Question

Consider the combustion of a single molecule of \(\mathrm{CH}_{4}(g)\) forming \(\mathrm{H}_{2} \mathrm{O}(l)\) as a product. (a) How much energy, in \(\mathrm{J},\) is produced during this reaction? (b) A typical X-ray light source has an energy of \(8 \mathrm{keV}\). How does the energy of combustion compare to the energy of the X-ray?

Short answer

(a) The energy produced during the combustion of a single molecule of CH4 is approximately -1.6 * 10^(-19) J, which is an exothermic reaction. (b) The energy of an X-ray photon (1.282 x 10^(-15) J) is about four orders of magnitude larger than the energy released during the combustion of a single molecule of CH4.

Step-by-step solution

Identify the chemical equation for the combustion of CH4(g) forming H2O(l)

The balanced chemical equation for the combustion of methane (CH4) in the presence of oxygen (O2) to produce carbon dioxide (CO2) and liquid water (H2O) is as follows: \[CH_{4(g)} + 2O_{2(g)} \rightarrow CO_{2(g)} + 2H_{2 O(l)}\]

Determine the energy of combustion using standard enthalpies of formation

We can calculate the energy produced during the combustion using standard enthalpies of formation for each compound involved (\(ΔH_f^0\)): ΔH_reaction = Σ [\(ΔH_f^0\) (products)] - Σ [\(ΔH_f^0\) (reactants)] From standard enthalpy of formation tables, we have (all values in kJ/mol): - \(ΔH_f^0\) for CH4(g) = -74.8 - \(ΔH_f^0\) for O2(g) = 0.0 (reference state) - \(ΔH_f^0\) for CO2(g) = -393.5 - \(ΔH_f^0\) for H2O(l) = -285.8 Plugging these values into the equation: ΔH_reaction = [1(-393.5) + 2(-285.8)] - [1(-74.8) + 2(0)]

Calculate the energy produced in the reaction

We can now calculate the energy produced in the reaction: ΔH_reaction = (-393.5 - 2 * 285.8) - (-74.8) ΔH_reaction = (-393.5 - 571.6) + 74.8 ΔH_reaction = -965.3 kJ/mol

Convert the energy produced for a single molecule of CH4

We need to find the energy produced for a single molecule of CH4. To do this, we'll divide the energy produced per mole by Avogadro's number (NA = 6.022 x 10^23 mol^(-1)). Energy per molecule (J) = ΔH_reaction (kJ/mol) * 1000 (J/kJ) / NA Energy per molecule (J) = -965.3 * 1000 / (6.022 * 10^23) Energy per molecule (J) ≈ -1.6 * 10^(-19) J (a) The energy produced during the combustion of a single molecule of CH4 is approximately -1.6 * 10^(-19) J. The negative sign indicates that the reaction is exothermic, meaning that the energy is released during the process.

Compare the combustion energy with the energy of an X-ray photon

Given that an X-ray photon has an energy of 8 keV, we need to convert this value to joules. 1 eV = 1.602 x 10^(-19) J Energy of X-ray photon (J) = 8 * 1000 * (1.602 x 10^(-19) J) Energy of X-ray photon (J) = 1.282 x 10^(-15) J (b) Comparing the energy of the combustion and the X-ray photon: Energy of combustion: -1.6 * 10^(-19) J Energy of X-ray photon: 1.282 x 10^(-15) J Thus, the energy of the X-ray photon is about four orders of magnitude larger than the energy released during the combustion of a single molecule of CH4.

Key concepts

Enthalpy of Combustion

The enthalpy of combustion refers to the amount of heat energy released when a substance completely reacts with oxygen under standard conditions. This concept is crucial in thermodynamics and energy calculations because it helps us understand how much energy can be harnessed from burning fuels. For example, during the combustion of methane (CH₄), which is a common household fuel, a significant amount of heat is released, making it a valuable source of energy. The combustion process is exothermic, meaning it gives off heat, which is why the calculated enthalpy change ((ΔH_reaction)) is negative.

Understanding the enthalpy of combustion can also help us in assessing environmental impacts, as it provides a measure of how much carbon dioxide (CO₂) is produced per unit of energy generated. As we strive for more sustainable energy sources, knowing the enthalpy of combustion for various fuels helps in evaluating their potential and optimizing their use, reducing the carbon footprint where possible.

Chemical Equations

Chemical equations are symbolic representations of chemical reactions, showing the reactants transforming into products. A balanced chemical equation has the same number of atoms of each element on both the reactant and product sides, reflecting the law of conservation of mass. For instance, when methane undergoes combustion, the balanced chemical equation is: (CH₄(g)) + 2O₂(g)) → CO₂(g)) + 2H₂O(l). Balancing equations is a fundamental skill in chemistry as it ensures that the stoichiometry of the reaction is correct, which is essential for calculating the energy changes involved and for practical applications such as determining the amount of reactants needed or predicting the yield of products.

When students encounter difficulties in solving chemistry problems, it's often due to misunderstanding the balancing of equations. One improvement would be to practice by methodically counting atoms on each side and using coefficients to balance the equation, starting with elements that appear in the least number of compounds and saving hydrogen and oxygen for last.

Standard Enthalpy of Formation

The standard enthalpy of formation ((ΔH_f^0)) is the change in enthalpy when one mole of a compound is formed from its elements in their standard states. For pure elements in their standard state, like O₂(g), the value is defined as zero. In our methane combustion example, we use these standard values to calculate the overall energy change of the reaction.

The data from standard enthalpy of formation tables is incredibly useful because it allows us to predict whether reactions are exothermic or endothermic, and thus, we can gauge the heat exchange without conducting the experiment in a lab. For students seeking to deepen their understanding of this concept, they should focus on how these thermodynamic quantities are measured and tabulated, exploring how experimental data leads to values that scientists use universally in calculations such as the ones seen in the combustion of methane.