Calculate the enthalpy of formation of ammonia (NH3)
Answer: -49.5kJmol-1
Did you get this answer, if so how? Thank you
So you know H2 + 0.5Os = H2O has a delta H value of -288, in the first equation there are 6 H2O. so 6 x -288 = -1728
therefore, 4NH3 = 2N2 where delta H now equals -1728 + 1530 this gives 4NH3 = 2N2 a delta H value of -198 since there are 4 NH3, -198 divided by 4 gives -49.5
So you know H2 + 0.5Os = H2O has a delta H value of -288, in the first equation there are 6 H2O. so 6 x -288 = -1728
therefore, 4NH3 = 2N2 where delta H now equals -1728 + 1530 this gives 4NH3 = 2N2 a delta H value of -198 since there are 4 NH3, -198 divided by 4 gives -49.5