Question

1. Combustion. Balance the Equation below for the combustion of n-butane. Show your work. Then, calculate the reaction enthalpy (per mol butane) at standard temperature and pressure, Allar for two cases: i) using liquid water; and ii) using water vapor. Show your work. Why are the two different and what do they mean? Is the reaction exothermic or endothermic?

Answer 1

C_4 gives 4Co_2 H_10 gives 5H_2_o} 80 50 } 13/2 O_2 for case i delta H degree_am = sigma n delta H degree_product - sigma n delta H degree_reactant = [5 delta H degree_H_2O(l) + 4 delta H degree_0.2(q)] - [delta H degree_f C_4H10(g) + 13/2 delta H degree_0.2] = [(5) (-285.83) + 4(-393.51)] - [(-125.6) + 0] delta H degree_lin = -2877.59 KJ/mol for case (i) for case ii delta H degree = 5 delta H degree_fH20(g) + 4 delta H degree_f(0.2) -delta H degree_f + 13/2 delta = 5 (-241.82) + 4(-393.51) + 125.6 + 0 delta H degree = -2658 .14KJ /mol for case ii delta H degree for H_2O (liquid) < delta H degree_r times mfor H_2O (gas) because H_2O (l) consume some energy to become H_2O (gas) \r\n delta H degree_rxm = -ve, I, e this reaction is exothermic