Chapter 12.5, Problem 12.7PSP

Interpretation Introduction

Interpretation:

The concentrations of ${\text{N}}_{\text{2}}{\text{O}}_{\text{4(g)}}$ and ${\text{NO}}_{\text{2(g)}}$ at equilibrium for the given reaction has to be calculated.

Concept Introduction:

Equilibrium constant:

At equilibrium, the ratio of products to reactants (each raised to the power corresponding to its stoichiometric coefficient) has a constant value, K

For a general reaction,

  $\text{aA+bB}\rightleftarrows \text{cC+dD}$

  $\text{Equilibrium}\text{constant}\text{K}\text{=}\frac{{\left[\text{C}\right]}^{\text{c}}{\left[\text{D}\right]}^{\text{d}}}{{\left[\text{A}\right]}^{\text{a}}{\left[\text{B}\right]}^{\text{b}}}$

Reaction Quotient:

Reaction quotient, Q is the ratio between product of the product concentration to the product of the reactant concentration with each term raised to the power of its balancing coefficient.

Consider a general equation,

  $a\text{A +}b\text{A}\stackrel{}{\rightleftharpoons }c\text{C}+d\text{D}$

Where a, b, c and d are the stoichiometric coefficients.  The reaction quotient is,

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

Comparison of ${\text{Q}}_{\text{c}}{\text{ and K}}_{\text{c}}$ :

When ${\text{Q}}_{\text{c}}{\text{ < K}}_{\text{c}}$, the reaction will proceed in right direction, $\text{reactants}\stackrel{}{\to }\text{products}$

When ${\text{Q}}_{\text{c}}{\text{ > K}}_{\text{c}}$, the reaction will proceed in left direction, $\text{reactants}\stackrel{}{\leftarrow }\text{products}$

When ${\text{Q}}_{\text{c}}{\text{ = K}}_{\text{c}}$, the reaction will remain unchanged, $\text{reactants}\stackrel{}{\rightleftharpoons }\text{products}$