Chapter 17, Problem 17.97P

(a)

Interpretation Introduction

Interpretation:

For the following reaction, the percent yield of ${\text{H}}_{\text{2}}$ when an equimolar mixture of ${\text{CH}}_{\text{4}}{\text{ and CO}}_{\text{2}}$ with a total pressure of $\text{20}\text{.0 atm}$ reaches equilibrium at $\text{1200}\text{.K}$, at which ${\text{K}}_{\text{p}}\text{ = 3}{\text{.548×10}}^{\text{6}}$ has to be found.

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,

    ${\text{P}}_{{\text{CH}}_{\text{4}}}{\text{(init) = P}}_{{\text{CO}}_{\text{2}}}\text{(init) = 10}\text{.0atm}$

$\begin{array}{l}{\text{Pressure(atm) CH}}_{\text{4}}{\text{(g) + CO}}_{\text{2}}\text{(g) }\underset{}{\rightleftharpoons }{\text{ 2CO(g) + 2H}}_{\text{2}}\text{(g)}\\ \text{Initial 10}\text{.0 10}\text{.0 0 0}\\ \text{Change -x -x +2x +2x}\\ \text{Equilibrium 10}\text{.0-x 10}\text{.0-x 2x 2x}\\ \\ {\text{K}}_{\text{p}}\text{=}\frac{{\text{P}}_{\text{CO}}^{\text{2}}{\text{P}}_{{\text{H}}_{\text{2}}}^{\text{2}}}{{\text{P}}_{{\text{CH}}_{\text{4}}}{\text{P}}_{{\text{CO}}_{\text{2}}}}\text{ = }\frac{{\text{(2x)}}^{\text{2}}{\text{(2x)}}^{\text{2}}}{\text{(10}\text{.0-x)(10}\text{.0-x)}}\text{= 3}{\text{.548×10}}^{\text{6}}\\ \text{Taking the square root on both side}\\ \frac{{\text{(2x)}}^{\text{2}}}{\text{(10}\text{.0-x)}}\text{ = 1}{\text{.8836135×10}}^{\text{3}}\\ \text{Quadratic equation:}\\ {\text{ax}}^{\text{2}}\text{+ bx + c = 0}\\ {\text{4x}}^{\text{2}}\text{ + (1}{\text{.8836135×10}}^{\text{3}}\text{x) -1}{\text{.8836135×10}}^{\text{4}}\text{= 0}\\ \text{a = 4, b = 1}{\text{.8836135×10}}^{\text{3}}\text{, c = -1}{\text{.8836135×10}}^{\text{4}}\\ \text{x = }\frac{\text{-b±}\sqrt{{\text{b}}^{\text{2}}\text{-4ac}}}{\text{2a}}\text{ = }\frac{\text{-1}{\text{.8836135×10}}^{\text{3}}\text{±}\sqrt{{\text{(1}{\text{.8836135×10}}^{\text{3}}\text{)}}^{\text{2}}\text{-4(4)(-1}{\text{.883615×10}}^{\text{4}}\text{)}}}{\text{2(4)}}\\ \text{x = 9}\text{.796209}\\ {\text{P}}_{{\text{H}}_{\text{2}}}\text{= 2x = 2(9}\text{.796209) = 19}\text{.592419 atm}\\ {\text{If the reaction proceeded to completion, the partial pressure of H}}_{\text{2}}\text{wouldbe20}\text{.0atm}\\ \text{The percent yield = }\frac{\text{19}\text{.592418atm}}{\text{20}\text{.0atm}}\text{(100\%) = 97096209 = 98}\text{.0\%}\\ \end{array}$

The percent yield = $\text{98}\text{.0\%}$

(b)

Interpretation Introduction

Interpretation:

For the following reaction, the percent yield of ${\text{H}}_{\text{2}}$ when an equimolar mixture of ${\text{CH}}_{\text{4}}{\text{ and CO}}_{\text{2}}$ with a total pressure of $\text{20}\text{.0 atm}$ reaches equilibrium at $\text{1300}\text{.K}$, at which ${\text{K}}_{\text{p}}\text{ = 2}{\text{.626 ×10}}^7$ has to be found.

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,

    ${\text{P}}_{{\text{CH}}_{\text{4}}}{\text{(init) = P}}_{{\text{CO}}_{\text{2}}}\text{(init) = 10}\text{.0atm}$

$\begin{array}{l}{\text{Pressure(atm) CH}}_{\text{4}}{\text{(g) + CO}}_{\text{2}}\text{(g) }\underset{}{\rightleftharpoons }{\text{ 2CO(g) + 2H}}_{\text{2}}\text{(g)}\\ \text{Initial 10}\text{.0 10}\text{.0 0 0}\\ \text{Change -x -x +2x +2x}\\ \text{Equilibrium 10}\text{.0-x 10}\text{.0-x 2x 2x}\\ \\ {\text{K}}_{\text{p}}\text{=}\frac{{\text{P}}_{\text{CO}}^{\text{2}}{\text{P}}_{{\text{H}}_{\text{2}}}^{\text{2}}}{{\text{P}}_{{\text{CH}}_{\text{4}}}{\text{P}}_{{\text{CO}}_{\text{2}}}}\text{ = }\frac{{\text{(2x)}}^{\text{2}}{\text{(2x)}}^{\text{2}}}{\text{(10}\text{.0-x)(10}\text{.0-x)}}\text{= 2}{\text{.626×10}}^7\\ \text{Taking the square root on both side}\\ \frac{{\text{(2x)}}^{\text{2}}}{\text{(10}\text{.0-x)}}\text{ = 5}{\text{.124451×10}}^{\text{3}}\\ \text{Quadratic equation:}\\ {\text{ax}}^{\text{2}}\text{+ bx + c = 0}\\ {\text{4x}}^{\text{2}}\text{ + (5}{\text{.124451×10}}^{\text{3}}\text{x) - 5}{\text{.124451×10}}^{\text{4}}\text{= 0}\\ \text{a = 4, b = 5}{\text{.124451×10}}^{\text{3}}\text{, c = -5}{\text{.124451×10}}^{\text{4}}\\ \text{x = }\frac{\text{-b±}\sqrt{{\text{b}}^{\text{2}}\text{-4ac}}}{\text{2a}}\text{ = }\frac{\text{-5}{\text{.124451×10}}^{\text{3}}\text{±}\sqrt{{\text{(5}{\text{.124451×10}}^{\text{3}}\text{)}}^{\text{2}}\text{-4(4)(-5}{\text{.124451×10}}^{\text{4}}\text{)}}}{\text{2(4)}}\\ \text{x = 9}\text{.923144}\\ {\text{P}}_{{\text{H}}_{\text{2}}}\text{= 2x = 2(9}\text{.923144) = 19}\text{.84629 atm}\\ {\text{If the reaction proceeded to completion, the partial pressure of H}}_{\text{2}}\text{wouldbe20}\text{.0atm}\\ \text{The percent yield = }\frac{\text{19}\text{.84629atm}}{\text{20}\text{.0atm}}\text{(100\%) = 99}\text{.23144 = 99}\text{.0\%}\\ \end{array}$

The percent yield = $\text{99}\text{.0\%}$

(c)

Interpretation Introduction

Interpretation:

${\text{ΔH}}_{\text{rxn}}^{\text{o}}$ has to be found using the van’t Hoff equation.

Concept Introduction:

van’t Hoff equation gives the effect of equilibrium constant by change in temperature.

The van’t Hoff equation is given as

  $\begin{array}{l}\text{ln}\frac{{\text{K}}_{\text{2}}}{{\text{K}}_{\text{1}}}\text{ = -}\frac{{\text{ΔH}}_{\text{r×n}}^{\text{°}}}{\text{R}}\left(\frac{\text{1}}{{\text{T}}_{\text{2}}}\text{-}\frac{\text{1}}{{\text{T}}_{\text{1}}}\right)\\ \\ \text{where, K = Rate constant}\\ \\ \text{R = Universal gas constant = 8}\text{.314 J/mol}\text{.K}\\ \\ \text{T = Temperature}\\ \\ {\text{ΔH}}_{\text{r×n}}^{\text{°}}\text{= enthalpy of the reaction}\end{array}$

Explanation of Solution

  $\begin{array}{l}{\text{K}}_{\text{1}}\text{ = 3}{\text{.548×10}}^{\text{6}}{\text{ T}}_{\text{1}}\text{=1200}{\text{.K ΔH}}_{\text{rxn}}^{\text{o}}\text{=?}\\ {\text{K}}_{\text{2}}\text{ = 2}{\text{.626×10}}^{\text{7}}{\text{ T}}_{\text{2}}\text{=1300}\text{.K R=8}\text{.314J/mol}\text{.K}\\ \text{ln}\frac{{\text{K}}_{\text{2}}}{{\text{K}}_{\text{1}}}\text{ = -}\frac{{\text{ΔH}}_{\text{rxn}}^{\text{o}}}{\text{R}}\text{(}\frac{\text{1}}{{\text{T}}_{\text{2}}}\text{-}\frac{\text{1}}{{\text{T}}_{\text{1}}}\text{)}\\ \text{ln}\frac{\text{2}{\text{.626×10}}^{\text{7}}}{\text{3}{\text{.548×10}}^{\text{6}}}\text{ = }\frac{{\text{ΔH}}_{\text{rxn}}^{\text{o}}}{\text{(8}\text{.314}\frac{\text{J}}{\text{molK}}\text{)}}\text{(}\frac{\text{1}}{\text{1200}\text{.K}}\text{-}\frac{\text{1}}{\text{1300}\text{.K}}\text{)}\\ {\text{ΔH}}_{\text{rxn}}^{\text{o}}\text{ = }\frac{\text{2}\text{.0016628}}{\text{7}{\text{.710195×10}}^{\text{-6}}}\text{ = 2}{\text{.5961247×10}}^{\text{5}}\text{ = 2}{\text{.60×10}}^{{}^{\text{5}}}\text{J/mol}\\ \end{array}$

${\text{ΔH}}_{\text{rxn}}^{\text{o}}$ for the reaction is $\text{2}{\text{.60×10}}^{\text{5}}\text{J/mol}$