Chapter 9.8, Problem 2QQ

Interpretation Introduction

Interpretation:

From the given options, the equilibrium constant expression for the reaction has to be chosen.

    ${\text{3C}}_{\text{(g)}}\underset{}{\to }{\text{A}}_{\text{(g)}}{\text{+2B}}_{\text{(g)}}$

Concept Introduction:

Law of Chemical Equilibrium:

The equilibrium constant is the product of molar concentrations of the product which is raised to its stoichiometric coefficients divided by the product of molar concentrations of the reactant which is raised to its stoichiometric coefficients.

Equilibrium Constant:

Consider a reaction,

    $\text{aA+bB}\iff \text{cC+dD}$

    Forward reaction rate ${\text{K}}_{\text{f}}$= ${\text{[A]}}^{\text{a}}{\text{[B]}}^{\text{B}}$

    Backward reaction rate ${\text{K}}_{\text{b}}$= ${\text{[C]}}^{\text{c}}{\text{[D]}}^{\text{d}}$

    At equilibrium, the rate of forward reaction = rate of backward reaction

    $K_{\begin{array}{l}C\\ \end{array}}=\frac{{\text{K}}_{\text{f}}}{{\text{K}}_{\text{b}}}$ $\frac{{\text{K}}_{\text{f}}}{{\text{K}}_{\text{b}}}\text{=}\frac{{\text{[C]}}^{\text{c}}{\text{[D]}}^{\text{d}}}{{\text{[A]}}^{\text{a}}{\text{[B]}}^{\text{B}}}$

    ${\text{K}}_{\text{c}}\text{=}\frac{{\text{K}}_{\text{f}}}{{\text{K}}_{\text{b}}}\text{=}\frac{{\text{[C]}}^{\text{c}}{\text{[D]}}^{\text{d}}}{{\text{[A]}}^{\text{a}}{\text{[B]}}^{\text{B}}}$

${\text{K}}_{\text{C}}$ is the equilibrium constant.