Chemistry - Thermochemistry Question with Solution | TestHub
At 300 K, the standard enthalpies of formation of C6H5COOH(s), CO2(g) and H2O(l) are -408, -393 and -286 kJ/mol, respectively. The enthalpy of combustion of benzoic acid(s) at 300 K is (one or more than one correct)
Options:(select one or more)
Answer:
Solution:
The combustion of benzoic acid is represented by the following reaction:
C6H5COOH(s) + 15/2O2(g) tarrow 7CO2(g) + 3H2O(l)
The enthalpy change (Δ H) for this reaction is calculated as:
Δ H = 7(-393) + 3(-286) - (-408) = -2751 - 858 + 408 = -3201 kJ/mol
This value represents the enthalpy change at constant pressure.
To find the internal energy change (Δ U) at constant volume, we first calculate the change in the number of moles of gas (Δ ng):
Δ ng = 7 - 7.5 = -0.5
Using the relationship between Δ H and Δ U:
Δ U = Δ H - Δ ng RT
Substituting the given values (assuming R = 8.314 J/(mol·K) and T = 300 K):
Δ U = -3201 + 0.5 × (8.314 × 300)/1000 = -3201 + 1.25 = -3199.75 ≈ -3199.76 kJ/mol
This is the internal energy change at constant volume.
Therefore, the correct options are (a) and (d).
