Question

The gas ethane, C2H6(g), can be used in welding. When ethane is burned in oxygen, the reaction is: 2 C2H6(g) + 7 O2(g)4 CO2(g) + 6 H2O(g)

(a) Using the following data, calculate ΔH° for this reaction. ΔH°f kJ mol-1: C2H6(g) = -84.0 ; CO2(g) = -393.5 ; H2O(g) = -241.8 ΔH° = kJ

(b) Calculate the total heat capacity of 4 mol of CO2(g) and 6 mol of H2O(g), using CCO2(g) = 37.1 J K-1 mol-1 and CH2O(g) = 33.6 J K-1 mol-1. C = J K-1

(c) When this reaction is carried out in an open flame, almost all the heat produced in part (a) goes to raise the temperature of the products. Assuming that the reactants are at 25°C, calculate the maximum flame temperature that is attainable in an open flame burning ethane in oxygen. The actual flame temperature would be lower than this because heat is lost to the surroundings. Maximum temperature = °C

Answer 1

(a) H_f = H_f( of reactant ) - H_f( of product)

= ( 2* -84.0) - ( 4* -393.5 + 6* -241.8)

= (-168) - ( -1574-1450.8)

= ( -168) - ( -3024.8)

= -168 + 3024.8 = 2856.8 KJ/mol

H_f OF O₂ not considered since in standard sttate heat of formation of any element is zero.

(b) total heat capacity = no of moles of CO₂ * Heat capacity of CO₂ + No of moles of H₂O* heat capacity of H₂O

= 4*37.1 + 6* 33.6

= 148.4 + 201.6

= 350 J K^-1 mol^-1