Chapter 17, Problem 17.99P

(a)

Interpretation Introduction

Interpretation:

Overall reaction for the formation of carbon dioxide and hydrogen from methane, steam and oxygen has to be given.

Answer to Problem 17.99P

Overall reaction is ${\text{2CH}}_{\text{4}}\left(g\right){\text{ + O}}_{\text{2}}\left(g\right){\text{ + 2H}}_{\text{2}}\text{O}\left(g\right)\underset{}{\rightleftharpoons }\text{ 2CO}\left(g\right){\text{ + 6H}}_{\text{2}}\left(g\right)$.

Explanation of Solution

Given reaction is shown below:

  $\begin{array}{l}{\text{2CH}}_{\text{4}}\left(g\right){\text{ + O}}_{\text{2}}\left(g\right)\underset{}{\rightleftharpoons }\text{ 2CO}\left(g\right){\text{ + 4H}}_{\text{2}}\left(g\right)\left(1\right)\\ \\ \text{CO}\left(g\right){\text{ + H}}_{\text{2}}\text{O}\left(g\right)\underset{}{\rightleftharpoons }{\text{ CO}}_{\text{2}}\left(g\right){\text{ + H}}_{\text{2}}\left(g\right)\left(2\right)\end{array}$

To get the overall reaction for the formation of carbon dioxide and hydrogen from methane, steam and oxygen, reaction (1) and (2) has to be added. Reaction (2) is multiplied with $2$ for cancelling $CO\left(g\right)$ from both reactions.

Therefore,

Overall reaction for the formation of carbon dioxide and hydrogen from methane, steam and oxygen follows as,

  $\begin{array}{l}{\text{2CH}}_{\text{4}}\left(g\right){\text{ + O}}_{\text{2}}\left(g\right)\underset{}{\rightleftharpoons }\bcancel{\text{2CO}\left(g\right)}{\text{ + 4H}}_{\text{2}}\left(g\right)\\ \\ \bcancel{\text{2CO}\left(g\right)}{\text{ + 2H}}_{\text{2}}\text{O}\left(g\right)\underset{}{\rightleftharpoons }{\text{ 2CO}}_{\text{2}}\left(g\right){\text{ + 2H}}_{\text{2}}\left(g\right)\\ \underset{\_}{}\\ \\ {\text{2CH}}_{\text{4}}\left(g\right){\text{ + O}}_{\text{2}}\left(g\right){\text{ + 2H}}_{\text{2}}\text{O}\left(g\right)\underset{}{\rightleftharpoons }\text{ 2CO}\left(g\right){\text{ + 6H}}_{\text{2}}\left(g\right)\end{array}$

Overall reaction is ${\text{2CH}}_{\text{4}}\left(g\right){\text{ + O}}_{\text{2}}\left(g\right){\text{ + 2H}}_{\text{2}}\text{O}\left(g\right)\underset{}{\rightleftharpoons }\text{ 2CO}\left(g\right){\text{ + 6H}}_{\text{2}}\left(g\right)$.

(b)

Interpretation Introduction

Interpretation:

${\text{K}}_{\text{p}}$ for the overall reaction has to be determined.

Concept Introduction:

Equilibrium constant: It is the ratio of products to reactants has a constant value when the reaction is in equilibrium at a certain temperature. And it is represented by the letter K.

For a reaction,

$\text{aA}\text{+}\text{bB}\underset{}{\rightleftharpoons }\text{cC}\text{+}\text{dD}$

The equilibrium constant in terms of  partial pressure is, $K=\frac{{\left(P_{\text{C}}\right)}^{\text{c}}{\left(P_D\right)}^{\text{d}}}{{\left(P_A\right)}^{\text{a}}{\left(P_{\text{B}}\right)}^{\text{b}}}$

where,

a, b, c and d are the stoichiometric coefficients of reactant and product in the reactions

Answer to Problem 17.99P

$K_p$ for the overall reaction is $1.76\times 10^{29}.$

Explanation of Solution

Given reaction is shown below:

  $\begin{array}{l}{\text{2CH}}_{\text{4}}\left(g\right){\text{ + O}}_{\text{2}}\left(g\right)\underset{}{\rightleftharpoons }\text{ 2CO}\left(g\right){\text{ + 4H}}_{\text{2}}\left(g\right)\left(1\right)\\ \\ \text{CO}\left(g\right){\text{ + H}}_{\text{2}}\text{O}\left(g\right)\underset{}{\rightleftharpoons }{\text{ CO}}_{\text{2}}\left(g\right){\text{ + H}}_{\text{2}}\left(g\right)\left(2\right)\\ \\ K_{p1}\text{ = 9}\text{.34}\times {\text{10}}^{28}\\ \\ K_{p2}\text{ = 1}\text{.374}\end{array}$

To get the overall reaction for the formation of carbon dioxide and hydrogen from methane, steam and oxygen, reaction (1) and (2) has to be added. Reaction (2) is multiplied with $2$ for cancelling $CO\left(g\right)$ from both reactions.

Hence, $K_{p2}$ becomes $K_{p2}\text{ = }{\left(1.374\right)}^2\text{ = 1}\text{.888}$

Therefore,

$K_p$ for the overall reaction is given below,

  $\begin{array}{c}{K}_p\text{ = }{K}_{p1}{K}_{p2}\\ \\ =\left(\text{9}\text{.34}\times {\text{10}}^{28}\right)\left(1.888\right)\\ \\ =\text{ 1}\text{.76}\times {\text{10}}^{29}\end{array}$

$K_p$ for the overall reaction is $1.76\times 10^{29}.$

(c)

Interpretation Introduction

Interpretation:

${\text{K}}_{\text{C}}$ for the overall reaction has to be calculated.

Concept Introduction:

The relation between and ${\text{K}}_{\text{p}}$ and ${\text{K}}_{\text{c}}$ is given by the following equation.

${\text{K}}_{\text{p}}{\text{ = K}}_{\text{c}}{\text{(RT)}}^{\Delta {\text{n}}_{\text{gas}}}\boxed{\begin{array}{l}{\text{K}}_{\text{p}}\text{ = }Equilibriumcons\tan tintermsofpartialpressure\\ {\text{K}}_{\text{c}}\text{ = }Equilibriumcons\tan tintermsofconcentration\\ \Delta {\text{n}}_{\text{gas}}\text{ = moles of gaseous product }-\text{ moles of gaseous reactant}\end{array}}$

Only moles of gaseous products and reactants are used for calculating $\Delta {\text{n}}_{\text{gas}}.$

Answer to Problem 17.99P

${\text{K}}_{\text{C}}$ for the overall reaction is $3.19\times 10^{23}.$

Explanation of Solution

Given data is shown below:

  $\begin{array}{c}Overallreaction:{\text{2CH}}_{\text{4}}\left(g\right){\text{ + O}}_{\text{2}}\left(g\right){\text{ + 2H}}_{\text{2}}\text{O}\left(g\right)\underset{}{\rightleftharpoons }\text{ 2CO}\left(g\right){\text{ + 6H}}_{\text{2}}\left(g\right)\\ \\ {\text{K}}_P\text{ = 1}\text{.76}\times {\text{10}}^{29}\\ \\ T\text{ = 1000 K}\end{array}$

• Determine number of moles of reactants and products:

$\begin{array}{c}No.\text{ of moles of reactants = 2}{\text{.0 mol CH}}_4\text{ + 1}{\text{.0 mol O}}_2\text{ + 2}{\text{.0 mol H}}_{2}O\\ \\ =\text{ 5}\text{.0 mol}\\ \\ No.\text{ of moles of products = 2}{\text{.0 mol CO}}_2\text{ + 6}{\text{.0 mol H}}_2\\ \\ =\text{ 8}\text{.0 mol}\end{array}$

• Determine $\Delta n_{gas}$:

$\Delta n_{gas}$ is determined as follows,

  $\begin{array}{l}\Delta {\text{n}}_{\text{gas}}\text{= moles of gaseous product - moles of gaseous reactant}\\ \\ =8-5\\ \\ =3\end{array}$

• Determine ${\text{K}}_{\text{C}}$:

${\text{K}}_{\text{C}}$ of the reaction can be determined from given ${\text{K}}_{\text{p}}$ as below,

  $\begin{array}{c}{\text{K}}_{\text{p}}{\text{ = K}}_{\text{c}}{\text{(RT)}}^{\Delta {\text{n}}_{\text{gas}}}\\ \\ \text{1}\text{.76}\times {\text{10}}^{29}={\text{ K}}_{\text{c}}{\left[\left(0.0821\text{ atm}\text{.L/mol}\text{.K}\right)\left(1000\text{ K}\right)\right]}^3\\ \\ {\text{K}}_{\text{c}}=\text{ 3}\text{.19}\times {10}^{23}\end{array}$

Therefore,

${\text{K}}_{\text{C}}$ for the overall reaction is $3.19\times 10^{23}.$

(d)

Interpretation Introduction

Interpretation:

Final pressure for the given reaction has to be calculated using the given data.

Concept Introduction:

The relation between pressure and number of moles of gas is given by the following equation.

  $\begin{array}{c}\frac{n_{initial}}{P_{initial}}\text{ = }\frac{n_{final}}{P_{final}}\\ \\ where,\\ n\text{ = No}\text{. of moles}\\ P=\text{ Pressure}\end{array}$

Answer to Problem 17.99P

Final pressure for the given reaction is $48atm.$

Explanation of Solution

Given data is shown below:

  $\begin{array}{c}Overallreaction:{\text{2CH}}_{\text{4}}\left(g\right){\text{ + O}}_{\text{2}}\left(g\right){\text{ + 2H}}_{\text{2}}\text{O}\left(g\right)\underset{}{\rightleftharpoons }\text{ 2CO}\left(g\right){\text{ + 6H}}_{\text{2}}\left(g\right)\\ \\ P_{initial}\text{ = 30}\text{.0 atm reactant}\end{array}$

• Determine initial number of moles of gas:

$\begin{array}{c}Total\text{ mol of gas initially = }No.\text{ of moles of reactants }\\ \\ \text{= 2}{\text{.0 mol CH}}_4\text{ + 1}{\text{.0 mol O}}_2\text{ + 2}{\text{.0 mol H}}_{2}O\\ \\ =\text{ 5}\text{.0 mol}\end{array}$

• Determine initial number of moles of gas:

$\begin{array}{c}Total\text{ mol of gas finally = }No.\text{ of moles of products}\\ \\ \text{ = 2}{\text{.0 mol CO}}_2\text{ + 6}{\text{.0 mol H}}_2\\ \\ =\text{ 8}\text{.0 mol}\end{array}$

• Determine final pressure::

Final pressure for the given reaction is determined as follows,

  $\begin{array}{c}\frac{n_{initial}}{P_{initial}}\text{ = }\frac{n_{final}}{P_{final}}\\ \\ \frac{5.0\text{ mol}}{30.0\text{ atm}}\text{ = }\frac{8.0\text{ mol}}{P_{final}}\\ \\ P_{final}\text{ = }\left(30.0\text{ atm}\right)\left(\frac{8.0\text{ mol}}{5.0\text{ mol}}\right)\\ \\ =\text{ 48 atm}\end{array}$

Final pressure for the given reaction is $48atm.$