(C) Given reactions:$(1)$ ${H_2(g)} + \frac{1}{2}{S_2(s)} \rightleftharpoons {H_2S(g)}$; ${K_1} = \frac{[{H_2S}]}{[{H_2}][{S_2}]^{1/2}}$$(2)$ ${H_2(g)

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(C) Given reactions:$(1)$ ${H_2(g)} + \frac{1}{2}{S_2(s)} \rightleftharpoons {H_2S(g)}$; ${K_1} = \frac{[{H_2S}]}{[{H_2}][{S_2}]^{1/2}}$$(2)$ ${H_2(g)} + {Br_2(g)} \rightleftharpoons 2{HBr(g)}$; ${K_2} = \frac{[{HBr}]^2}{[{H_2}][{Br_2}]}$Target reaction: ${Br_2(g)} + {H_2S(g)} \rightleftharpoons 2{HBr(g)} + \frac{1}{2}{S_2(s)}$To obtain the target reaction,we perform: (Reaction $2$) - (Reaction $1$).Equilibrium constant for the target reaction $K_3 = \frac{K_2}{K_1}$.$K_3 = \frac{[{HBr}]^2}{[{H_2}][{Br_2}]} \times \frac{[{H_2}][{S_2}]^{1/2}}{[{H_2S}]} = \frac{[{HBr}]^2[{S_2}]^{1/2}}{[{Br_2}][{H_2S}]}$.

Class 11 Chemistry 6-1.Equilibrium (Chemical Equilibrium)Law of equilibrium and Equilibrium constant

Difficult

If for ${H_2(g)} + \frac{1}{2}{S_2(s)} \rightleftharpoons {H_2S(g)}$ and ${H_2(g)} + {Br_2(g)} \rightleftharpoons 2{HBr(g)}$ the equilibrium constants are $K_1$ and $K_2$ respectively,the reaction ${Br_2(g)} + {H_2S(g)} \rightleftharpoons 2{HBr(g)} + \frac{1}{2}{S_2(s)}$ would have equilibrium constant

A
${K_1} \times {K_2}$
B
${K_1}/{K_2}$
C
${K_2}/{K_1}$
D
${K_2^2}/{K_1}$

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6-1.Equilibrium (Chemical Equilibrium)

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Law of equilibrium and Equilibrium constant

The figure shows the change in concentration of species $A$ and $B$ as a function of time. The equilibrium constant $K_C$ for the reaction $2A_{(g)} \rightleftharpoons B_{(g)}$ is

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$2 NOCl_{(g)} \rightleftharpoons 2 NO_{(g)} + Cl_{2(g)}$
In an experiment,$2.0 \ mol$ of $NOCl$ was placed in a $1 \ L$ flask and the concentration of $NO$ after equilibrium was established,was found to be $0.4 \ mol/L$. The equilibrium constant at $30^{\circ} C$ is $....... \times 10^{-4}$.

DifficultJEE MAIN 2022

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For the given reaction,choose the correct expression of $K_{C}$ from the following: $Fe_{(aq)}^{3+} + SCN_{(aq)}^{-} \rightleftharpoons (FeSCN)_{(aq)}^{2+}$

DifficultJEE MAIN 2024

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The value of $K_{C}$ is $64$ at $800 \ K$ for the reaction $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$.
The value of $K_{C}$ for the following reaction is:
$NH_{3(g)} \rightleftharpoons \frac{1}{2}N_{2(g)} + \frac{3}{2}H_{2(g)}$

MediumJEE MAIN 2020

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In a reversible reaction $A \underset{k_2}{\overset{k_1}{\longleftrightarrow}} B$,the initial concentrations of $A$ and $B$ are $a$ and $b$ in moles per litre,and the equilibrium concentrations are $(a - x)$ and $(b + x)$ respectively. Express $x$ in terms of $k_1, k_2, a,$ and $b$.

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The figure shows the change in concentration of species $A$ and $B$ as a function of time. The equilibrium constant $K_C$ for the reaction $2A_{(g)} \rightleftharpoons B_{(g)}$ is

$2 NOCl_{(g)} \rightleftharpoons 2 NO_{(g)} + Cl_{2(g)}$In an experiment,$2.0 \ mol$ of $NOCl$ was placed in a $1 \ L$ flask and the concentration of $NO$ after equilibrium was established,was found to be $0.4 \ mol/L$. The equilibrium constant at $30^{\circ} C$ is $....... \times 10^{-4}$.

For the given reaction,choose the correct expression of $K_{C}$ from the following: $Fe_{(aq)}^{3+} + SCN_{(aq)}^{-} \rightleftharpoons (FeSCN)_{(aq)}^{2+}$

The value of $K_{C}$ is $64$ at $800 \ K$ for the reaction $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$.The value of $K_{C}$ for the following reaction is:$NH_{3(g)} \rightleftharpoons \frac{1}{2}N_{2(g)} + \frac{3}{2}H_{2(g)}$

In a reversible reaction $A \underset{k_2}{\overset{k_1}{\longleftrightarrow}} B$,the initial concentrations of $A$ and $B$ are $a$ and $b$ in moles per litre,and the equilibrium concentrations are $(a - x)$ and $(b + x)$ respectively. Express $x$ in terms of $k_1, k_2, a,$ and $b$.

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$2 NOCl_{(g)} \rightleftharpoons 2 NO_{(g)} + Cl_{2(g)}$
In an experiment,$2.0 \ mol$ of $NOCl$ was placed in a $1 \ L$ flask and the concentration of $NO$ after equilibrium was established,was found to be $0.4 \ mol/L$. The equilibrium constant at $30^{\circ} C$ is $....... \times 10^{-4}$.

The value of $K_{C}$ is $64$ at $800 \ K$ for the reaction $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$.
The value of $K_{C}$ for the following reaction is:
$NH_{3(g)} \rightleftharpoons \frac{1}{2}N_{2(g)} + \frac{3}{2}H_{2(g)}$