1. Tardigrade
  2. Question
  3. Chemistry
  4. For the reaction:- N2(g) +3H2(g) → 2NH3(g) Under certain conditions of temperature and partial pressure of the reactants, the rate of formation of NH 3 is 10-3 kg hr -1. The rate of consumption of H 2 under same condition is:

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Q. For the reaction:-
$N_{2\left(g\right)} +3H_{2\left(g\right)} \to 2NH_{3\left(g\right)}$
Under certain conditions of temperature and partial pressure of the reactants, the rate of formation of $NH _{3}$ is $10^{-3}\, kg\, hr ^{-1}.$ The rate of consumption of $H _{2}$ under same condition is:

Chemical Kinetics

Solution:

$N _{2( g )}+3 H _{2( g )} \rightarrow 2 NH _{3( g )}$

Rate of rxn $=\frac{- d }{ dt }\left( N _{2}\right)=\frac{-1}{3} \frac{ d }{ dt }\left( H _{2}\right)$

$=+\frac{1}{2} \frac{ d }{ dt }\left( NH _{3}\right)$

Rate of formation of $NH _{3}=+\frac{ d }{ dt }\left( N _{2}\right)=\frac{10^{-3} \times 10^{3}}{17}$

$=0.0588\, mol\, hr ^{-1}$

$\frac{-1}{3} \frac{d}{d t}\left(H_{2}\right)=+\frac{1}{2} \frac{d}{d t}\left(N H_{3}\right)$

$\frac{- d }{ dt }\left( H _{2}\right)=\frac{3}{2} \times \frac{ d }{ dt }\left( NH _{3}\right)=\frac{3}{2} \times 0.0588$

$=0.0882\, mol\, ltr ^{-1}\, hr ^{-1}$

$H _{2}$ 1 mole $\rightarrow 2g$ or $0.002\, kg$

0.0882 moles $\rightarrow 0.002 \times 0.0882=0.0001764\, kg\, hr ^{-1}$