Problem 109 - Chemistry The Central Science

Question

Burning methane in oxygen can produce three different carbon-containing products: soot (very fine particles of graphite), CO(g), and \(\mathrm{CO}_{2}(g) .\) (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 equations for the combustion of methane to produce soot, CO(g), and CO2(g) are: 1. \( CH_{4}(g) + 2O_{2}(g) \rightarrow C(graphite) + 2H_{2}O(l) \) 2. \( CH_{4}(g) + O_{2}(g) \rightarrow CO(g) + 2H_{2}O(l) \) 3. \( CH_{4}(g) + 2O_{2}(g) \rightarrow CO_{2}(g) + 2H_{2}O(l) \) The standard enthalpies for these reactions are: 1. ΔH° = -497.6 kJ/mol 2. ΔH° = -607.3 kJ/mol 3. ΔH° = -890.3 kJ/mol CO2(g) is the predominant product of methane combustion when the oxygen supply is adequate because its formation has the lowest standard enthalpy, making it the most exothermic and thermodynamically favorable reaction.

Step 1: Balanced equations for methane combustion reactions

We need to write the balanced equations for the reaction of methane (CH4) with oxygen (O2) to produce: 1. Soot (graphite particles) 2. CO(g) 3. CO2(g) (a) Reaction 1: Methane + Oxygen → Soot (graphite) + water \( CH_{4}(g) + O_{2}(g) \rightarrow C(graphite) + 2H_{2}O(l) \) To balance this equation, we insert a coefficient in front of the oxygen molecule: \( CH_{4}(g) + 2O_{2}(g) \rightarrow C(graphite) + 2H_{2}O(l) \) (b) Reaction 2: Methane + Oxygen → CO(g) + water \( CH_{4}(g) + O_{2}(g) \rightarrow CO(g) + 2H_{2}O(l) \) This equation is already balanced. (c) Reaction 3: Methane + Oxygen → CO2(g) + water \( CH_{4}(g) + O_{2}(g) \rightarrow CO_{2}(g) + 2H_{2}O(l) \) To balance this equation, we insert a coefficient in front of the oxygen molecule: \( CH_{4}(g) + 2O_{2}(g) \rightarrow CO_{2}(g) + 2H_{2}O(l) \)

Step 2: Calculate the standard enthalpies for each reaction

The standard enthalpies for each reaction can be calculated using the standard enthalpies of formation, which can be looked up in a reference table. We will denote the standard enthalpy of formation as ΔHf°. For methane (CH4(g)): ΔHf° = -74.8 kJ/mol For oxygen (O2(g)): ΔHf° = 0 kJ/mol (it's an element in its standard state) For graphite (C): ΔHf° = 0 kJ/mol (it's an element in its standard state) For CO(g): ΔHf° = -110.5 kJ/mol For CO2(g): ΔHf° = -393.5 kJ/mol For water (H2O(l)): ΔHf° = -285.8 kJ/mol Using Hess's Law, the standard enthalpy of a reaction (ΔH°) can be calculated as the sum of the standard enthalpies of formation of the products minus the sum of the standard enthalpies of formation of the reactants: (a) Reaction 1: ΔH° = [ΔHf°(C) + 2ΔHf°(H2O(l))] - [ΔHf°(CH4(g)) + 2ΔHf°(O2(g))] ΔH° = [0 + 2(-285.8)] - [(-74.8) + 0] ΔH° = -497.6 kJ/mol (b) Reaction 2: ΔH° = [ΔHf°(CO) + 2ΔHf°(H2O(l))] - [ΔHf°(CH4(g)) + ΔHf°(O2(g))] ΔH° = [(-110.5) + 2(-285.8)] - [(-74.8) + 0] ΔH° = -607.3 kJ/mol (c) Reaction 3: ΔH° = [ΔHf°(CO2) + 2ΔHf°(H2O(l))] - [ΔHf°(CH4(g)) + 2ΔHf°(O2(g))] ΔH° = [(-393.5) + 2(-285.8)] - [(-74.8) + 0] ΔH° = -890.3 kJ/mol

Step 3: Explain why CO2(g) is the predominant product when the oxygen supply is adequate

When the oxygen supply is adequate, CO2(g) is the predominant product of methane combustion because the formation of CO2(g) results in the greatest release of energy (has the lowest standard enthalpy of the three reactions). In other words, the reaction that produces CO2(g) is the most exothermic and thermodynamically favorable among the three reactions under analysis. Thus, when there is enough oxygen, the reaction will predominantly proceed to form CO2(g) to minimize the overall energy of the system and maximize stability.

Key Concepts

Balanced Chemical Equations

Balanced chemical equations are fundamental for understanding chemical reactions because they represent the conservation of mass. Every chemical reaction equation needs to have the same number of each type of atom on both sides. This reflects the law of conservation of mass, indicating that matter cannot be created or destroyed in a closed system.

When we balance reactions for methane (CH_4g) combustion, we make sure that the number of carbon, hydrogen, and oxygen atoms are equal on both the reactant and product sides. For instance, to produce CO_2g, we start with:

• Reactants: Methane (CH_4g) and Oxygen (O_2g)
• Products: Carbon Dioxide (CO_2g) and Water (H₂O(l))

To balance this, coefficients are added, turning the equation into:\[ CH_{4}(g) + 2O_{2}(g) \rightarrow CO_{2}(g) + 2H_{2}O(l) \]Balancing ensures that both sides have equal numbers of C, H,and O atoms, thus respecting the conservation laws.

Standard Enthalpies of Formation

Standard enthalpies of formation provide a way to quantify the heat change when one mole of a compound forms from its elements under standard conditions. This quantity is denoted \( \Delta H_f^\degree \)and it can be found in reference tables for most substances.

For instance, for methane (CH_4g), \( \Delta H_f^\degree \)is -74.8 kJ/mol, indicating the heat released when methane is formed. To calculate the enthalpy change of a reaction, we use the formula:\[\Delta H^\degree = \sum \Delta H_f^\degree(products) - \sum \Delta H_f^\degree(reactants)\]
This method allows determining the overall energy change, which tells us whether a reaction is endothermic or exothermic. Enthalpies of formation are crucial for understanding energy changes during chemical transformations.

Hess's Law

Hess's Law states that the total enthalpy change of a chemical reaction is the same, regardless of how many steps the reaction is carried out in. This principle allows us to calculate enthalpy changes that may not be directly measurable.

When we apply Hess's Law to the combustion of methane, we use the standard enthalpies of formation for each reactant and product. This approach involves calculating the sum of the enthalpies of the products and subtracting the sum of the enthalpies of the reactants.

If the enthalpy calculated is negative, the reaction is exothermic, meaning it releases heat. Hess's Law is an essential tool in thermodynamics, simplifying the process of energy accounting in chemical reactions.

Exothermic Reactions

Exothermic reactions are those that release energy, typically in the form of heat, to the surroundings. These reactions have a negative enthalpy change (\( \Delta H < 0 \)).

Combustion of methane (CH_4g) is a classic example of an exothermic reaction. When methane burns completely in an excess of oxygen to form carbon dioxide and water, it releases a significant amount of energy (-890.3 kJ/mol for CO_2g formation).

This energy release is why CO_2g is the primary product of methane combustion when oxygen is adequate. Exothermic reactions are crucial in energy production, providing the energy needed for various industrial processes and everyday applications like heating and powering vehicles.