Chapter 17, Problem 17.109P

(a)

Interpretation Introduction

Interpretation:

The ${\text{K}}_{\text{c}}$ for the system at temperature ${\text{900}}^{\text{o}}\text{ C}$ 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

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

Explanation of Solution

The reaction for the given system will be:

    ${\text{CO(g) + H}}_{\text{2}}\text{O(g) }\underset{}{\rightleftharpoons }{\text{ CO}}_{\text{2}}{\text{(g) + H}}_{\text{2}}\text{(g)}$

Initial ${\text{[CO] and initial [H}}_{\text{2}}\text{O] = 0}\text{.100mol/20}\text{.00L = 0}\text{.00500M}$

$\begin{array}{l}{\text{ CO H}}_{\text{2}}\text{O }\underset{}{\rightleftharpoons }{\text{ CO}}_{\text{2}}{\text{ H}}_{\text{2}}\\ \text{Initial 0}\text{.00500M 0}\text{.00500M 0 0}\\ \text{Change -x -x +x +x}\\ \text{-----------------------------------------------------------------------------------}\\ \text{Equilibrium 0}\text{.00500-x 0}\text{.00500-x x x }\\ {\text{[CO]}}_{\text{equlibrium}}\text{= 0}\text{.00500-x = 2}{\text{.24×10}}^{\text{-3}}{\text{M = [H}}_{\text{2}}\text{O]}\\ \text{ x = 0}{\text{.00276 M = [CO}}_{\text{2}}{\text{] = [H}}_{\text{2}}\text{]}\\ {\text{ K}}_{\text{c}}\text{= }\frac{{\text{[CO}}_{\text{2}}{\text{][H}}_{\text{2}}\text{]}}{{\text{[CO][H}}_{\text{2}}\text{O]}}\text{ = }\frac{\text{[0}\text{.00276][0}\text{.00276]}}{\text{[0}\text{.00224][0}\text{.00224]}}\text{ = 1}\text{.518176 = 1}\text{.52}\end{array}$

(b)

Interpretation Introduction

Interpretation:

${\text{P}}_{\text{total}}$ in the flask at equilibrium has to be calculated.

Explanation of Solution

The reaction for the given system will be:

    ${\text{CO(g) + H}}_{\text{2}}\text{O(g) }\underset{}{\rightleftharpoons }{\text{ CO}}_{\text{2}}{\text{(g) + H}}_{\text{2}}\text{(g)}$

    $\begin{array}{l}{\text{M}}_{\text{total}}{\text{= [CO] + [H}}_{\text{2}}{\text{O] + [CO}}_{\text{2}}{\text{] + [H}}_{\text{2}}\text{]}\\ \\ \text{ = (0}\text{.00224 M) + (0}\text{.00224 M) + (0}\text{.00276 M) + (0}\text{.00276 M)}\\ \\ \text{ = 0}\text{.01000 MM}\\ \\ {\text{n}}_{\text{total}}{\text{ = (M}}_{\text{total}}\text{)(V) = (0}\text{.01000 mol/L)(20}\text{.00L) = 0}\text{.2000 mol total}\\ \\ \text{PV = nRT (ideal gas equation)}\\ \\ {\text{P}}_{\text{total}}{\text{= n}}_{\text{total}}\text{RT/V = }\frac{\text{(0}\text{.2000mol)(0}\text{.0806}\frac{\text{Latm}}{\text{molK}}\text{)((273+900)K)}}{\text{(20}\text{.00L)}}\\ \text{ = 0}\text{.9625638 = 0}\text{.9626atm}\end{array}$

(c)

Interpretation Introduction

Interpretation:

The number of moles that must be added to double pressure has to be found.

Explanation of Solution

According to ideal gas equation $\text{PV = nRT}$

Pressure is directly proportional to number of moles.

Then equal number of moles must be added $\text{= 0}\text{.2000 mol CO}$

(d)

Interpretation Introduction

Interpretation:

After ${\text{P}}_{\text{total}}$ is doubled and the system remains equilibrium, ${\text{[CO]}}_{\text{eq}}$ has to be found.

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

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

Explanation of Solution

The reaction for the given system will be:

    ${\text{CO(g) + H}}_{\text{2}}\text{O(g) }\underset{}{\rightleftharpoons }{\text{ CO}}_{\text{2}}{\text{(g) + H}}_{\text{2}}\text{(g)}$

  $\begin{array}{l}{\text{Initial [CO] and initial [H}}_{\text{2}}\text{O] = 0}\text{.100mol/20}\text{.00L = 0}\text{.00500M}\\ \\ \text{To that add 0}\text{.2000molCO/20}\text{.00L = 0}\text{.01000 M to compensate for the added CO}\\ \end{array}$

  $\begin{array}{l}{\text{ CO H}}_{\text{2}}\text{O }\underset{}{\rightleftharpoons }{\text{ CO}}_{\text{2}}{\text{ H}}_{\text{2}}\\ \text{Initial 0}\text{.00224 M 0}\text{.00224M 0}\text{.00276 M 0}\text{.00276 M}\\ \text{Added CO 0}\text{.01000 M }\\ \text{Change -x -x +x +x}\\ \text{-----------------------------------------------------------------------------------}\\ \text{Equilibrium 0}\text{.01224-x 0}\text{.00224-x 0}\text{.00276+x 0}\text{.00276+x }\\ \\ {\text{ K}}_{\text{c}}\text{= }\frac{{\text{[CO}}_{\text{2}}{\text{][H}}_{\text{2}}\text{]}}{{\text{[CO][H}}_{\text{2}}\text{O]}}\text{ = }\frac{\text{[0}\text{.00276+x][0}\text{.00276+x]}}{\text{[0}\text{.01224-x][0}\text{.00224-x]}}\text{ = 1}\text{.518176 }\\ \\ \frac{\text{[7}{\text{.6176×10}}^{\text{-6}}\text{+5}{\text{.52×10}}^{\text{-3}}{\text{x+x}}^{\text{2}}\text{]}}{\text{[2}{\text{.7417×10}}^{\text{-5}}\text{-1}{\text{.448×10}}^{\text{-2}}{\text{x+x}}^{\text{2}}\text{]}}\text{ = 1}\text{.518176}\\ \\ \text{ 7}{\text{.6176×10}}^{\text{-6}}\text{+5}{\text{.52×10}}^{\text{-3}}{\text{x+x}}^{\text{2}}\text{= (1}\text{.518176)(2}{\text{.7417×10}}^{\text{-5}}\text{-1}{\text{.448×10}}^{\text{-2}}{\text{x+x}}^{\text{2}}\text{)}\\ \\ \text{ 0}{\text{.51817x}}^{\text{2}}\text{-0}\text{.027503x + 3}{\text{.400714×10}}^{\text{-5}}\text{= 0}\\ \\ \text{a = 0}\text{.518176 , b = -0}\text{.027503 , c = 3}{\text{.400714×10}}^{\text{-5}}\\ \\ \text{x = }\frac{\text{-b±}\sqrt{{\text{b}}^{\text{2}}\text{-4ac}}}{\text{2a}}\\ \\ \text{x = }\frac{\text{-(-0}\text{.027503) ± }\sqrt{{\text{(-0}\text{.027503)}}^{\text{2}}\text{-4(0}\text{.518176)(3}{\text{.400714×10}}^{\text{-5}}\text{)}}}{\text{2(0}\text{.518176)}}\\ \\ \text{x = 1}{\text{.31277×10}}^{\text{-3}}\\ \\ \text{[CO] = 0}\text{.01224-x = 0}\text{.01224-(1}{\text{.31277×10}}^{\text{-3}}\text{) = 0}\text{.01092723 = 0}\text{.01093 M}\end{array}$