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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Write the expression for the equilibrium constant,$K_{c}$ for each of the following reactions:
$(i)$ $2 NOCl_{(g)} \longleftrightarrow 2 NO_{(g)} + Cl_{2(g)}$
$(ii)$ $2 Cu(NO_{3})_{2(s)} \longleftrightarrow 2 CuO_{(s)} + 4 NO_{2(g)} + O_{2(g)}$
$(iii)$ $CH_{3}COOC_{2}H_{5(aq)} + H_{2}O_{(l)} \longleftrightarrow CH_{3}COOH_{(aq)} + C_{2}H_{5}OH_{(aq)}$
$(iv)$ $Fe^{3+}_{(aq)} + 3 OH^{-}_{(aq)} \longleftrightarrow Fe(OH)_{3(s)}$
$(v)$ $I_{2(s)} + 5 F_{2(g)} \longleftrightarrow 2 IF_{5(g)}$

In the equilibrium $AB \rightleftharpoons A + B$; if the equilibrium concentration of $A$ is doubled,the equilibrium concentration of $B$ would become:

Write the equilibrium constant (equilibrium law) for the following reactions:
$(i)$ $4NH_{3(g)} + 5O_{2(g)} \rightleftharpoons 4NO_{(g)} + 6H_{2}O_{(g)}$
$(ii)$ $H_{2(g)} + I_{2(g)} \rightleftharpoons 2HI_{(g)}$
$(iii)$ $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$

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