$N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}; K_1$
$NH_{3(g)} \rightleftharpoons \frac{1}{2} N_{2(g)} + \frac{3}{2} H_{2(g)}; K_2$
$\frac{1}{2} N_{2(g)} + \frac{3}{2} H_{2(g)} \rightleftharpoons NH_{3(g)}; K_3$
$2NH_{3(g)} \rightleftharpoons N_{2(g)} + 3H_{2(g)}; K_4$
If $K_1 = K_2^x = K_3^y = K_4^z$,then the correct values of $x, y,$ and $z$ are respectively:

  • A
    $-2, 2, -1$
  • B
    $-1, 2, -2$
  • C
    $-2, 2, 1$
  • D
    $2, 1, -2$

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For the reaction $P_{4(s)} + 5O_{2(g)} \rightleftharpoons P_4O_{10(s)}$,the equilibrium constant expression $K_c$ is:

The equilibrium constant $K$ for the synthesis of $HI$ is $50$. The equilibrium constant $K$ for the dissociation of $HI$ is:

The following concentrations were obtained for the formation of $NH_{3}$ from $N_{2}$ and $H_{2}$ at equilibrium at $500\, K$: $[N_{2}] = 1.5 \times 10^{-2}\, M$,$[H_{2}] = 3.0 \times 10^{-2}\, M$ and $[NH_{3}] = 1.2 \times 10^{-2}\, M$. Calculate the equilibrium constant.

$K_1, K_2$ and $K_3$ are the equilibrium constants of the following reactions $(I), (II)$ and $(III)$ respectively.
$(I) \, N_2 + 2O_2 \rightleftharpoons 2NO_2$
$(II) \, 2NO_2 \rightleftharpoons N_2 + 2O_2$
$(III) \, NO_2 \rightleftharpoons \frac{1}{2} N_2 + O_2$
The correct relation from the following is

If an equilibrium process equation is multiplied by a factor $n$,how does the equilibrium constant change? Explain with an example.

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