Problem 102 - Chemistry: The Central Science

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?

Step-by-Step Solution

Answer

The balanced chemical equations for the three reactions are: 1. \(CH_{4}(g) + O_{2}(g) \rightarrow C(s) + 2H_{2}O(l)\) 2. \(CH_{4}(g) + \dfrac{3}{2}O_{2}(g) \rightarrow CO(g) + 2H_{2}O(l)\) 3. \(CH_{4}(g) + 2O_{2}(g) \rightarrow CO_{2}(g) + 2H_{2}O(l)\) The standard enthalpies for the reactions are: 1. \(\Delta H^{\circ}_{rxn,1} = -882.0\, kJ/mol\) 2. \(\Delta H^{\circ}_{rxn,2} = -802.3\, kJ/mol\) 3. \(\Delta H^{\circ}_{rxn,3} = -802.7\, kJ/mol\) When the oxygen supply is adequate, CO\(_2\) is the predominant carbon-containing product because its formation is the most exothermic reaction among the three, making it more thermodynamically favorable.

Step 1: Write the balanced chemical equations for each reaction

We have three main reactions: 1. Methane reacting with oxygen to form soot (graphite) and water: \[CH_{4}(g) + O_{2}(g) \rightarrow C(s) + 2H_{2}O(l)\] 2. Methane reacting with oxygen to form carbon monoxide and water: \[CH_{4}(g) + \dfrac{3}{2} O_{2}(g) \rightarrow CO(g) + 2H_{2}O(l)\] 3. Methane reacting with oxygen to form carbon dioxide and water: \[CH_{4}(g) + 2O_{2}(g) \rightarrow CO_{2}(g) + 2H_{2}O(l)\]

Step 2: Calculate the standard enthalpies for each reaction

For this part, we will use tabulated values of standard enthalpies of formation for each species. These values are: - \(\Delta H^{\circ}_{f}[CH_4(g)]\) = -74.6 kJ/mol - \(\Delta H^{\circ}_{f}[O_2(g)]\) = 0 kJ/mol (by convention, since O2 is the reference state for oxygen) - \(\Delta H^{\circ}_{f}[C(s)]\) = 0 kJ/mol (by convention, since graphite is the reference state for carbon) - \(\Delta H^{\circ}_{f}[CO(g)]\) = -110.5 kJ/mol - \(\Delta H^{\circ}_{f}[CO_2(g)]\) = -393.5 kJ/mol - \(\Delta H^{\circ}_{f}[H_{2}O(l)]\) = -285.8 kJ/mol To find the standard enthalpy of each reaction, we use the equation: \[\Delta H^{\circ}_{rxn} = \Delta H^{\circ}_{f}(\text{products}) - \Delta H^{\circ}_{f}(\text{reactants})\] 1. For the first reaction (formation of soot): \[\Delta H^{\circ}_{rxn,1} = [C(s) + 2H_{2}O(l)] - [CH_{4}(g) + O_{2}(g)]\] \[\Delta H^{\circ}_{rxn,1} = (0 - 2 \times 285.8) - (-74.6 + 0)\] \[\Delta H^{\circ}_{rxn,1} = -882.0\, kJ/mol\] 2. For the second reaction (formation of CO): \[\Delta H^{\circ}_{rxn,2} = [CO(g) + 2H_{2}O(l)] - [CH_{4}(g) + \dfrac{3}{2} O_{2}(g)]\] \[\Delta H^{\circ}_{rxn,2} = (-110.5 - 2 \times 285.8) - (-74.6)\] \[\Delta H^{\circ}_{rxn,2} = -802.3\, kJ/mol\] 3. For the third reaction (formation of CO2): \[\Delta H^{\circ}_{rxn,3} = [CO_{2}(g) + 2H_{2}O(l)] - [CH_{4}(g) + 2 O_{2}(g)]\] \[\Delta H^{\circ}_{rxn,3} = (-393.5 - 2 \times 285.8) - (-74.6)\] \[\Delta H^{\circ}_{rxn,3} = -802.7\, kJ/mol\]

Step 3: Explain why CO\(_2\) is the predominant product when the oxygen supply is adequate

When the oxygen supply is adequate, CO\(_2\) is the predominant product because the formation of CO\(_2\) is the most exothermic reaction among the three. The reaction releases more energy, which makes it more thermodynamically favorable. In other words, the CO\(_2\)-forming reaction has the highest negative value of standard enthalpy, which corresponds to a more stable system after the reaction occurs. Under limited oxygen supply, the formation of CO or soot might occur. However, once we have enough oxygen, the system tends to a more stable state by releasing more energy, which favors CO\(_2\) as the predominant carbon-containing product.

Key Concepts

Enthalpy Changes

Enthalpy changes are crucial aspects of understanding combustion reactions. Enthalpy, denoted as \( \Delta H \), measures the heat change in a reaction at constant pressure. When we talk about standard enthalpy changes, we're considering the reaction under standard conditions. This measure helps us see how much energy is released or absorbed during the process.
In combustion reactions, enthalpy changes determine which product will be predominant. For methane's combustion, converting it to carbon dioxide is highly exothermic. This means it releases a lot of energy, making the reaction favorable. The general formula to calculate the standard enthalpy change for a reaction is:

\[ \Delta H^{\circ}_{rxn} = \sum \Delta H^{\circ}_{f}(\text{products}) - \sum \Delta H^{\circ}_{f}(\text{reactants}) \]

• Exothermic reactions: These reactions have a negative \( \Delta H \), indicating energy release.
• Endothermic reactions: These reactions have a positive \( \Delta H \), showing energy is absorbed.

By analyzing the enthalpy, we can predict which combustion path (carbon to soot, carbon monoxide, or carbon dioxide) is most favorable given adequate oxygen.

Stoichiometry

Stoichiometry is the science of calculating the amounts of reactants and products involved in a chemical reaction. It's essential in understanding how combustion reactions play out.
Simplified, stoichiometry involves using the balanced chemical equation to determine the proportions of reactants needed and the products formed. For example, in methane combustion, if you know how many moles of methane you have, stoichiometry allows us to calculate moles of oxygen needed and carbon dioxide or other products formed.
For the reaction of methane producing carbon dioxide:

\[ CH_{4}(g) + 2 O_{2}(g) \rightarrow CO_{2}(g) + 2 H_{2}O(l) \]

• One mole of \( CH_4 \) reacts with two moles of \( O_2 \).
• This results in one mole of \( CO_2 \) and two moles of \( H_2O \) as products.

The key to mastering stoichiometry is recognizing the mole ratios from the balanced equation. This helps ensure no reactants are wasted, and maximum efficiency is achieved in reactions.
Stoichiometry guides us on how various factors like limiting reactants can affect reaction outcomes, especially in combustion processes.

Chemical Equations

Chemical equations are symbolic representations of chemical reactions, showing how reactants transform into products. A balanced chemical equation shows the exact proportions of each element, respecting the Law of Conservation of Mass.
In combustion reactions, chemical equations help us visualize the process of burning fuels like methane. For instance, the combustion of methane can result in different products based on oxygen availability:

• Formation of soot (graphite): \[ CH_{4}(g) + O_{2}(g) \rightarrow C(s) + 2 H_{2}O(l) \]
• Formation of carbon monoxide: \[ CH_{4}(g) + \dfrac{3}{2} O_{2}(g) \rightarrow CO(g) + 2 H_{2}O(l) \]
• Formation of carbon dioxide: \[ CH_{4}(g) + 2 O_{2}(g) \rightarrow CO_{2}(g) + 2 H_{2}O(l) \]

Balancing equations requires ensuring the same number of each type of atom on both sides of the equation. This balance reflects the reality of nature where atoms are neither created nor destroyed in chemical reactions.
Understanding chemical equations is foundational for analyzing reactions, predicting products, and calculating energy changes, as seen in combustion.