Question

solve question 2 based on the result of question 1

Question 1

Acetone (C3H6O (l)) burns in oxygen according to the equation:

                         4 O₂(g) + C₃H₆O(l) = 3 CO₂(g) + 3H₂O(l

At 25 °C the enthalpy change for this reaction is -1789.9 kJ. The standard enthalpy of formation ∆_fH^o of CO₂ (g) and H₂O (l) are -393.5 1 and -285.83 kJ/mol, respectively. Determine the enthalpy of formation of C₃H₆O(l) at 25 °C. Please enter your answer (unit: kJ/mol) with 2 decimal as 101.23. Please pay attention to the sign of your answer.

the result that we got : - 248.09

Question 2

Using the result from last question to calculate the reaction enthalpy Δ_rH at 0^oC (instead of 25^oC, as given in the last question), if the heat capacities at constant pressure (C_p) of C₃H₆O(l), O₂(g), CO₂(g), and H₂O(l) are124.7J K^-1mol^-1, 29.36J K^-1mol^-1, 29.10J K^-1mol^-1, and 75.29J K^-1mol^-1, respectively. Please enter your answer with 2 decimals as 23.22. Using KJ/mol as the unit. Pay attention to the sign.

Answer 1

Outline to solve the given problem:

Δ_fH​ = Cp * ​∆​T

Δ_rH = {3*(Δ_fH)_CO2(g) + 3*(Δ_fH)_H2O(l)} - {(Δ_fH)_C3H6O(l) + 4*(Δ_fH)_O2(g)}

Coming to the problem:

Here, ​∆​T = (0 - 25) ^oC = -25 ^oC = (-25+273) K = 248 K

Now, (Δ_fH)_CO2(g) = 29.10J K-1mol-1 * 248 K = 7.2168 KJ/mol

(Δ_fH)_H2O(l) = 75.29J K-1mol-1 * 248 K = 18.67192 KJ/mol

(Δ_fH)_C3H6O(l) = 124.7J K-1mol-1 * 248 K = 30.9256 KJ/mol

(Δ_fH)_O2(g) = 29.36J K-1mol-1 * 248 K = 7.28128 KJ/mol

Therefore, Δ_rH = (3*7.2168​ + 3*18.67192​) - (30.9256​ + 4*7.28128​)

= 17.62 KJ/mol