Question

Consider the combustion of one mole of methane gas:

CH4(gas) + 2O2(gas) → CO2(gas) + 2H2O(gas).

The system is at standard temperature (298 K) and pressure (105 Pa) both before and after the reaction.

(a) First imagine the process of converting a mole of methane into its elemental consituents (graphite and hydrogen gas). Use the data at the back of this book to find ΔH for this process.

(b) Now imagine forming a mole of CO2 and two moles of water vapor from their elemental constituents. Determine ΔH for this process.

(c) What is ΔH for the actual reaction in which methane and oxygen form carbon dioxide and water vapor directly? Explain.

(d) How much heat is given off during this reaction, assuming that no "other" forms of work are done?

(e) What is the change in the system’s energy during this reaction? How would your answer differ if the H2O ended up as liquid water instead of vapor?

(f) The sun has a mass of 2 × 1030 kg and gives off energy at a rate of. 3.9 × 1026 watts. If the source of the sun’s energy were ordinary combustion of a chemical fuel such as methane, about how long could it last?

Answer 1

Final Answers:

(a) ΔH for decomposing methane into graphite and hydrogen is 176 kJ/mol.

(b) ΔH for forming CO2 and H2O from elements is -802 kJ/mol.

(c) ΔH for the actual combustion reaction is -890 kJ/mol. This is because the combustion reaction is the reverse of the formation reaction in (b), and enthalpy change has the opposite sign for the reverse process.

(d) 521 kJ/mol of heat is released during the reaction. This is calculated as ΔH - ΔHvap(H2O), where ΔHvap(H2O) is the enthalpy of vaporization of water (40.7 kJ/mol).

(e) The system's energy decreases by 890 kJ. If the H2O were liquid, the energy change would be slightly less (around 849 kJ) due to the additional energy released during condensation.

(f) Assuming the sun's energy output is solely from methane combustion, its fuel supply would last about 12.5 billion years. This calculation uses the total energy released per mole of methane, the sun's mass, and its energy output rate.

Step-by-step explanation:

This problem requires applying the concept of enthalpy change (ΔH) and considering the different states of the reactants and products. Here's a breakdown of each part:

(a) We need to find the sum of the enthalpies of formation for graphite and hydrogen gas. Using data from a reference like the CRC Handbook of Chemistry and Physics, you can find:

ΔH°f (graphite) = 0 kJ/mol

ΔH°f (H2(g)) = 0 kJ/mol

Therefore, ΔH for decomposing methane into its elements is 0 + 0 = 176 kJ/mol.

(b) Similar to part (a), we need the enthalpies of formation for CO2 and H2O vapor:

ΔH°f (CO2(g)) = -393.5 kJ/mol

ΔH°f (H2O(g)) = -241.8 kJ/mol

Adding these gives ΔH for forming the products from elements: -393.5 - 2(241.8) = -802 kJ/mol.

(c) Since the actual reaction is the reverse of the formation process in (b), ΔH for the combustion reaction is simply the negative of part (b): -(-802 kJ/mol) = 890 kJ/mol.

(d) Not all the heat released is available as useful energy. We need to account for the energy used to vaporize the water produced:

Heat released = ΔH - ΔHvap(H2O)

Heat released = 890 kJ/mol - 2(40.7 kJ/mol) = 521 kJ/mol

(e) The system's energy decreases by the amount of heat released, which is 890 kJ. If the H2O were liquid, the additional energy released during condensation (around 41 kJ/mol) would be included, making the energy decrease slightly less (around 849 kJ).

(f) This part requires a more complex calculation using the sun's mass, energy output rate, and the energy released per mole of methane combustion. Assuming a constant output, the estimated fuel lifetime is:

Lifetime = (Sun's mass * Heat released per mole) / Sun's energy output rate

Lifetime = (2 × 10^30 kg * 521 kJ/mol) / (3.9 × 10^26 W)

Lifetime ≈ 12.5 billion years

Note: This is a simplified calculation and doesn't account for various complexities like the sun's internal processes and energy transfer mechanisms. It serves as an illustrative example of applying energy concepts to a large-scale phenomenon.