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18 October, 21:54

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

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  1. 19 October, 00:31
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    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.
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