Chemistry - Thermochemistry Question with Solution | TestHub
ΔHf° for CO2(g) and H2O(l) are -94.0 and -68.0 kcal/mol. ΔHf° for the propanoic acid is one-third of the enthalpy of combustion of the acid and has the same sign as the latter. ΔHf° of propanoic acid is
Options:
Answer:
Solution:
The combustion reaction for propanoic acid is:
C2H5COOH(l) + 7/2O2(g) tarrow 3CO2(g) + 3H2O(l)
Let the standard enthalpy of formation, Δ H°f, for C2H5COOH(l) be x.
The standard enthalpy of combustion, Δ H°c, can be calculated using the formula:
Δ H°c = Σ (products) - Σ (reactants)
Given the standard enthalpies of formation for CO2(g) as -94.0 kcal/mol and for H2O(l) as -68.0 kcal/mol, and knowing that the standard enthalpy of formation for O2(g) is 0 kcal/mol:
Δ H°c = [3(-94.0) + 3(-68.0)] - [x + 7/2(0)]
Δ H°c = [-282.0 - 204.0] - x
Δ H°c = -486.0 - x
We are given that x = (Δ H°c)/3 (with the same sign). Substituting this into the equation for Δ H°c:
3x = -486.0 - x
4x = -486.0
x = (-486.0)/4
x = -121.5 kcal/mol
This corresponds to option (c).
