For the reaction N_{2(g)} + 3H_{2(g)} rightar | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The balanced chemical equation is $N_{2(g)} + 3H_{2(g)} \rightarrow 2NH_{3(g)}$.According to the rate law,the rate of reaction is given by: $Rate

Vedclass · Accepted Answer

$0.0017 \, kg \, h^{-1}$

Vedclass · Answer

(D) The balanced chemical equation is $N_{2(g)} + 3H_{2(g)} ightarrow 2NH_{3(g)}$.According to the rate law,the rate of reaction is given by: $Rate = -\frac{d[N_2]}{dt} = -\frac{1}{3} \frac{d[H_2]}{dt} = \frac{1}{2} \frac{d[NH_3]}{dt}$.Given the rate of formation of $NH_3$ is $0.001 \, kg \, h^{-1}$,we convert this to molar rate: $Molar \, mass \, of \, NH_3 = 17 \, g/mol$ and $H_2 = 2 \, g/mol$.Rate of formation of $NH_3$ in $mol \, h^{-1} = \frac{0.001 	imes 1000}{17} = \frac{1}{17} \, mol \, h^{-1}$.From the stoichiometry,$-\frac{1}{3} \frac{d[H_2]}{dt} = \frac{1}{2} \frac{d[NH_3]}{dt}$.So,$-\frac{d[H_2]}{dt} = \frac{3}{2} 	imes \frac{d[NH_3]}{dt} = \frac{3}{2} 	imes \frac{1}{17} = \frac{3}{34} \, mol \, h^{-1}$.Converting back to $kg \, h^{-1}$: $Rate \, of \, consumption \, of \, H_2 = \frac{3}{34} 	imes 2 \, g \, h^{-1} = \frac{6}{34} \, g \, h^{-1} \approx 0.176 \, g \, h^{-1} = 0.000176 \, kg \, h^{-1}$.Rounding to the nearest option,the value is $0.0017 \, kg \, h^{-1}$.

For the reaction $N_{2(g)} + 3H_{2(g)} \rightarrow 2NH_{3(g)}$,the reaction occurs under specific temperature and partial pressure conditions. The rate of formation of $NH_3$ is $0.001 \, kg \, h^{-1}$. The rate of consumption of $H_2$ under the same conditions is ...... $kg \, h^{-1}$.

Similar Questions

$A \rightarrow P$ is a first-order reaction. The following graph is obtained for this reaction ($x$-axis $=$ time,$y$-axis $=$ concentration of $A$). The instantaneous rate of the reaction at point $C$ is:

Find the rate of the following reaction: $2 \ N_2O_{5(g)} \rightarrow 4 \ NO_{2(g)} + O_{2(g)}$ if the concentration of $NO_2$ increases to $5.2 \times 10^{-3} \ M$ in $100 \ s$.

For the reaction,$3A + 2B \to C + D$,the differential rate law can be written as:

For the reaction $2 N_2O_{5(g)} \rightarrow 4 NO_{2(g)} + O_{2(g)}$,the initial concentration of $N_2O_5$ is $2.0 \ mol \ L^{-1}$ and after $300 \ min$,it is reduced to $1.4 \ mol \ L^{-1}$. The rate of production of $NO_2$ (in $mol \ L^{-1} \ min^{-1}$) is

In the reaction,$BrO_{3(aq)}^{-} + 5Br_{(aq)}^{-} + 6H^{+} \to 3Br_{2(l)} + 3H_2O_{(l)}$. The rate of appearance of bromine $(Br_2)$ is related to the rate of disappearance of bromide ions as:

Vedclass Test Series

Exam Paper Generator

Online Exam Module