NH3​]​=−31​dtd[H2​]​

+dtd[NH3​]​=−32​dtd[H2​]​

+dtd[NH3​]​=−23​dtd[H2​]​

dtd[NH3​]​=−dtd[H2​]​