The following equilibrium constants are given: $N_2 + 3 H_2 \rightleftharpoons 2 NH_3$ $(k_1)$,$N_2 + O_2 \rightleftharpoons 2 NO$ $(k_2)$,$H_2 + 1/2 O_2 \rightleftharpoons H_2 O$ $(k_3)$. The equilibrium constant for the oxidation of $1 \text{ mole } NH_3$ by oxygen to give $NO$ according to the reaction $NH_3 + 5/4 O_2 \rightleftharpoons NO + 3/2 H_2 O$ is:

  • A
    $\frac{k_2^{1/2} k_3^{3/2}}{k_1^{1/2}}$
  • B
    $\frac{k_2^2 k_3}{k_1}$
  • C
    $\frac{k_1 k_2}{k_3}$
  • D
    $\frac{k_2 k_3^3}{k_1}$

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For the reaction $H_{2(g)} + I_{2(g)} \rightleftharpoons 2HI_{(g)}$,the value of $K_p$ changes with:

The value of $K_{c}$ for the reaction $3 O_{2(g)} \longleftrightarrow 2 O_{3(g)}$ is $2.0 \times 10^{-50}$ at $25^{\circ} C$. If the equilibrium concentration of $O_{2}$ in air at $25^{\circ} C$ is $1.6 \times 10^{-2} \, M$,what is the concentration of $O_{3} ?$

For the gaseous reactions $(I)$ and $(II)$,the equilibrium constants are $X$ and $Y$,respectively.
$I. \frac{1}{2} N_{2(g)} + O_{2(g)} \rightleftharpoons NO_{2(g)}$
$II. 2 NO_{2(g)} \rightleftharpoons N_2O_{4(g)}$
Using the above reactions,the equilibrium constant $Z$ for the reaction $(III)$ given below is:
$III. N_2O_{4(g)} \rightleftharpoons N_{2(g)} + 2 O_{2(g)}$

$A$ reaction is $A + B \rightleftharpoons C + D$. Initially,we start with equal concentrations of $A$ and $B$. At equilibrium,the number of moles of $C$ is two times that of $A$. What is the equilibrium constant $(K_c)$ of the reaction?

For the reaction $2NO_{2(g)} \rightleftharpoons 2NO_{(g)} + O_{2(g)}$,$K_c = 1.8 \times 10^{-6}$ at $185\,^{\circ}C$. At $185\,^{\circ}C$,the value of $K_c$ for the reaction $NO_{(g)} + \frac{1}{2}O_{2(g)} \rightleftharpoons NO_{2(g)}$ is

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