Consider the following intermediate chemical equations.

2 equations. First: upper P subscript 4 (s) plus 6 upper C l subscript 2 (g) right arrow 4 upper P upper C l subscript 3 (g) delta H 1 equals negative 4,439 kilojoules. Second: 4 upper P upper C l subscript 3 (g) right arrow upper P subscript 4 (s) plus 10 upper C l subscript 2 (g) Delta H 2 equals 3,438 kilojoules.

What is the enthalpy of the overall chemical reaction Upper P upper C l subscript 5 (g) right arrow upper P upper C l subscript 3 (g) plus Uper C l subscript 2 (g).?

-999 kJ

-250. kJ

250. kJ

999 kJ

Question

Cullen

Chemistry

4 December, 17:42

Consider the following intermediate chemical equations.

2 equations. First: upper P subscript 4 (s) plus 6 upper C l subscript 2 (g) right arrow 4 upper P upper C l subscript 3 (g) delta H 1 equals negative 4,439 kilojoules. Second: 4 upper P upper C l subscript 3 (g) right arrow upper P subscript 4 (s) plus 10 upper C l subscript 2 (g) Delta H 2 equals 3,438 kilojoules.

What is the enthalpy of the overall chemical reaction Upper P upper C l subscript 5 (g) right arrow upper P upper C l subscript 3 (g) plus Uper C l subscript 2 (g).?

-999 kJ

-250. kJ

250. kJ

999 kJ

+3

Answers (1)

Know the Answer?

Renee Bowers · Answer 1

-250.3kJ

Explanation:

Based in the reactions and using - Hess's law-:

(1) P₄ (s) + 6 Cl₂ (g) → 4PCl₃ (g) ΔH₁ = - 4439kJ

(2) 4PCl₅ (g) → P₄ (s) + 10Cl₂ ΔH₂ = 3438kJ

The sum of (1) + (2) is:

4PCl₅ (g) → 4PCl₃ (g) + 4 Cl₂ ΔH = - 4439kJ + 3438kJ = - 1001kJ

Dividing this reaction in 4:

PCl₅ (g) → PCl₃ (g) + Cl₂ ΔH = - 1001kJ / 4 = - 250.3kJ