(C) The balanced chemical equation is: $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$Initial moles:$n(N_2) = \frac{56 \ g}{28 \ g/mol} = 2 \ mol

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(C) The balanced chemical equation is: $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$Initial moles:$n(N_2) = \frac{56 \ g}{28 \ g/mol} = 2 \ mol$$n(H_2) = \frac{8 \ g}{2 \ g/mol} = 4 \ mol$$n(NH_3) = 0 \ mol$At equilibrium:Let $x$ be the extent of reaction.$n(N_2) = 2 - x$$n(H_2) = 4 - 3x$$n(NH_3) = 2x$Given that at equilibrium,$34 \ g$ of $NH_3$ is present:$n(NH_3) = \frac{34 \ g}{17 \ g/mol} = 2 \ mol$Therefore,$2x = 2$,which gives $x = 1$.Equilibrium moles:$n(N_2) = 2 - 1 = 1 \ mol$$n(H_2) = 4 - 3(1) = 1 \ mol$$n(NH_3) = 2 \ mol$The equilibrium moles are $1, 1, 2$ respectively.

Class 11 Chemistry 6-1.Equilibrium (Chemical Equilibrium)Mix Examples- 6-1.Equilibrium (Chemical Equilibrium)

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$N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$
$56 \ g$ of nitrogen and $8 \ g$ of hydrogen gas are heated in a closed vessel. At equilibrium,$34 \ g$ of ammonia are present. The equilibrium number of moles of nitrogen,hydrogen and ammonia are respectively:

A
$1, 2, 2$
B
$2, 2, 1$
C
$1, 1, 2$
D
$2, 1, 2$

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6-1.Equilibrium (Chemical Equilibrium)

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Mix Examples- 6-1.Equilibrium (Chemical Equilibrium)

If the reaction is started with $NH_4COONH_{2(s)}$ and the equilibrium mixture has a total pressure of $3 \ atm$,then the $K_P$ for the reaction $NH_4COONH_{2(s)} \rightleftharpoons 2NH_{3(g)} + CO_{2(g)}$ is ..... $atm^3$.

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In a closed vessel of $1 \ L$ capacity,$2 \ mol$ of $N_2$ and $6 \ mol$ of $H_2$ are mixed. If at equilibrium $50\% \ N_2$ is converted into $NH_3$,then the value of $K_c$ for the reaction $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$ will be:

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When $20 \ g$ of $CaCO_3$ are subjected to decomposition at $227 \ ^oC$ in a closed container of $10 \ L$ capacity,$50 \%$ of $CaCO_3$ remained unreacted at equilibrium. Calculate $K_P$ for $CaCO_{3(s)} \rightleftharpoons CaO_{(s)} + CO_{2(g)}$ in $atm$.

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Given the equilibrium constants for the following reactions:
$(1) \ CO_{2(g)} + H_{2(g)} \rightleftharpoons CO_{(g)} + H_2O_{(g)} \ ; \ K_1$
$(2) \ CO_{(g)} + H_2O_{(g)} \rightleftharpoons CO_{2(g)} + H_{2(g)} \ ; \ K_2$
What is the relationship between $K_1$ and $K_2$?

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$N_2O_{4(g)}$ at $300 \ K$ is kept in a closed container under $1 \ atm$. At equilibrium,$20\%$ of $N_2O_{4(g)}$ is converted to $NO_{2(g)}$.
$N_2O_{4(g)} \rightleftharpoons 2NO_{2(g)}$
Hence,the resultant pressure is: (in $atm$)

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$N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$$56 \ g$ of nitrogen and $8 \ g$ of hydrogen gas are heated in a closed vessel. At equilibrium,$34 \ g$ of ammonia are present. The equilibrium number of moles of nitrogen,hydrogen and ammonia are respectively:

Similar Questions

If the reaction is started with $NH_4COONH_{2(s)}$ and the equilibrium mixture has a total pressure of $3 \ atm$,then the $K_P$ for the reaction $NH_4COONH_{2(s)} \rightleftharpoons 2NH_{3(g)} + CO_{2(g)}$ is ..... $atm^3$.

In a closed vessel of $1 \ L$ capacity,$2 \ mol$ of $N_2$ and $6 \ mol$ of $H_2$ are mixed. If at equilibrium $50\% \ N_2$ is converted into $NH_3$,then the value of $K_c$ for the reaction $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$ will be:

When $20 \ g$ of $CaCO_3$ are subjected to decomposition at $227 \ ^oC$ in a closed container of $10 \ L$ capacity,$50 \%$ of $CaCO_3$ remained unreacted at equilibrium. Calculate $K_P$ for $CaCO_{3(s)} \rightleftharpoons CaO_{(s)} + CO_{2(g)}$ in $atm$.

Given the equilibrium constants for the following reactions:$(1) \ CO_{2(g)} + H_{2(g)} \rightleftharpoons CO_{(g)} + H_2O_{(g)} \ ; \ K_1$$(2) \ CO_{(g)} + H_2O_{(g)} \rightleftharpoons CO_{2(g)} + H_{2(g)} \ ; \ K_2$What is the relationship between $K_1$ and $K_2$?

$N_2O_{4(g)}$ at $300 \ K$ is kept in a closed container under $1 \ atm$. At equilibrium,$20\%$ of $N_2O_{4(g)}$ is converted to $NO_{2(g)}$.$N_2O_{4(g)} \rightleftharpoons 2NO_{2(g)}$Hence,the resultant pressure is: (in $atm$)

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$N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$
$56 \ g$ of nitrogen and $8 \ g$ of hydrogen gas are heated in a closed vessel. At equilibrium,$34 \ g$ of ammonia are present. The equilibrium number of moles of nitrogen,hydrogen and ammonia are respectively:

Given the equilibrium constants for the following reactions:
$(1) \ CO_{2(g)} + H_{2(g)} \rightleftharpoons CO_{(g)} + H_2O_{(g)} \ ; \ K_1$
$(2) \ CO_{(g)} + H_2O_{(g)} \rightleftharpoons CO_{2(g)} + H_{2(g)} \ ; \ K_2$
What is the relationship between $K_1$ and $K_2$?

$N_2O_{4(g)}$ at $300 \ K$ is kept in a closed container under $1 \ atm$. At equilibrium,$20\%$ of $N_2O_{4(g)}$ is converted to $NO_{2(g)}$.
$N_2O_{4(g)} \rightleftharpoons 2NO_{2(g)}$
Hence,the resultant pressure is: (in $atm$)