For the gaseous reaction,equilibrium constant is given:
$XeF_6 + H_2O \rightleftharpoons XeOF_4 + 2HF, K_1$
$XeO_4 + XeF_6 \rightleftharpoons XeOF_4 + XeO_3F_2, K_2$
The equilibrium constant for the reaction:
$XeO_4 + 2HF \rightleftharpoons XeO_3F_2 + H_2O$ will be

  • A
    $\frac{K_1}{K_2}$
  • B
    $\frac{K_1}{K_2^2}$
  • C
    $\frac{K_1^2}{K_2}$
  • D
    $\frac{K_2}{K_1}$

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For the balanced reaction $A + B \rightleftharpoons 2C$,if the equilibrium concentrations of both $A$ and $B$ are $0.20 \ mol/L$,and the concentration of $C$ is $0.60 \ mol/L$,then the equilibrium constant for this reaction will be:

If $K_1, K_2, K_3$ are equilibrium constants for the formation of $AD, AD_2, AD_3$ respectively as follows: $A + D \rightleftharpoons AD$,$AD + D \rightleftharpoons AD_2$,$AD_2 + D \rightleftharpoons AD_3$. Then the equilibrium constant '$K$' for $A + 3D \rightleftharpoons AD_3$ is related as:

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

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What is the equilibrium constant $(K_C)$ for the given reaction?
$N_2 + O_2 \rightleftharpoons 2 NO$
Where the equilibrium concentrations of $N_2$,$O_2$ and $NO$ are found to be $4 \times 10^{-3} \ M$,$3 \times 10^{-3} \ M$ and $3 \times 10^{-3} \ M$ respectively.

The value of $K_{c}$ for the reaction $3 O_{2(g)} \longleftrightarrow 2 O_{3(g)}$ is $2.0 \times 10^{-50}$ at $25^{\circ} C$. If the equilibrium concentration of $O_{2}$ in air at $25^{\circ} C$ is $1.6 \times 10^{-2} \, M$,what is the concentration of $O_{3} ?$

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