The reaction,$2A_{(g)} + B_{(g)} \rightleftharpoons 3C_{(g)} + D_{(g)}$ is begun with the concentrations of $A$ and $B$ both at an initial value of $1.00 \ M$. When equilibrium is reached,the concentration of $D$ is measured and found to be $0.25 \ M$. The value for the equilibrium constant for this reaction is given by the expression:

  • A
    $[(0.75)^3(0.25)] \div [(1.00)^2(1.00)]$
  • B
    $[(0.75)^3(0.25)] \div [(0.50)^2(0.75)]$
  • C
    $[(0.75)^3(0.25)] \div [(0.50)^2(0.25)]$
  • D
    $[(0.75)^3(0.25)] \div [(0.75)^2(0.25)]$

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The following equilibrium constants are given:
$N_{2} + 3 H_{2} \rightleftharpoons 2 NH_{3} ; K_{1}$
$N_{2} + O_{2} \rightleftharpoons 2 NO ; K_{2}$
$H_{2} + \frac{1}{2} O_{2} \rightleftharpoons H_{2} O ; K_{3}$
The equilibrium constant for the oxidation of $2 \text{ mole}$ of $NH_{3}$ to give $NO$ is

For the reactions $A \rightleftharpoons B; K_c = 2$,$B \rightleftharpoons C; K_c = 4$,and $C \rightleftharpoons D; K_c = 6$,the value of $K_c$ for the reaction $A \rightleftharpoons D$ is:

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

Explain the use of the equilibrium constant to predict the extent of a reaction with an example.

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At $T \ K$,the $K_C$ value for the reaction $\frac{1}{3} N_{2(g)} + H_{2(g)} \rightleftharpoons \frac{2}{3} NH_{3(g)}$ is $50$. The $K_C$ value for the reaction $2 NH_{3(g)} \rightleftharpoons N_{2(g)} + 3 H_{2(g)}$ at the same temperature is:

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