Ca(s)+N₂(g)+3O₂(g)⟶Ca(NO₃)₂(s)
(1) 8Ca(s)+Ca(NO₃)₂(s)→Ca₃N₂(s)+6CaO(s) ΔH = -3304 kJ
(2) Ca₃N₂(s)⟶3Ca(s)+N₂(g) ΔH = +432 kJ
(3) 2CaO(s)⟶2Ca(s)+O₂(g) ΔH = +1270 kJ
Flip all three equations: (4) Ca₃N₂(s)+6CaO(s) → 8Ca(s)+Ca(NO₃)₂(s) ΔH = +3304 kJ (5) 3Ca(s)+N₂(g) ⟶ Ca₃N₂(s) ΔH = -432 kJ (6) 2Ca(s)+O₂(g) ⟶ 2CaO(s) ΔH = -1270 kJ Since we reversed the equations, the delta H will change signs.
Now, multiply the third equation by 3 (7) 6Ca(s)+3O₂(g) ⟶ 6CaO(s) ΔH = -3810 kJ Since you multiplied all the coefficients by 3, the delta H will also be multiplied by 3.
Now add equations 4, 5, and 7 together. Since we’re adding reactions, the delta H will all be added together(8) Ca₃N₂(s) + 6CaO(s) + 3Ca(s)+ N₂(g) + 6Ca(s)+ 3O₂(g) ⟶ 8Ca(s) + Ca(NO₃)₂(s) + Ca₃N₂(s) + 6CaO(s) ΔH = 3304 + (-432) + (-3810) kJ = -938 kJ
Simplify the equation. You will notice Ca₃N₂(s) and 6CaO(s) will cancel out.3Ca(s)+ N₂(g) + 6Ca(s)+ 3O₂(g) ⟶ 8Ca(s) + Ca(NO₃)₂(s)Simplify the Ca(s) 9Ca(s)+ N₂(g) + 3O₂(g) ⟶ 8Ca(s) + Ca(NO₃)₂(s)We can remove 8 Ca(s) from both sidesCa(s)+ N₂(g) + 3O₂(g) ⟶ Ca(NO₃)₂(s) ΔH = -938 kJ
This is the equation we wanted the ΔH for. The ΔH for this reaction is -938 kJ.