Chemistry - Thermochemistry Question with Solution | TestHub
Determine the standard enthalpy of reaction: C3H8(g) + H2(g) → C2H6(g) + CH4(g), using the given enthalpies under standard conditions:
Compound | H2(g) | CH4(g) | C2H6(g) | C (graphite)
ΔH° (kJ/mol) | -285.8 | -890.0 | -1560.0 | -393.5
The standard enthalpy of formation of C3H8(g) is -103.8 kJ/mol
Options:
Answer:
Solution:
The tabulated values represent the enthalpies of combustion for H₂ to H₂O(l), CH₄ and C₂H₆ to CO₂ and H₂O(l), and C to CO₂.
The enthalpy of formation for methane, CH₄, is calculated as:
Δf H(CH4) = -393.5 + 2(-285.8) - (-890.0) = -75.1 kJ
The enthalpy of formation for ethane, C₂H₆, is calculated as:
Δf H(C2H6) = 2(-393.5) + 3(-285.8) - (-1560.0) = -84.4 kJ
The overall enthalpy change, Δ H, is:
Δ H = [-84.4 + (-75.1)] - [-103.8 + 0] = -55.7 kJ
This corresponds to option (a).
