Chapter 17, Problem 17.87P

(a)

Interpretation Introduction

Interpretation:

The oxidation of ${\text{SO}}_{\text{2}}$ in the manufacture of sulfuric acid determines that ${\text{K}}_{\text{c}}\text{ = 1}{\text{.7×18}}^{\text{8}}\text{ at 600}\text{.K}$:

  ${\text{2SO}}_{\text{2(g)}}{\text{+ O}}_{\text{2(g)}}\rightleftharpoons {\text{2SO}}_{\text{3(g)}}$

At equilibrium, ${\text{P}}_{{\text{SO}}_{\text{3}}}\text{= 300}{\text{.atm and P}}_{{\text{O}}_{\text{2}}}\text{= 100}\text{.atm}$.  ${\text{P}}_{{\text{SO}}_{\text{2}}}$ has to be calculated.

Concept Introduction:

Equilibrium constant using partial pressure:

The equilibrium constant of partial pressure can be defined as the ratio of products and reactants concentration at equilibrium in terms of partial pressure.

For a reaction,

  ${\text{aA}}_{\text{(g)}}{\text{+ bB}}_{\text{(g)}}\rightleftharpoons {\text{cC}}_{\text{(g)}}{\text{+ dD}}_{\text{(g)}}$

The expression of ${\text{K}}_{\text{p}}$ can be given as

  ${\text{K}}_{\text{p}}\text{ = }\frac{{{\text{(P}}_{\text{C}}\text{)}}^{\text{c}}{{\text{(P}}_{\text{D}}\text{)}}^{\text{d}}}{{{\text{(P}}_{\text{A}}\text{)}}^{\text{a}}{{\text{(P}}_{\text{B}}\text{)}}^{\text{b}}}$

Explanation of Solution

Given, ${\text{K}}_{\text{c}}\text{ = 1}{\text{.7×18}}^{\text{8}}\text{ at 600}\text{.K}$.

  $\begin{array}{l}{\text{K}}_{\text{p}}{\text{= K}}_{\text{c}}{\text{(RT)}}^{\text{Δn}}\\ \\ \text{where, Δn = moles of gaseous products - moles of gaseous reactants}\\ \\ \Delta \text{n = 2 - 3 = -1}\end{array}$

  $\begin{array}{l}{\text{K}}_{\text{p}}\text{= (1}{\text{.7×10}}^{\text{8}}\text{)[(0}\text{.0821 L}\text{.atm/mol}\text{.K)(600}{\text{.K)]}}^{\text{-1}}\\ \\ {\text{K}}_{\text{p}}\text{ = 3}{\text{.451×10}}^{\text{6}}\end{array}$

At equilibrium,

  $\begin{array}{l}{\text{K}}_{\text{p}}\text{= }\frac{{\text{P}}_{{\text{SO}}_{\text{3}}}^{\text{2}}}{{\text{P}}_{{\text{SO}}_{\text{2}}}^{\text{2}}{\text{P}}_{{\text{O}}_{\text{2}}}}\\ \\ {\text{K}}_{\text{p}}\text{= }\frac{{\text{(300)}}^{\text{2}}}{{\text{P}}_{{\text{SO}}_{\text{2}}}^{\text{2}}}\\ \\ {\text{K}}_{\text{p}}\text{= 3}{\text{.451×10}}^{\text{6}}\\ \\ {\text{P}}_{{\text{SO}}_{\text{2}}}\text{= 0}\text{.016149 atm}\end{array}$

${\text{P}}_{{\text{SO}}_{\text{2}}}$ is calculated as $\text{0}\text{.016149 atm}$.

(b)

Interpretation Introduction

Interpretation:

An engineer places a mixture of $\text{0}{\text{.0040 mol of SO}}_{\text{2(g)}}$ and $\text{0}{\text{.0028 mol of O}}_{\text{2(g)}}$ in a $\text{1}\text{.0-L}$ container and raises the temperature to $\text{1000 K}$.  At equilibrium, $\text{0}{\text{.0020 mol of SO}}_{\text{3(g)}}$ is present.  For this reaction, at $\text{1000 K}$, ${\text{K}}_{\text{c}}{\text{ and P}}_{{\text{SO}}_{\text{2}}}$ has to be calculated.

Concept Introduction:

Equilibrium constant:

The relationship between the concentration of products and concentration of reactants in a chemical reaction at equilibrium is said to be equilibrium constant.  It is denoted by $\text{K}$.

For a reaction,

  $\text{xX + yY }\rightleftharpoons \text{ zZ}$

The expression of $\text{K}$ can be given as

  $\begin{array}{l}{\text{K}}_{\text{c}}\text{ = }\frac{{\text{[Z]}}^{\text{z}}}{{\text{[X]}}^{\text{x}}{\text{[Y]}}^{\text{y}}}\\ \\ \text{where,}\text{[X] = equilibrium concentration of X}\\ \\ \text{[Y] = equilibrium concentration of Y}\\ \\ \text{[Z] = equilibrium concentration of Z}\end{array}$

Pressure can be calculated using ideal gas equation.

  $\begin{array}{l}\text{PV =nRT}\\ \\ \text{P =}\frac{\text{nRT}}{\text{V}}\\ \\ \text{where,}\text{P = Pressure}\\ \\ \text{V = Volume}\\ \\ \text{R = universal gas constant}\\ \\ \text{T = temperature}\\ \\ \text{n = number of moles}\end{array}$

Explanation of Solution

The given reaction is

  ${\text{2SO}}_{\text{2(g)}}{\text{+ O}}_{\text{2(g)}}\rightleftharpoons {\text{2SO}}_{\text{3(g)}}$

Initial:

Concentration of ${\text{2SO}}_{\text{2(g)}}$ is $\text{0}\text{.0040}$.

Concentration of ${\text{O}}_{\text{2(g)}}$ is $\text{0}\text{.0028}$.

Concentration of ${\text{2SO}}_{\text{3(g)}}$ is zero.

Change:

Concentration of ${\text{2SO}}_{\text{2(g)}}$ is $\text{-2x}$.

Concentration of ${\text{O}}_{\text{2(g)}}$ is $\text{-x}$.

Concentration of ${\text{2SO}}_{\text{3(g)}}$ is $\text{+2x}$.

At equilibrium:

Concentration of ${\text{2SO}}_{\text{2(g)}}$ is $\text{0}\text{.0040-2x}$.

Concentration of ${\text{O}}_{\text{2(g)}}$ is $\text{0}\text{.0028-x}$.

Concentration of ${\text{2SO}}_{\text{3(g)}}$ is $\text{2x}$.

Given, at equilibrium, $\text{0}{\text{.0020 mol of SO}}_{\text{3(g)}}$.

  $\begin{array}{l}\text{2x = 0}\text{.0020}\\ \text{x = 0}\text{.0010}\end{array}$

By substituting the $\text{x}$ value, concentrations of reactants and products at equilibrium are calculated.

Concentration of ${\text{SO}}_{\text{2}}$:

  $\begin{array}{l}{\text{[SO}}_{\text{2}}\text{]}\text{= 0}\text{.0040-2x}\\ \\ \text{= 0}\text{.0040-2(0}\text{.0010)}\\ \\ \text{= 0}\text{.0020 M}\end{array}$

Concentration of ${\text{O}}_{\text{2}}$:

  $\begin{array}{l}{\text{[O}}_{\text{2}}\text{]}\text{= 0}\text{.0028-x}\\ \\ \text{= 0}\text{.0028-0}\text{.0010}\\ \\ \text{= 0}\text{.0018 M}\end{array}$

Concentration of ${\text{SO}}_{\text{3}}$:

  $\begin{array}{l}{\text{[SO}}_{\text{3}}\text{]}\text{= 2x}\\ \\ \text{= 2(0}\text{.0010)}\\ \\ \text{= 0}\text{.0020 M}\end{array}$

By using the obtained concentration values at equilibrium, ${\text{K}}_{\text{c}}$ is calculated.

  $\begin{array}{l}{\text{K}}_{\text{c}}\text{ = }\frac{{{\text{[SO}}_{\text{3}}\text{]}}^{\text{2}}}{{{\text{[SO}}_{\text{2}}\text{]}}^{\text{2}}{\text{[O}}_{\text{2}}\text{]}}\\ \\ {\text{K}}_{\text{c}}\text{ = }\frac{{\text{[0}\text{.0020]}}^{\text{2}}}{{\text{[0}\text{.0020]}}^{\text{2}}\text{[0}\text{.0018]}}\\ \\ {\text{K}}_{\text{c}}\text{ = 555}\text{.5556}\end{array}$

At $\text{1000}\text{.K}$, ${\text{P}}_{{\text{SO}}_{\text{2}}}$ can be calculated by ideal gas equation.

  $\begin{array}{l}{\text{P}}_{{\text{SO}}_{\text{2}}}\text{ = }\frac{\text{nRT}}{\text{V}}\\ \\ {\text{P}}_{{\text{SO}}_{\text{2}}}\text{ = }\frac{\text{(0}\text{.0020 mol)}\left(\text{0}\text{.0821 L}\text{.atm/mol}\text{.K}\right)\text{(1000}\text{.K)}}{\text{(1}\text{.0 L)}}\\ \\ {\text{P}}_{{\text{SO}}_{\text{2}}}\text{ = 0}\text{.1642 atm}\end{array}$

At $\text{1000 K}$, ${\text{K}}_{\text{c}}{\text{ and P}}_{{\text{SO}}_{\text{2}}}$ are

  $\begin{array}{l}{\text{K}}_{\text{c}}\text{ = 555}\text{.5556}\\ \\ {\text{P}}_{{\text{SO}}_{\text{2}}}\text{= 0}\text{.1642 atm}\end{array}$