Question

Many power plants produce energy by burning carbon-based fuels, which also produces carbon dioxide. Carbon dioxide is a greenhouse gas, so over-production can have negative effects on the environment. Use enthalpy of formation data to calculate the number of moles of CO2(g) produced per megajoule of heat released from the combustion of each fuel under standard conditions (1 atm and 25 �C).

Each must be in mol * MJ^-1

(a) coal, C(s, graphite)
(b) natural gas, CH4(g);
(c) propane, C3H8(g);
(d) octane, C8H18(l) (?Hf� = �250.1 kJ �mol^�1).

Answer 1

Concepts and reason

The problem is based on the concept of enthalpy change of a reaction. For a reaction, standard enthalpy change is equal to difference in change in enthalpy of product and reactant. Generally, for combustion reaction, standard enthalpy change is negative that is heat evolved during the reaction.

Fundamentals

For a general combustion reaction as follows:

${\mathop{\rm Reactant}\nolimits} \to {\rm{product}}$

Standard enthalpy change of the reaction is as follows:

$\Delta {H_{{\rm{reaction}}}} = \sum {\Delta {H_{{\rm{product}}}}} - \sum {\Delta {H_{{\rm{reactant}}}}}$

Here, $\sum {\Delta {H_{{\rm{product}}}}}$ sum of heat of formation of all the products, $\sum {\Delta {H_{{\rm{reactant}}}}}$ is sum of heat of formation of all the reactants and $\Delta {H_{{\rm{reaction}}}}$ enthalpy of reaction of combustion of fuel.

Part a

Combustion reaction of coal is as follows:

${\rm{C}}\left( s \right) + {{\rm{O}}_{\rm{2}}}\left( g \right) \to {\rm{C}}{{\rm{O}}_{\rm{2}}}\left( g \right)$

Calculate standard enthalpy change of the reaction as follows:

$\Delta {H_{{\rm{reaction}}}} = \Delta {H_{{\rm{C}}{{\rm{O}}_{\rm{2}}}\left( g \right)}} - \Delta {H_{{\rm{C}}\left( s \right)}} - \Delta {H_{{{\rm{O}}_{\rm{2}}}\left( g \right)}}$

Substitute -393.5 kJ for $\Delta {H_{{\rm{C}}{{\rm{O}}_{\rm{2}}}\left( g \right)}}$ , 0 kJ for $\Delta {H_{{\rm{C}}\left( s \right)}}$ and 0 kJ for $\Delta {H_{{{\rm{O}}_{\rm{2}}}\left( g \right)}}$ thus,

$\begin{array}{c}\\\Delta {H_{{\rm{reaction}}}} = - 393.5{\rm{ kJ}} - 0{\rm{ kJ}} - 0{\rm{ kJ}}\\\\{\rm{ = }} - 393.5{\rm{ kJ}}\\\end{array}$

Here, negative sign indicates heat is produced in the reaction.

From the reaction, number of moles of ${\rm{C}}{{\rm{O}}_{\rm{2}}}$ produced is 1 mol so, calculate number of moles of ${\rm{C}}{{\rm{O}}_{\rm{2}}}$ produced per MJ, ${n_{{\rm{C}}{{\rm{O}}_{\rm{2}}}}}$ of heat as follows:

${n_{{\rm{C}}{{\rm{O}}_{\rm{2}}}}} = \frac{{1{\rm{ mol C}}{{\rm{O}}_2}}}{{\Delta {H_{{\rm{reaction}}}}}}\left( {\frac{{{{10}^3}{\rm{kJ}}}}{{1{\rm{ MJ}}}}} \right)$

Substitute 393.5 kJ for $\Delta {H_{{\rm{reaction}}}}$ thus,

$\begin{array}{c}\\{n_{{\rm{C}}{{\rm{O}}_{\rm{2}}}}} = \frac{{1{\rm{ mol C}}{{\rm{O}}_2}}}{{393.5{\rm{ kJ}}}}\left( {\frac{{{{10}^3}{\rm{kJ}}}}{{1{\rm{ MJ}}}}} \right)\\\\ = 2.54{\rm{ mol M}}{{\rm{J}}^{ - 1}}\\\end{array}$

Part b

Combustion reaction of natural gas is as follows:

${\rm{C}}{{\rm{H}}_4}\left( g \right) + 2{{\rm{O}}_{\rm{2}}}\left( g \right) \to {\rm{C}}{{\rm{O}}_{\rm{2}}}\left( g \right) + 2{{\mathop{\rm H}\nolimits} _2}O\left( g \right)$

Calculate standard enthalpy change of the reaction as follows:

$\Delta {H_{{\rm{reaction}}}} = \Delta {H_{{\rm{C}}{{\rm{O}}_{\rm{2}}}\left( g \right)}} + 2\Delta {H_{{{\rm{H}}_2}{\rm{O}}\left( g \right)}} - 2\Delta {H_{{{\rm{O}}_{\rm{2}}}\left( g \right)}} - \Delta {H_{{\rm{C}}{{\rm{H}}_4}\left( g \right)}}$

Substitute -393.5 kJ for $\Delta {H_{{\rm{C}}{{\rm{O}}_{\rm{2}}}\left( g \right)}}$ , -241.8 kJ for $\Delta {H_{{{\rm{H}}_{\rm{2}}}{\rm{O}}}}$ , -74.85 kJ for $\Delta {H_{{\rm{C}}{{\rm{H}}_4}\left( g \right)}}$ and 0 kJ for $\Delta {H_{{{\rm{O}}_{\rm{2}}}\left( g \right)}}$ thus,

$\begin{array}{c}\\\Delta {H_{{\rm{reaction}}}} = - 393.5{\rm{ kJ}} + 2\left( { - 241.8{\rm{ kJ}}} \right) - 2\left( {0{\rm{ kJ}}} \right) - \left( { - 74.85{\rm{ kJ}}} \right)\\\\ = - 802.3{\rm{ kJ}}\\\end{array}$

Here, negative sign indicates heat is produced in the reaction.

From the reaction, number of moles of ${\rm{C}}{{\rm{O}}_{\rm{2}}}$ produced is 1 mol so, calculate number of moles of ${\rm{C}}{{\rm{O}}_{\rm{2}}}$ produced per MJ, ${n_{{\rm{C}}{{\rm{O}}_{\rm{2}}}}}$ of heat as follows:

${n_{{\rm{C}}{{\rm{O}}_{\rm{2}}}}} = \frac{{1{\rm{ mol C}}{{\rm{O}}_2}}}{{\Delta {H_{{\rm{reaction}}}}}}\left( {\frac{{{{10}^3}{\rm{kJ}}}}{{1{\rm{ MJ}}}}} \right)$

Substitute 802.3 KJ for $\Delta {H_{reaction}}$ thus,

$\begin{array}{c}\\{n_{{\rm{C}}{{\rm{O}}_{\rm{2}}}}} = \frac{{1{\rm{ mol C}}{{\rm{O}}_2}}}{{802.3{\rm{ kJ}}}}\left( {\frac{{{{10}^3}{\rm{kJ}}}}{{1{\rm{ MJ}}}}} \right)\\\\ = 1.24{\rm{ mol M}}{{\rm{J}}^{ - 1}}\\\end{array}$

Part c

Combustion reaction of propane is as follows:

${{\rm{C}}_3}{{\rm{H}}_8}\left( g \right) + 5{{\rm{O}}_{\rm{2}}}\left( g \right) \to 3{\rm{C}}{{\rm{O}}_{\rm{2}}}\left( g \right) + 4{{\mathop{\rm H}\nolimits} _2}O\left( g \right)$

Calculate standard enthalpy change of the reaction as follows:

$\Delta {H_{{\rm{reaction}}}} = 3\Delta {H_{{\rm{C}}{{\rm{O}}_{\rm{2}}}\left( g \right)}} + 4\Delta {H_{{{\rm{H}}_2}{\rm{O}}\left( g \right)}} - 5\Delta {H_{{{\rm{O}}_{\rm{2}}}\left( g \right)}} - \Delta {H_{{{\rm{C}}_3}{{\rm{H}}_8}\left( g \right)}}$

Substitute -393.5 kJ for $\Delta {H_{{\rm{C}}{{\rm{O}}_{\rm{2}}}\left( g \right)}}$ , -241.8 kJ for $\Delta {H_{{{\rm{H}}_{\rm{2}}}{\rm{O}}\left( g \right)}}$ , -103.8 kJ for $\Delta {H_{{{\rm{C}}_3}{{\rm{H}}_8}\left( g \right)}}$ and 0 kJ for $\Delta {H_{{{\rm{O}}_{\rm{2}}}\left( g \right)}}$ thus,

$\begin{array}{c}\\\Delta {H_{{\rm{reaction}}}} = 3\left( { - 393.5{\rm{ kJ}}} \right) + 4\left( { - 241.8{\rm{ kJ}}} \right) - 5\left( {0{\rm{ kJ}}} \right) - \left( { - 103.8{\rm{ kJ}}} \right)\\\\ = - 2043.9{\rm{ kJ}}\\\end{array}$

Here, negative sign indicates heat is produced in the reaction.

From the reaction, number of moles of ${\rm{C}}{{\rm{O}}_{\rm{2}}}$ produced is 3 mol so, calculate number of moles of ${\rm{C}}{{\rm{O}}_{\rm{2}}}$ produced per MJ, ${n_{{\rm{C}}{{\rm{O}}_{\rm{2}}}}}$ of heat as follows:

${n_{{\rm{C}}{{\rm{O}}_{\rm{2}}}}} = \frac{{3{\rm{ mol C}}{{\rm{O}}_2}}}{{\Delta {H_{{\rm{reaction}}}}}}\left( {\frac{{{{10}^3}{\rm{kJ}}}}{{1{\rm{ MJ}}}}} \right)$

Substitute 2043.9 kJ for $\Delta {H_{reaction}}$ thus,

$\begin{array}{c}\\{n_{{\rm{C}}{{\rm{O}}_{\rm{2}}}}} = \frac{{3{\rm{ mol C}}{{\rm{O}}_2}}}{{2043.9{\rm{ kJ}}}}\left( {\frac{{{{10}^3}{\rm{kJ}}}}{{1{\rm{ MJ}}}}} \right)\\\\ = 1.46{\rm{ mol M}}{{\rm{J}}^{ - 1}}\\\end{array}$

Part d

Combustion reaction of octane is as follows:

${{\rm{C}}_8}{{\rm{H}}_{18}}\left( g \right) + \frac{{25}}{2}{{\rm{O}}_{\rm{2}}}\left( g \right) \to 8{\rm{C}}{{\rm{O}}_{\rm{2}}}\left( g \right) + 9{{\mathop{\rm H}\nolimits} _2}O\left( g \right)$

Or,

${\rm{2}}{{\rm{C}}_8}{{\rm{H}}_{18}}\left( g \right) + 25{{\rm{O}}_{\rm{2}}}\left( g \right) \to 16{\rm{C}}{{\rm{O}}_{\rm{2}}}\left( g \right) + 18{{\mathop{\rm H}\nolimits} _2}O\left( g \right)$

Calculate standard enthalpy change of the reaction as follows:

$\Delta {H_{{\rm{reaction}}}} = 16\Delta {H_{{\rm{C}}{{\rm{O}}_{\rm{2}}}\left( g \right)}} + 18\Delta {H_{{{\rm{H}}_2}{\rm{O}}\left( g \right)}} - 25\Delta {H_{{{\rm{O}}_{\rm{2}}}\left( g \right)}} - 2\Delta {H_{{{\rm{C}}_8}{{\rm{H}}_{18}}\left( g \right)}}$

Substitute -393.5 kJ for $\Delta {H_{{\rm{C}}{{\rm{O}}_{\rm{2}}}\left( g \right)}}$ , -241.8 kJ for $\Delta {H_{{{\rm{H}}_{\rm{2}}}{\rm{O}}\left( g \right)}}$ , -250.1 kJ for $\Delta {H_{{{\rm{C}}_3}{{\rm{H}}_8}\left( g \right)}}$ and 0 kJ for $\Delta {H_{{{\rm{O}}_{\rm{2}}}\left( g \right)}}$ thus,

$\begin{array}{c}\\\Delta {H_{{\rm{reaction}}}} = 16\left( { - 393.5{\rm{ kJ}}} \right) + 18\left( { - 241.8{\rm{ kJ}}} \right) - 25\left( {0{\rm{ kJ}}} \right) - 2\left( { - 250.1{\rm{ kJ}}} \right)\\\\ = - 10148.2{\rm{ kJ}}\\\end{array}$

Here, negative sign indicates heat is produced in the reaction.

From the reaction, number of moles of ${\rm{C}}{{\rm{O}}_{\rm{2}}}$ produced is 16 mol so, calculate number of moles of ${\rm{C}}{{\rm{O}}_{\rm{2}}}$ produced per MJ, ${n_{{\rm{C}}{{\rm{O}}_{\rm{2}}}}}$ of heat as follows:

${n_{{\rm{C}}{{\rm{O}}_{\rm{2}}}}} = \frac{{16{\rm{ mol C}}{{\rm{O}}_2}}}{{\Delta {H_{{\rm{reaction}}}}}}\left( {\frac{{{{10}^3}{\rm{kJ}}}}{{1{\rm{ MJ}}}}} \right)$

Substitute $10148.2{\rm{ kJ}}$ for $\Delta {H_{reaction}}$ thus,

$\begin{array}{c}\\{n_{{\rm{C}}{{\rm{O}}_{\rm{2}}}}} = \frac{{16{\rm{ mol C}}{{\rm{O}}_2}}}{{10148.2{\rm{ kJ}}}}\left( {\frac{{{{10}^3}{\rm{kJ}}}}{{1{\rm{ MJ}}}}} \right)\\\\ = 1.57{\rm{ mol M}}{{\rm{J}}^{ - 1}}\\\end{array}$

Ans: Part a

Number of moles of ${\rm{C}}{{\rm{O}}_{\rm{2}}}$ produced per MJ of heat is $2.54{\rm{ mol M}}{{\rm{J}}^{ - 1}}$ .

Part b

Number of moles of ${\rm{C}}{{\rm{O}}_{\rm{2}}}$ produced per MJ of heat is $1.24{\rm{ mol M}}{{\rm{J}}^{ - 1}}$ .

Part c

Number of moles of ${\rm{C}}{{\rm{O}}_{\rm{2}}}$ produced per MJ of heat is $1.46{\rm{ mol M}}{{\rm{J}}^{ - 1}}$ .

Part d

Number of moles of ${\rm{C}}{{\rm{O}}_{\rm{2}}}$ produced per MJ of heat is $1.57{\rm{ mol M}}{{\rm{J}}^{ - 1}}$ .