Question

The complete combustion of methane, \(\mathrm{CH}_{4}(g)\), to form \(\mathrm{H}_{2} \mathrm{O}(l)\) and \(\mathrm{CO}_{2}(g)\) at constant pressure releases \(890 \mathrm{~kJ}\) of heat per mole of \(\mathrm{CH}_{4}\). (a) Write a balanced thermochemical equation for this reaction. (b) Draw an enthalpy diagram for the reaction.

Short answer

(a) The balanced thermochemical equation for the complete combustion of methane is: \( \mathrm{CH}_{4}(g) + 2\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g) + 2\mathrm{H}_{2}\mathrm{O}(l) \hspace{10mm} \Delta H = -890 \mathrm{~kJ} \). (b) To draw the enthalpy diagram, follow these steps: 1. Label the y-axis as "Enthalpy, H". 2. Draw a horizontal line representing the reactants, and label it "Reactants: \(\mathrm{CH}_4(g) + 2\mathrm{O}_2(g)\)". 3. Draw a lower horizontal line representing the products, and label it "Products: \(\mathrm{CO}_2(g) + 2\mathrm{H}_2\mathrm{O}(l)\)". 4. Connect the reactants and products lines with an arrow pointing downwards to illustrate that heat is released during the reaction. 5. Label the arrow with the heat released: "\(\Delta H = -890 \mathrm{~kJ}\)".

Step-by-step solution

(a) Balanced Thermochemical Equation

The complete combustion of methane involves the reaction of methane with oxygen to produce carbon dioxide and water. We start by writing the unbalanced equation: \( \mathrm{CH}_{4}(g) + \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g) + \mathrm{H}_{2}\mathrm{O}(l) \) In order to balance the equation, we need to make sure that the number of atoms of each element on the reactant and product sides matches. We have 1 carbon atom, 4 hydrogen atoms, and 2 oxygen atoms on the reactant side, and 1 carbon atom, 2 hydrogen atoms, and 3 oxygen atoms on the product side. To balance the equation, we will perform the following steps: 1. Balance the hydrogen atoms by placing a coefficient of 2 in front of \(\mathrm{H}_{2}\mathrm{O}(l)\): \( \mathrm{CH}_{4}(g) + \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g) + 2\mathrm{H}_{2}\mathrm{O}(l) \) 2. Balance the oxygen atoms by placing a coefficient of 2 in front of \(\mathrm{O}_{2}(g)\): \( \mathrm{CH}_{4}(g) + 2\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g) + 2\mathrm{H}_{2}\mathrm{O}(l) \) Now that the chemical equation is balanced, we can include the enthalpy change information: \( \mathrm{CH}_{4}(g) + 2\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g) + 2\mathrm{H}_{2}\mathrm{O}(l) \hspace{10mm} \Delta H = -890 \mathrm{~kJ} \)

(b) Enthalpy Diagram

To draw the enthalpy diagram, we will represent the reactants and products with horizontal lines, where the heights of the lines represent their respective enthalpy levels. The difference between the heights of the reactants and products lines reflects the amount of heat released during the reaction. 1. Label the y-axis as "Enthalpy, H". 2. Draw a horizontal line representing the reactants, and label it "Reactants: \(\mathrm{CH}_4(g) + 2\mathrm{O}_2(g)\)". 3. Draw a lower horizontal line representing the products, and label it "Products: \(\mathrm{CO}_2(g) + 2\mathrm{H}_2\mathrm{O}(l)\)". 4. Connect the reactants and products lines with an arrow pointing downwards to illustrate that heat is released during the reaction. 5. Label the arrow with the heat released: "\(\Delta H = -890 \mathrm{~kJ}\)". The enthalpy diagram should look like: [![Enthalpy Diagram][1]][1] [1]: https://i.stack.imgur.com/k5SRk.png

Key concepts

Combustion Reaction

A combustion reaction is a type of chemical reaction where a substance combines with oxygen to produce heat and usually light. The combustion of methane (\( \mathrm{CH}_4(g) \)) is a classic example of this process. Here, methane burns in oxygen \( \mathrm{O}_2(g) \) to form carbon dioxide \( \mathrm{CO}_2(g) \) and water \( \mathrm{H}_2O(l) \).
In combustion, the substance that burns is called the fuel. The products of complete combustion of hydrocarbons like methane are carbon dioxide and water. Incomplete combustion might produce carbon monoxide or carbon (soot) when there isn't enough oxygen.
Understanding this concept helps in multiple applications, from energy production to transportation. For thermochemical calculations, the key feature is the heat released, which can be measured as part of the energy output of the reaction.

Enthalpy Change

Enthalpy change, denoted by \( \Delta H \), represents the heat absorbed or released during a chemical reaction at constant pressure. In thermochemistry, it is crucial to know whether a reaction releases or absorbs heat.
For the complete combustion of methane, the reaction is exothermic, meaning it releases heat. The enthalpy change is negative: \( \Delta H = -890 \, \mathrm{kJ/mol} \). This indicates the reaction releases 890 kJ of energy per mole of methane combusted, showcasing its energy output.

Enthalpy change is essential for understanding energy management in chemical processes. For instance, power plants rely on burning fuels like methane, and knowing the \( \Delta H \) helps calculate efficiency and optimize energy production.

Balanced Chemical Equation

A balanced chemical equation provides a clear representation of the chemical reaction, ensuring that the number of atoms for each element are equal on both sides of the equation. This follows the law of conservation of mass where matter is neither created nor destroyed.

To balance the combustion reaction of methane:

• Write the initial equation: \( \mathrm{CH}_4(g) + \mathrm{O}_2(g) \longrightarrow \mathrm{CO}_2(g) + \mathrm{H}_2O(l) \).
• Balance the hydrogen atoms by adding a coefficient of 2 before \( \mathrm{H}_2O \), matching the 4 hydrogen atoms from methane: \( \mathrm{CH}_4(g) + \mathrm{O}_2(g) \longrightarrow \mathrm{CO}_2(g) + 2\mathrm{H}_2O(l) \).
• Balance the oxygen by placing a coefficient of 2 in front of \( \mathrm{O}_2 \), aligning with the total 4 oxygen atoms needed for carbon dioxide and water: \( \mathrm{CH}_4(g) + 2\mathrm{O}_2(g) \longrightarrow \mathrm{CO}_2(g) + 2\mathrm{H}_2O(l) \).

Including the enthalpy change in the balanced equation helps provide a complete picture of the reaction:\( \Delta H = -890 \, \mathrm{kJ/mol} \) reflects the energy dynamics involved.