Chapter 14, Problem 14.30QE

Interpretation Introduction

Interpretation:

The reaction quotient $\left(Q\right)$ and direction of the following reaction has to be determined. Also, the number line for $Q$ and $K_{\text{c}}$ has to be drawn.

  ${\text{2SO}}_2\left(\text{g}\right)+{\text{O}}_{\text{2}}\left(\text{g}\right)\rightleftharpoons 2{\text{SO}}_3\left(\text{g}\right)$

Concept Introduction:

The condition of equilibrium is a state of balance of processes that runs in opposite directions. At equilibrium, the formation of a product from the reactant balances the formation of reactant from the product. Also, the change in concentration of reaction and product seems to be negligible at equilibrium state.

A reaction quotient $Q$ is an algebraic form of equilibrium constant $\left(K_{\text{eq}}\right)$ for all concentrations including reaction concentrations at equilibrium. The reaction quotient $Q$ can also help in prediction of direction of reaction when compared with the equilibrium constant $\left(K_{\text{eq}}\right)$.

The general equilibrium reaction is as follows:

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

Here,

$\text{A}$ and $\text{B}$ are the reactants.

$\text{C}$ and $\text{D}$ are products.

$a$ and $b$ are the stoichiometric coefficients of reactants.

$c$ and $d$ are the stoichiometric coefficients of products.

The expression of the reaction quotient for the above reaction is as follows:

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

Here,

$K_{\text{c}}$ is the equilibrium constant.

$\left[\text{C}\right]$ is the concentration of $\text{C}$.

$\left[\text{D}\right]$ is the concentration of $\text{D}$.

$\left[\text{A}\right]$ is the concentration of $\text{A}$.

$\left[\text{B}\right]$ is the concentration of $\text{B}$.

The concentration of reactants and product changes in order to bring reaction quotient and equilibrium constant closer. Therefore, the direction of reaction can be predicted as follows:

(1) If $Q$ is less than $K_{\text{eq}}$ then the reaction moves to increase the concentration of products, which means reaction moves towards right direction.

(2)If $Q$ is greater than $K_{\text{eq}}$ then the reaction moves to decrease the concentration of products, that means reaction moves towards the left direction.

(3)If $Q$ is equal to $K_{\text{eq}}$ then the reaction is in equilibrium.