(B) The reaction is $N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g).$ The reaction quotient $Q_P$ is given by the expression: $Q_P = \frac{(P_{NH_3})^2}

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(B) The reaction is $N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g).$ The reaction quotient $Q_P$ is given by the expression: $Q_P = \frac{(P_{NH_3})^2}{(P_{N_2})(P_{H_2})^3}.$ Substituting the given partial pressures: $Q_P = \frac{(3)^2}{(1)(2)^3} = \frac{9}{8} = 1.125 \ atm^{-2}.$ Comparing $Q_P$ with $K_P$: $Q_P = 1.125$ and $K_P = 4.28 \times 10^{-5}.$ Since $Q_P > K_P,$ the reaction will proceed in the backward direction to reach equilibrium.

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

Medium

$A$ reaction mixture containing $H_2, N_2$ and $NH_3$ has partial pressures of $2 \ atm, 1 \ atm$ and $3 \ atm$ respectively at $725 \ K.$ If the value of $K_P$ for the reaction,$N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$ is $4.28 \times 10^{-5} \ atm^{-2}$ at $725 \ K,$ in which direction will the net reaction proceed?

A
Forward
B
Backward
C
No net reaction
D
Direction of reaction cannot be predicted

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Class 11

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Chapter

6-1.Equilibrium (Chemical Equilibrium)

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

Consider the following two equilibrium reactions:
$i$. $2NH_{3(g)} \rightleftharpoons N_{2(g)} + 3H_{2(g)}$
$ii$. $2ND_{3(g)} \rightleftharpoons N_{2(g)} + 3D_{2(g)}$
What is the difference in their equilibrium constants $(K_c)$?

Medium

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One mole of $N_2O_4(g)$ is taken in a closed container at $1 \ atm$ and $300 \ K$. When it is heated to $600 \ K$,$20 \%$ of $N_2O_4(g)$ dissociates into $NO_2(g)$. The resulting pressure is .......... $atm$.

Difficult

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In a $13 \ L$ vessel at $1027 \ ^oC$,the reaction $C_{(s)} + S_{2_{(g)}} \rightleftharpoons CS_{2_{(g)}}$ is carried out with $12 \ g$ of $C$,$64 \ g$ of $S_2$,and $76 \ g$ of $CS_2$. What is the total pressure in terms of $R$ (in $R$)?

Medium

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$3.00 \ mol$ of $PCl_5$ kept in $1 \ L$ closed reaction vessel was allowed to attain equilibrium at $380 \ K$. If $1.59 \ mol$ of reactant was converted into the product at equilibrium,then $K_c$ is:

Medium

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For the reaction $SO_{2(g)} + NO_{2(g)} \rightleftharpoons SO_{3(g)} + NO_{(g)}$,the equilibrium constant $K_c$ is $16$. If $1 \ mol$ of each gas is taken in a $1 \ dm^3$ vessel,the equilibrium concentration of $NO$ will be ....

Medium

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Consider the following two equilibrium reactions:$i$. $2NH_{3(g)} \rightleftharpoons N_{2(g)} + 3H_{2(g)}$$ii$. $2ND_{3(g)} \rightleftharpoons N_{2(g)} + 3D_{2(g)}$What is the difference in their equilibrium constants $(K_c)$?

One mole of $N_2O_4(g)$ is taken in a closed container at $1 \ atm$ and $300 \ K$. When it is heated to $600 \ K$,$20 \%$ of $N_2O_4(g)$ dissociates into $NO_2(g)$. The resulting pressure is .......... $atm$.

In a $13 \ L$ vessel at $1027 \ ^oC$,the reaction $C_{(s)} + S_{2_{(g)}} \rightleftharpoons CS_{2_{(g)}}$ is carried out with $12 \ g$ of $C$,$64 \ g$ of $S_2$,and $76 \ g$ of $CS_2$. What is the total pressure in terms of $R$ (in $R$)?

$3.00 \ mol$ of $PCl_5$ kept in $1 \ L$ closed reaction vessel was allowed to attain equilibrium at $380 \ K$. If $1.59 \ mol$ of reactant was converted into the product at equilibrium,then $K_c$ is:

For the reaction $SO_{2(g)} + NO_{2(g)} \rightleftharpoons SO_{3(g)} + NO_{(g)}$,the equilibrium constant $K_c$ is $16$. If $1 \ mol$ of each gas is taken in a $1 \ dm^3$ vessel,the equilibrium concentration of $NO$ will be ....

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Consider the following two equilibrium reactions:
$i$. $2NH_{3(g)} \rightleftharpoons N_{2(g)} + 3H_{2(g)}$
$ii$. $2ND_{3(g)} \rightleftharpoons N_{2(g)} + 3D_{2(g)}$
What is the difference in their equilibrium constants $(K_c)$?