Question

Given the following data:

P4(s) + 6 Cl2(g) → 4 PCl3(g) ΔH = −1225.6 kJ

P4(s) + 5 O2(g) → P4O10(s) ΔH = −2967.3 kJ

PCl3(g) + Cl2(g) → PCl5(g) ΔH = −84.2 kJ

PCl3(g) + 1/2 O2(g) → Cl3PO(g) ΔH = −285.7 kJ

Calculate ΔH for the reaction P4O10(s) + 6 PCl5(g) → 10 Cl3PO(g).

Answer 1

Answer:

ΔH = -610.1 kJ

Explanation:

By the Hess Law, when a reaction is formed by various steps, the enthalpy change (ΔH) of the global reaction is the sum of the enthalpy change of the steps reactions. Besides, if it's necessary for a change in the reaction, ΔH will suffer the same change. If the reaction multiplied by a number, ΔH will be multiplied by the same number, and if the reaction is inverted, the signal of ΔH is inverted.

P₄(s) + 6Cl₂(g) → 4 PCl₃(g) ΔH = -1225.6 kJ

P₄(s) + 5O₂(g) → P₄O₁₀(s) ΔH = -2967.3 kJ (inverted)

PCl₃(g) + Cl₂(g) → PCl₅(g) ΔH = -84.2 kJ (inverted and multiplied by 6)

PCl₃(g) + 1/2O₂(g) → Cl₃PO(g) ΔH = -285.7 kJ (multiplied by 10)

P₄(s) + 6Cl₂(g) → 4 PCl₃(g) ΔH = -1225.6 kJ

P₄O₁₀(s) → P₄(s) + 5O₂(g) ΔH = +2967.3 kJ

6PCl₅(g) → 6PCl₃(g) + 6Cl₂(g) ΔH = +505.2 kJ

10PCl₃(g) + 5O₂(g) → 10Cl₃PO(g) ΔH = -2857.0 kJ

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The bolded substances will be eliminated because have the same amount in product and reactant:

P₄O₁₀(s) + 6PCl₅(g) → 10Cl₃PO(g)

ΔH = -1225.6 + 2967.3 + 505.2 -2857.0

ΔH = -610.1 kJ

Answer 2

Based on the data provided, the enthalpy change ΔH for the given reaction is -610.1 kJ.

What is enthalpy change of a reaction?

The enthalpy change of a reaction is the energy evolved or absorbed when reactant molecules react to form products.

From Hess' Law of constant heat summation, the enthalpy change (ΔH) of the reaction is the sum of the enthalpy change of the several reaction steps. reactions.

Considering the sum of the intermediate reaction steps:

1. P₄(s) + 6Cl₂(g) → 4 PCl₃(g) ΔH = -1225.6 kJ
2. P₄(s) + 5O₂(g) → P₄O₁₀(s) ΔH = -2967.3 kJ (reversing)
3. PCl₃(g) + Cl₂(g) → PCl₅(g) ΔH = -84.2 kJ (reversed and multiplied by 6)
4. PCl₃(g) + 1/2O₂(g) → Cl₃PO(g) ΔH = -285.7 kJ (multiplied by 10)

The reactions and enthalpy changes become:

1. P₄(s) + 6Cl₂(g) → 4 PCl₃(g) ΔH = -1225.6 kJ
2. P₄O₁₀(s) → P₄(s) + 5 O₂(g) ΔH = +2967.3 kJ
3. 6 PCl₅(g) → 6 PCl₃(g) + 6 Cl₂(g) ΔH = +505.2 kJ
4. 10 PCl₃(g) + 5 O₂(g) → 10 Cl₃PO(g) ΔH = -2857.0 kJ

Summing 1, 2, 3 and 4 gives:

• P₄O₁₀(s) + 6 PCl₅(g) → 10Cl₃PO(g)

Enthalpy change, ΔH is then calculated thus:

ΔH = -1225.6 + 2967.3 + 505.2 -2857.0

ΔH = -610.1 kJ

Therefore, the enthalpy change ΔH for the given reaction is -610.1 kJ.

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