Chapter 9.7, Problem 1QQ

Interpretation Introduction

Interpretation:

From the given options, one which give the correct statement for a system at chemical equilibrium has to be chosen.

Concept Introduction:

Chemical Equilibrium:

It is a point at which the rate of the forward reaction equals to the rate of backward reaction.  That is, there will be no net change in concentrations.  Chemical equilibrium can also be called as the state at which the rate of forward and backward reactions occur simultaneously at the same rate.  If the rates are equal and there is no net change in the concentration of the reactants and the products, then that state is said to be in dynamic equilibrium.

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.