Combine the reactions: P₄(s)+ 6 Cl₂(g)→ 4 PCl₃(g)Δ_rH°₂₉₈ = -1225.0 kJ/mol PCl₃(g)+ 3 H₂O(l)→ H₃PO₃(aq)+ 3 HCl(g)= -853.5 kJ/mol 2 H₂(g)+ O₂(g)→ 2 H₂O(l)= -571.5 kJ/mol H₂(g)+ Cl₂(g)→ 2 HCl(g)= -184.9 kJ/mol To obtain the reaction: P₄(s)+ 6 H₂(g)+ 6 O₂(g)→ 4 H₃PO₃(aq). What is the value of Δ_rH°₂₉₈ for this reaction?

Question

Quizplus · Accepted Answer

To obtain the desired reaction, we can add the given reactions as follows: P₄(s) + 6 Cl₂(g) → 4 PCl₃(g) 4 PCl₃(g) + 12 H₂O(l) → 4 H₃PO₃(aq) + 12 HCl(g) 6 H₂(g) + 6 O₂(g) → 6 H₂O(l) 6 H₂(g) + 6 Cl₂(g) → 12 HCl(g) When we add these reactions, we obtain: P₄(s) + 6 H₂(g) + 6 O₂(g) → 4 H₃PO₃(aq) + 6 HCl(g) The value of Δ_rH°₂₉₈ for this reaction can be calculated by taking the sum of the Δ_rH°₂₉₈ values of the individual reactions. When we do this, we get: Δ_rH°₂₉₈ = (-1225.0 kJ/mol) + (-853.5 kJ/mol) + (-571.5 kJ/mol) + (-184.9 kJ/mol) = -2825.9 kJ/mol Therefore, the answer is B.

Combine the Reactions: P4(s)+ 6 Cl2(g)→ 4 PCl3(g)ΔrH°298 = -1225.0 KJ/mol

Combine the Reactions: P₄(s)+ 6 Cl₂(g)→ 4 PCl₃(g)Δ_rH°₂₉₈ = -1225.0 KJ/mol