Calculate the enthalpy change for the reaction Ca (s) + N2 (g) + 3O2 (g) ⟶Ca (NO3) 2 (s) from the following equations: 8Ca (s) + Ca (NO3) 2 (s) →Ca3N2 (s) + 6CaO (s) ΔH = - 3304 kJ Ca3N2 (s) ⟶3Ca (s) + N2 (g) ΔH = + 432 kJ 2CaO (s) ⟶2Ca (s) + O2 (g) ΔH = + 1270 kJ

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) + N ₂ (g) + 6Ca (s) + 3O₂ (g) ⟶ 8Ca (s) + Ca (NO ₃) ₂ (s) + Ca ₃N₂ (s) + 6CaO (s)

ΔH = 3304 + (-432) + (-3810) kJ = - 938 kJ

Simplify the equation. You will notice Ca ₃N₂ (s) and 6CaO (s) will cancel out.

3Ca (s) + N ₂ (g) + 6Ca (s) + 3O₂ (g) ⟶ 8Ca (s) + Ca (NO ₃) ₂ (s)

Simplify the Ca (s)

9Ca (s) + N ₂ (g) + 3O₂ (g) ⟶ 8Ca (s) + Ca (NO ₃) ₂ (s)

We can remove 8 Ca (s) from both sides

Ca (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.