At $673 \ K$,for the reaction $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$,the equilibrium constant $K_c = 0.50$. Calculate $K_p$ at this temperature. (Given $R = 0.082 \ L \ atm \ K^{-1} \ mol^{-1}$)

  • A
    $1.64 \times 10^{-4}$
  • B
    $1.64 \times 10^{-3}$
  • C
    $1.64 \times 10^{-2}$
  • D
    $1.64 \times 10^{-5}$

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The equilibrium constant for the reversible reaction,$N_2 + 3H_2 \rightleftharpoons 2NH_3$ is $K$ and for the reaction $\frac{1}{2}N_2 + \frac{3}{2}H_2 \rightleftharpoons NH_3$ the equilibrium constant is $K'$. $K$ and $K'$ will be related as

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