Chapter 17, Problem 17.96P

Interpretation Introduction

Interpretation

Study of the water-gas shift reaction was made and the equilibrium was reached with ${\text{[CO] = [H}}_{\text{2}}{\text{O] = [H}}_{\text{2}}\text{] =0}{\text{.10M and [CO}}_{\text{2}}\text{] = 0}\text{.40M}$. The equilibrium is re-established after the addition of $\text{0}{\text{.60 mol of H}}_{\text{2}}$, then the new concentrations of all the components has to be determined.

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

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

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

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