Chapter 6, Problem 6.75P

(a)

Interpretation Introduction

Interpretation: An overall reaction for the production of methane needs to be determined.

Concept Introduction : The reaction of coal with water is as follows:

  $\text{C}\text{(coal)}\text{+}{\text{H}}_{\text{2}}\text{O}\text{(g)}\to \text{CO}\text{(g)}\text{+}{\text{H}}_{\text{2}}\text{(g)}$

Here, coal reacts with superheated steam to give CO and ${\text{H}}_{\text{2}}\text{(g)}$ .

Answer to Problem 6.75P

Overall reaction is $\text{2C}\text{(coal)}\text{+}2{\text{H}}_{\text{2}}\text{O}\text{(g)}\to {\text{CH}}_4\text{(g)}\text{+}{\text{CO}}_{\text{2}}\text{(g)}$

Explanation of Solution

  $\text{C}\text{(coal)}\text{+}{\text{H}}_{\text{2}}\text{O}\text{(g)}\to \text{CO(g)}\text{+}{\text{H}}_{\text{2}}\text{(g)}\Delta H^0{}_{rxn}\text{=}\text{129}\text{.7}\text{kJ}$ …… (1)

  $\text{C}\text{O}\text{+}{\text{H}}_{\text{2}}\text{O}\text{(g)}\to \text{CO(g)}\text{+}{\text{H}}_{\text{2}}\text{(g)}\Delta H^0{}_{rxn}\text{=}-41\text{kJ}$ …… (2)

  $\text{C}\text{O(g)}\text{+}3{\text{H}}_{\text{2}}\text{(g)}\to {\text{CH}}_4\text{(g)}\text{+}{\text{H}}_{\text{2}}\text{O(g)}\mathrm{.............}\Delta H^0{}_{rxn}\text{=}-206\text{kJ}$ …… (3)

Multiply equation (1) by 2 and add equation (1) and (2) and (3).

  $2\times \{\text{C}\text{(coal)}\text{+}{\text{H}}_{\text{2}}\text{O}\text{(g)}\to \cancel{\text{CO(g)}}\text{+}\cancel{{\text{H}}_{\text{2}}\text{(g)}}.\}\mathrm{............}\Delta H^0{}_{rxn}\text{=2}\times \text{(129}\text{.7}\text{kJ)}$ …… (1)

  $\cancel{\text{C}\text{O}(g)}\text{+}\cancel{{\text{H}}_{\text{2}}\text{O}\text{(g)}}\to {\text{CO}}_2\text{(g)}\text{+}\cancel{{\text{H}}_{\text{2}}\text{(g)}}\mathrm{.............}\Delta H^0{}_{rxn}=-41\text{kJ}$ …… (2)

  $\cancel{\text{C}\text{O}\text{(g)}}\text{+}\cancel{3{\text{H}}_{\text{2}}\text{(g)}}\to {\text{CH}}_4\text{(g)}\text{+}\cancel{{\text{H}}_{\text{2}}\text{O}\text{(g}}\text{)}\mathrm{.............}\Delta H^0{}_{rxn}\text{=}-206\text{kJ}$ ..... (3)

Overall reaction is $\text{2C}\text{(coal)}\text{+}2{\text{H}}_{\text{2}}\text{O}\text{(g)}\to {\text{CH}}_4\text{(g)}\text{+}{\text{CO}}_{\text{2}}\text{(g)}$

(b)

Interpretation Introduction

Interpretation: The value of $\Delta H^0{}_{rxn}$ for the overall reaction needs to be determined.

Concept Introduction: $\Delta H^0{}_{rxn}$ = Sum of $\Delta H^0{}_{rxn}$ of all the reactions.

  $\begin{array}{l}\text{Reactions for the production of mathane are as follows:}\\ \\ \text{ C}\text{(coal)}\text{+}{\text{H}}_{\text{2}}\text{O}\text{(g)}\to \text{CO(g)}\text{+}{\text{H}}_{\text{2}}\text{(g)}\mathrm{.............}\Delta H^0{}_{rxn}\text{=}\text{129}\text{.7}\text{kJ}\\ \text{C}\text{O}\text{+}{\text{H}}_{\text{2}}\text{O}\text{(g)}\to \text{CO(g)}\text{+}{\text{H}}_{\text{2}}\text{(g)}\mathrm{.............}\Delta H^0{}_{rxn}\text{=}-41\text{kJ}\\ \text{C}\text{O(g)}\text{+}3{\text{H}}_{\text{2}}\text{(g)}\to {\text{CH}}_4\text{(g)}\text{+}{\text{H}}_{\text{2}}\text{O(g)}\mathrm{.............}\Delta H^0{}_{rxn}\text{=}-206\text{kJ}\end{array}$

Answer to Problem 6.75P

  $\Delta H^0{}_{rxn}$ of the overall reaction = 12.4 kJ

Explanation of Solution

  $\text{C}\text{(coal)}\text{+}{\text{H}}_{\text{2}}\text{O}\text{(g)}\to \text{CO(g)}\text{+}{\text{H}}_{\text{2}}\text{(g)}\Delta H^0{}_{rxn}\text{=}\text{129}\text{.7}\text{kJ}$ …… (1)

  $\text{C}\text{O}\text{+}{\text{H}}_{\text{2}}\text{O}\text{(g)}\to \text{CO(g)}\text{+}{\text{H}}_{\text{2}}\text{(g)}\Delta H^0{}_{rxn}\text{=}-41\text{kJ}$ …… (2)

  $\text{C}\text{O(g)}\text{+}3{\text{H}}_{\text{2}}\text{(g)}\to {\text{CH}}_4\text{(g)}\text{+}{\text{H}}_{\text{2}}\text{O(g)}\mathrm{.............}\Delta H^0{}_{rxn}\text{=}-206\text{kJ}$ …… (3)

Multiply equation (1) by 2 and add equation (1) and (2) and (3).

  $2\times \{\text{C}\text{(coal)}\text{+}{\text{H}}_{\text{2}}\text{O}\text{(g)}\to \cancel{\text{CO(g)}}\text{+}\cancel{{\text{H}}_{\text{2}}\text{(g)}}.\}\mathrm{............}\Delta H^0{}_{rxn}\text{=2}\times \text{(129}\text{.7}\text{kJ)}$ …… (1)

  $\cancel{\text{C}\text{O}(g)}\text{+}\cancel{{\text{H}}_{\text{2}}\text{O}\text{(g)}}\to {\text{CO}}_2\text{(g)}\text{+}\cancel{{\text{H}}_{\text{2}}\text{(g)}}\mathrm{.............}\Delta H^0{}_{rxn}=-41\text{kJ}$ …… (2)

  $\cancel{\text{C}\text{O}\text{(g)}}\text{+}\cancel{3{\text{H}}_{\text{2}}\text{(g)}}\to {\text{CH}}_4\text{(g)}\text{+}\cancel{{\text{H}}_{\text{2}}\text{O}\text{(g}}\text{)}\mathrm{.............}\Delta H^0{}_{rxn}\text{=}-206\text{kJ}$ ..... (3)

  $\begin{array}{c}{(}_{\Delta }=\text{2}\times \text{(129}\text{.7}\text{kJ)}+(-41\text{kJ)}\text{+}(-206\text{kJ)}\\ \text{=}\text{12}\text{.4 kJ}\end{array}$

(c)

Interpretation Introduction

Interpretation: The total heat for gasifying 1.00 kg of coal needs to be determined.

Concept Introduction : The balanced reaction for combustion of methane is:

  ${\text{CH}}_{\text{4}}\text{(g)}\text{+}\text{2}{\text{O}}_{\text{2}}\text{(g)}\to {\text{CO}}_{\text{2}}\text{(g)}\text{+}{\text{2H}}_{\text{2}}\text{O}\text{(g)}$

Methane reacts with oxygen to give carbon dioxide and water.

  $\Delta H^0{}_{rxn}=\sum m\Delta H^0{}_f(\text{products})-\sum n\Delta H^0{}_f(\text{reactants})$

Here, m and n are moles of products and reactants.

Answer to Problem 6.75P

The total heat is $\text{=}-\text{3}\text{.29}\text{×}{\text{10}}^{\text{4}}\text{kJ}$

Explanation of Solution

The balanced reaction for combustion of methane is:

  ${\text{CH}}_{\text{4}}\text{(g)}\text{+}\text{2}{\text{O}}_{\text{2}}\text{(g)}\to {\text{CO}}_{\text{2}}\text{(g)}\text{+}{\text{2H}}_{\text{2}}\text{O}\text{(g)}$

  $\begin{array}{c}\Delta H^0{}_{rxn}=\left[1\text{mol}\Delta H^0{}_f{\text{(CO}}_{\text{2}}\text{(g)})+2\text{mol}\Delta H^0{}_f({\text{H}}_{\text{2}}\text{O(g}))\right]-\left[\begin{array}{l}1\text{mol}\Delta H^0{}_f({\text{CH}}_{\text{4}}\text{(g}))+\\ 2\text{mol}\Delta H^0{}_f({\text{O}}_{\text{2}}\text{(g)})\end{array}\right]\\ =\left[1\cancel{\text{mol}}\times \left(-393.5\frac{\text{kJ}}{\cancel{\text{mol}}}\right)+2\cancel{\text{mol}}\times \left(-241.8\frac{\text{kJ}}{\cancel{\text{mol}}}\right)\right]-\left[\begin{array}{l}1\cancel{\text{mol}}\times \left(-74.9\frac{\text{kJ}}{\cancel{\text{mol}}}\right)\\ +2\text{mol}\times 0\end{array}\right]\\ =\left[-\text{393}\text{.5}\text{kJ}-\text{483}\text{.6}\text{kJ}\right]-\left[-\text{74}\text{.9}\text{kJ}\right]\\ \text{=}-\text{877}\text{.1}\text{kJ}\text{+}\text{74}\text{.9}\text{kJ}\\ \text{=}-\text{802}\text{.2}\text{kJ}\end{array}$

  $\begin{array}{l}(\Delta H^0{}_{rxn})\text{for}\text{equation}\text{(1)}\text{=}-\text{657}\text{.0}\text{kJ}\\ (\Delta H^0{}_{rxn})\text{for}\text{equation}\text{(2)}\text{=}\text{32}\text{.9}\text{kJ}\end{array}$

  ${\text{Therefore, ΔH}}^{\text{0}}{}_{\text{rxn}}\text{for}\text{the}\text{combustion}\text{of}{\text{CH}}_{\text{4}}\text{per}\text{mole}\text{is}\text{-802}\text{.2}\text{kJ}$ .

  $\begin{array}{l}\text{Now,}\\ \text{1}\text{kg}\text{=}{\text{10}}^{\text{3}}\text{g}\\ \text{2}\text{mol}\text{C}\text{=}\text{12}\text{.4}\text{kJ}\\ \text{1}\text{mol}\text{C}\text{=}\text{12}\text{.00}\text{g}\\ \text{The}\text{conversion}\text{faactors}\text{are}\text{as}\text{follows:}\\ \frac{{10}^3\text{g}}{1\text{kg}},\frac{\text{1}\text{mol}\text{C}}{\text{12}\text{.00}\text{g}}\text{and}\frac{\text{12}\text{.4}\text{kJ}}{\text{2}\text{mol}\text{C}}\\ \text{Using}\text{the}\text{conversion}\text{factors}\text{to}\text{determine}\text{the}\text{heat}\text{for}\text{1}\text{.00}\text{kg}\text{of}\text{coal}\text{we}\text{get,}\\ \text{1}\text{.00}\text{kg}\text{×}\frac{{\text{10}}^{\text{3}}\text{g}}{\text{1}\text{kg}}\text{×}\frac{\text{1}\text{mol}\text{C}}{\text{12}\text{.00}\text{g}}\text{×}\frac{\text{12}\text{.4}\text{kJ}}{\text{2}\text{mol}\text{C}}\text{=}\text{517}\text{kJ}\\ \text{Thus, the heat for gasifying 1}\text{.00 kg of coal is 517 kJ}\end{array}$

Now use following conversion factors to determine the heat for combustion of 1.00 kg of methane.

  $\begin{array}{l}\text{1}\text{kg}\text{=}{\text{10}}^{\text{3}}\text{g}\\ 1\text{mol}{\text{CH}}_4\text{=}-802.2\text{kJ}\\ \text{1}\text{mol}{\text{CH}}_4\text{=}16.05\text{g}\end{array}$

The conversion factors are as follows:

  $\frac{{\text{10}}^{\text{3}}\text{g}}{\text{1}\text{kg}}\text{,}\frac{\text{1}\text{mol}\text{C}}{\text{12}\text{.00}\text{g}}\text{,}\frac{\text{1}\text{mol}\text{of}{\text{CH}}_{\text{4}}}{\text{2}\text{mol}\text{Coal}}\text{and}\frac{\text{-802}\text{.2}\text{kJ}}{\text{2}\text{mol}{\text{CH}}_{\text{4}}}$

Use the conversion factors to determine the heat for 1.00 kg of coal.

  $\text{1}\text{.00}\cancel{\text{kg}}\text{×}\frac{{\text{10}}^{\text{3}}\cancel{\text{g}}}{\text{1}\cancel{\text{kg}}}\text{×}\frac{\text{1}\text{mol}\text{Coal}}{\text{12}\text{.00}\text{g}}\text{×}\frac{\text{1}\cancel{\text{mol}\text{of}{\text{CH}}_{\text{4}}}}{\text{2}\text{mol}\text{Coal}}\times \frac{\text{-802}\text{.2}\text{kJ}}{\text{2}\text{mol}{\text{CH}}_{\text{4}}}\text{=}-33425\text{kJ}$

Thus, the heat for combustion of 1.00 kg of methane is $-33425\text{kJ}$ .

  $\begin{array}{c}\text{Total}\text{heat}\text{=}\text{Heat}\text{for}\text{gasifying}\text{of}\text{1}\text{.00}\text{kg}\text{coal}\text{+}\text{Combustion}\text{of}\text{1}\text{.0}\text{kg}\text{of}\text{methane}\\ \text{=}\text{517}\text{kJ}\text{+}\text{(}-\text{33425}\text{kJ)}\\ \text{=}-\text{32905}\text{kJ}\\ \text{=}-\text{3}\text{.29}\text{×}\text{10}{}^{\text{4}}\text{kJ}\end{array}$

Thus, the total heat is $-\text{3}\text{.29}\text{×}\text{10}{}^{\text{4}}\text{kJ}$ .