$HI$ was heated in a closed tube at $440\,^{\circ}C$ until equilibrium was obtained. At this temperature,$22\%$ of $HI$ was dissociated. The equilibrium constant for this dissociation will be:

  • A
    $0.282$
  • B
    $0.0796$
  • C
    $0.0199$
  • D
    $1.99$

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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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The reaction,$CO_{(g)} + 3H_{2(g)} \longleftrightarrow CH_{4(g)} + H_{2}O_{(g)}$ is at equilibrium at $1300 \, K$ in a $1 \, L$ flask. It also contains $0.30 \, mol$ of $CO$,$0.10 \, mol$ of $H_{2}$,and $0.02 \, mol$ of $H_{2}O$ and an unknown amount of $CH_{4}$ in the flask. Determine the concentration of $CH_{4}$ in the mixture. The equilibrium constant,$K_{c}$ for the reaction at the given temperature is $3.90$.

Give the equilibrium constant expression for the following reactions:
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For the equilibrium $N_2 + 3H_2 \rightleftharpoons 2NH_3$,$K_c$ at $1000 \ K$ is $2.37 \times 10^{-3}$. If at equilibrium $[N_2] = 2 \ M$ and $[H_2] = 3 \ M$,the concentration of $NH_3$ is: (in $M$)

The equilibrium constant for the reaction $SO_{2(g)} + \frac{1}{2} O_{2(g)} \rightleftharpoons SO_{3(g)}$ is $5 \times 10^{-2} \ atm^{-1/2}$. The equilibrium constant of the reaction $2 SO_{3(g)} \rightleftharpoons 2 SO_{2(g)} + O_{2(g)}$ would be

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