Question

Anhydrous ammonia is an ultra-clean and energy-dense alternative fuel that produces no greenhouse gases on combustion. In an experiment, gaseous ammonia is burned with oxygen in a container of fixed volume per the following equation: 4NH3(g) + 302(g) -> 2N2(g) 6H20(0) The initial and final states are at 298 K. After combustion with 12.80 g of O2, some NH3 remains unreacted. Calculate the enthalpy change during the process, given the following data. | ?,Ho (KJ mol-1) compound NH3 (g) H20 (1) -46.1 -285.8

Answer 1

4NH3(g) + 3O2(g) ----> 2N2(g) + 6H2O(l)

DH0rxn = (2*DH0f,N2(g) + 6*DH0f,H2O(l)) - (4*DH0f,NH3(g) + 3*DH0f,O2(g))

       = (2*0+6*-285.8) - (4*-46.1+3*0)

       = -1530.4 kj     

4 mol NH3(g) = 3 mol O2(g)

No of mol of O2 taken = 12.8/32 = 0.4 mol

3 mol O2 = 1530.4 Kj

0.4 mol O2 = 1530.4*0.4/3 = 204.053 kj

DH0rxn = -204.053 kj