Question

# solve question 2 based on the result of question 1 Question 1 Acetone (C3H6O (l)) burns...

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 ∆fHo 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 ΔrH at 0oC (instead of 25oC, as given in the last question), if the heat capacities at constant pressure (Cp) 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.

Outline to solve the given problem:

ΔfH​ = Cp * ​∆​T

ΔrH = {3*(ΔfH)CO2(g) + 3*(ΔfH)H2O(l)} - {(ΔfH)C3H6O(l) + 4*(ΔfH)O2(g)}

Coming to the problem:

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

Now, (ΔfH)CO2(g) = 29.10J K-1mol-1 * 248 K = 7.2168 KJ/mol

fH)H2O(l) = 75.29J K-1mol-1 * 248 K = 18.67192 KJ/mol

fH)C3H6O(l) = 124.7J K-1mol-1 * 248 K = 30.9256 KJ/mol

fH)O2(g) = 29.36J K-1mol-1 * 248 K = 7.28128 KJ/mol

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

= 17.62 KJ/mol