Question

Write a balanced equation for the combustion of gaseous methane?

Another potential future fuel is methanol (CH3OH). Write a balanced equation for the combustion of gaseous methanol?

Express your answer as a chemical formula.

 

Another potential future fuel is methanol (CH3OH). Use bond energies to calculate the enthalpy of combustion of methanol in kJ/mol.

Express your answer in kiloJoules to three significant figures.

 

Use bond energies to calculate AH rxn for this reaction: N2(g) + 3H2(g) rightarrow 2NH3(g).

Express your answer in kiloJoules to two significant figures.

Answer 1

Concepts and reason

The change in energy $\left( {\Delta H} \right)$ during phase change or during a reaction is known as Enthalpy. When energy is absorbed during reaction or phase change the sign for enthalpy is taken as positive and when energy is released during reaction or phase change the sign for enthalpy is taken as negative.

In the given question, you need to determine $\Delta H$ for the combustion reaction.

Fundamentals

When an organic substance is burnt in presence of air it always produces carbon dioxide and water. This reaction is exothermic in nature as heat is evolved during the reaction.

Part A

The reaction is as follows:

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

Part B

The reaction for the combustion of one mole of methanol is as follows:

${\rm{C}}{{\rm{H}}_3}{\rm{OH}}\left( g \right) + \frac{3}{2}{{\rm{O}}_2}\left( g \right) \to {\rm{C}}{{\rm{O}}_{\rm{2}}}\left( g \right) + 2{{\rm{H}}_{\rm{2}}}{\rm{O}}\left( g \right)$

Now, the enthalpy of the reaction is calculated as follows:

$\begin{array}{c}\\\Delta {H_{{\rm{rxn}}}} = {\rm{Bond}}\;{\rm{energies}}\;{\rm{of}}\;{\rm{all}}\;{\rm{the}}\;{\rm{reactants}} - {\rm{Bond}}\;{\rm{energies}}\;{\rm{of}}\;{\rm{all}}\;{\rm{the}}\;{\rm{products}}\\\\ = \left[ {3\left( {1\;{\rm{mol}}} \right)\left( {{\rm{C}} - {\rm{H}}} \right) + 1\left( {1\;{\rm{mol}}} \right)\left( {{\rm{C}} - {\rm{O}}} \right) + 1\left( {1\;{\rm{mol}}} \right)\left( {{\rm{O}} - {\rm{H}}} \right) + 1\left( {\frac{3}{2}\;{\rm{mol}}} \right)\left( {{\rm{O}} = {\rm{O}}} \right)} \right] - \\\\\left[ {2\left( {1\;{\rm{mol}}} \right)\left( {{\rm{C}} = {\rm{O}}} \right) + 2\left( {2\;{\rm{mol}}} \right)\left( {{\rm{O}} - {\rm{H}}} \right)} \right]\\\\ = \left[ {3\left( {1\;{\rm{mol}}} \right)\left( {414\;{\rm{kJ/mol}}} \right) + 1\left( {1\;{\rm{mol}}} \right)\left( {360\;{\rm{kJ/mol}}} \right) + 1\left( {1\;{\rm{mol}}} \right)\left( {{\rm{464}}\;{\rm{kJ/mol}}} \right) + 1\left( {\frac{3}{2}\;{\rm{mol}}} \right)\left( {498\;{\rm{kJ/mol}}} \right)} \right] - \\\\\left[ {2\left( {1\;{\rm{mol}}} \right)\left( {799\;{\rm{kJ/mol}}} \right) + 2\left( {2\;{\rm{mol}}} \right)\left( {464\;{\rm{kJ/mol}}} \right)} \right]\\\\ = - 641\;{\rm{kJ}}\\\end{array}$

Part C

The reaction is as follows:

${{\rm{N}}_2}\left( g \right) + 3{{\rm{H}}_2}\left( g \right) \to 2{\rm{N}}{{\rm{H}}_3}\left( g \right)$

Now, the enthalpy of the reaction is calculated as follows:

$\begin{array}{c}\\\Delta {H_{{\rm{rxn}}}} = {\rm{Bond}}\;{\rm{energies}}\;{\rm{of}}\;{\rm{all}}\;{\rm{the}}\;{\rm{reactants}} - {\rm{Bond}}\;{\rm{energies}}\;{\rm{of}}\;{\rm{all}}\;{\rm{the}}\;{\rm{products}}\\\\ = \left[ {1\left( {1\;{\rm{mol}}} \right)\left( {{\rm{N}} \equiv {\rm{N}}} \right) + 1\left( {3\;{\rm{mol}}} \right)\left( {{\rm{H}} - {\rm{H}}} \right)} \right] - \left[ {3\left( {2\;{\rm{mol}}} \right)\left( {{\rm{N}} - {\rm{H}}} \right)} \right]\\\\ = \left[ {1\left( {1\;{\rm{mol}}} \right)\left( {946\;{\rm{kJ/mol}}} \right) + 1\left( {3\;{\rm{mol}}} \right)\left( {436\;{\rm{kJ/mol}}} \right)} \right] - \left[ {3\left( {2\;{\rm{mol}}} \right)\left( {389\;{\rm{kJ/mol}}} \right)} \right]\\\\ = - 80\;{\rm{kJ}}\\\end{array}$

Ans: Part A

The balanced combustion reaction of methanol is as follows:

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

Part B

The enthalpy of the reaction is $- 641\;{\rm{kJ}}$ for one mole of methanol.

Part C

The enthalpy of the reaction is $- 80\;{\rm{kJ}}$ .