Question

Burning methane in oxygen can produce three different carbon-containing products: soot (very fine particles of graphite), \(\mathrm{CO}(\mathrm{g})\), and \(\mathrm{CO}_{2}(\mathrm{~g}) .\) (a) Write three balanced equations for the reaction of methane gas with oxygen to produce these three products. In each case assume that \(\mathrm{H}_{2} \mathrm{O}(l)\) is the only other product. (b) Determine the standard enthalpies for the reactions in part (a). (c) Why, when the oxygen supply is adequate, is \(\mathrm{CO}_{2}(g)\) the predominant carbon-containing product of the combustion of methane?

Short answer

The balanced equations for the combustion reactions of methane with oxygen to produce the three carbon-containing products are: 1. \( CH_{4}(g) + \frac{1}{2} O_{2}(g) \rightarrow C(graphite) + 2H_{2}O(l) \) 2. \( CH_{4}(g) + \frac{3}{2} O_{2}(g) \rightarrow CO(g) + 2H_{2}O(l) \) 3. \( CH_{4}(g) + 2 O_{2}(g) \rightarrow CO_{2}(g) + 2H_{2}O(l) \) The standard enthalpies of these reactions can be calculated by finding the difference between the sum of the enthalpies of the products and the sum of the enthalpies of the reactants. When the oxygen supply is adequate, CO2 is the predominant carbon-containing product because its formation releases more energy (higher negative ∆H), making it more thermodynamically favorable and more likely to occur under adequate oxygen supply.

Step-by-step solution

(a) Balanced Reactions

: For the combustion of methane, we have three reactions producing different carbon-containing products. 1. Methane reacts with oxygen to form soot (graphite) and water: \( \mathrm{CH}_{4}(\mathrm{g}) \;+\; \frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \;\rightarrow\; \mathrm{C(graphite)} \;+\; 2\mathrm{H}_{2}\mathrm{O}(l) \) 2. Methane reacts with oxygen to form carbon monoxide (CO) and water: \( \mathrm{CH}_{4}(\mathrm{g}) \;+\; \frac{3}{2} \mathrm{O}_{2}(\mathrm{g}) \;\rightarrow\; \mathrm{CO}(\mathrm{g}) \;+\; 2\mathrm{H}_{2}\mathrm{O}(l) \) 3. Methane reacts with oxygen to form carbon dioxide (CO2) and water: \( \mathrm{CH}_{4}(\mathrm{g}) \;+\; 2 \mathrm{O}_{2}(\mathrm{g}) \;\rightarrow\; \mathrm{CO}_{2}(\mathrm{g}) \;+\; 2\mathrm{H}_{2}\mathrm{O}(l) \)

(b) Standard Enthalpies

: To determine the standard enthalpies of each reaction, we find the difference between the sum of the enthalpies of the products and the sum of the enthalpies of the reactants. We will refer the standard enthalpies of formation from a reference table. Let ∆Hf for CH4, O2, C(graphite), CO, CO2, and H2O be ∆H1, ∆H2, ∆H3, ∆H4, ∆H5, and ∆H6, respectively. 1. For the first reaction (CH4 + 0.5 O2 -> C(graphite) + 2 H2O): ∆H = [1 * (∆H3) + 2 * (∆H6)] - [1 * (∆H1) + 0.5* (∆H2)] 2. For the second reaction (CH4 + 1.5 O2 -> CO + 2 H2O): ∆H = [1 * (∆H4) + 2 * (∆H6)] - [1 * (∆H1) + 1.5 * (∆H2)] 3. For the third reaction (CH4 + 2 O2 -> CO2 + 2 H2O): ∆H = [1 * (∆H5) + 2 * (∆H6)] - [1 * (∆H1) + 2 * (∆H2)] Insert the values from the reference table for enthalpy of formation and calculate the standard enthalpies for each reaction.

(c) Predominant Carbon-containing Product

: When the oxygen supply is adequate, CO2 is the predominant carbon-containing product because the formation of CO2 releases more energy (higher negative ∆H) as compared to the formation of CO or soot. In other words, the reaction that forms CO2 is more exothermic, making it more thermodynamically favorable, and thus, more likely to occur under adequate oxygen supply.

Key concepts

Methane Combustion

When methane (\(\text{CH}_4\)) combusts or burns in the presence of oxygen \((\text{O}_2)\), it can generate various carbon-containing products. The three primary products are soot (graphite), carbon monoxide \((\text{CO})\), and carbon dioxide \((\text{CO}_2)\). Each outcome depends on the oxygen availability during the combustion process.
To grasp the fundamental chemistry of methane combustion, it's crucial to remember that this is a type of chemical reaction known as oxidation. During this process, oxygen molecules react with the carbon and hydrogen elements found in methane.

• With limited oxygen, methane might produce soot, which is simply fine particles of carbon \((\text{C})\).
• With slightly more oxygen, it produces carbon monoxide \((\text{CO})\). This is often the case in more "incomplete" combustion situations.
• Adequate oxygen ensures that methane transforms entirely into carbon dioxide \((\text{CO}_2)\), a more complete combustion outcome.

Understanding these pathways helps us see how variances in oxygen availability affect what carbon product we receive during the burning of methane.

Enthalpy Calculations

In chemical reactions, enthalpy changes indicate how much energy is either absorbed or released as the reaction takes place. Enthalpy calculations are essential for understanding combustion reactions of methane, as they help us quantify energy outcomes.
To compute the enthalpy change \((\Delta H)\) of each methane combustion pathway, we start by using a reference table that lists "enthalpies of formation" for each substance involved in the reaction. This involves the following formula:
\[ \Delta H = \left( \Sigma \ \text{Enthalpies of products} \right) - \left( \Sigma \ \text{Enthalpies of reactants} \right) \]

• For each equation, sum up the enthalpies of formation of all products, being sure to multiply each by their stoichiometric coefficients from the balanced equation.
• Then, subtract the total enthalpies of reactants, also considering stoichiometry.
• The result: a positive \(\Delta H\) suggests an endothermic reaction (absorbs energy), while a negative \(\Delta H\) signifies an exothermic reaction (releases energy).

This process will demonstrate why carbon dioxide formation in methane combustion is favored when oxygen is adequate, as the reaction tends to release the most energy, making it energetically favorable.

Balanced Chemical Equations

Balanced chemical equations are fundamental to correctly understanding and depicting chemical reactions such as the combustion of methane. They ensure that the law of conservation of mass is honored, meaning the number of atoms for each element is equal on both sides of the equation.
For the combustion reactions of methane, writing balanced equations means accounting for every atom of carbon, hydrogen, and oxygen:

• With soot formation: \(\mathrm{CH}_{4}(\mathrm{g}) + \frac{1}{2} \mathrm{O}_2(\mathrm{g}) \rightarrow \mathrm{C(graphite)} + 2\mathrm{H}_2\mathrm{O}(l)\)
• With carbon monoxide formation: \(\mathrm{CH}_{4}(\mathrm{g}) + \frac{3}{2} \mathrm{O}_2(\mathrm{g}) \rightarrow \mathrm{CO}(\mathrm{g}) + 2\mathrm{H}_2\mathrm{O}(l)\)
• And with carbon dioxide formation: \(\mathrm{CH}_{4}(\mathrm{g}) + 2 \mathrm{O}_2(\mathrm{g}) \rightarrow \mathrm{CO}_2(\mathrm{g}) + 2\mathrm{H}_2\mathrm{O}(l)\)

By balancing these equations, we ensure that every atom is accounted for, preserving the identities and amounts of reactants and products. This balance is crucial for accurate enthalpy calculations and effectively predicting reaction behaviors.