The standard equilibrium constant,$K_p$ at $298 \, K$ for the reaction,$N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$ is $5.8 \times 10^5$. The value of standard equilibrium constant,if the concentration of gases is expressed in terms of $mol/L$,will be:
[Given : $R = 0.08314 \, L \, bar \, K^{-1} \, mol^{-1}$]

  • A
    $3.51 \times 10^6$
  • B
    $3.84 \times 10^7$
  • C
    $3.56 \times 10^8$
  • D
    $3.99 \times 10^9$

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$(1) \ N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)} \ ; \ K_1$
$(2) \ N_{2(g)} + O_{2(g)} \rightleftharpoons 2NO_{(g)} \ ; \ K_2$
$(3) \ H_{2(g)} + \frac{1}{2}O_{2(g)} \rightleftharpoons H_2O_{(g)} \ ; \ K_3$
The equation for the equilibrium constant of the reaction
$2NH_{3(g)} + \frac{5}{2}O_{2(g)} \rightleftharpoons 2NO_{(g)} + 3H_2O_{(g)}$
$(K_4)$ in terms of $K_1$,$K_2$,and $K_3$ is

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