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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

Respuesta :

TSO
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 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.


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