A 8.28 g sample of IBr(g) is placed in a container of capacity 164.2 ml and heated to 500 K. The equilibrium pressure of Br₂(g) in the system is 4.0 atm. The value of K_p for the reaction: 2IBr(g) ⇌ I

Question

A 8.28 g sample of IBr(g) is placed in a container of capacity 164.2 ml and heated to 500 K. The equilibrium pressure of Br₂(g) in the system is 4.0 atm. The value of K_p for the reaction: 2IBr(g) ⇌ I₂(g) + Br₂(g) is (Given: Atomic masses: I = 127, Br = 80)

Prashant Kotian · Accepted Answer

Correct answer: 4. Molar mass of IBr = 127 + 80 = 207 g/mol. Initial moles of IBr = 8.28 / 207 = 0.04 mol. Initial pressure of IBr = nRT/V = (0.04 × 0.0821 × 500) / 0.1642 = 10 atm. For 2IBr ⇌ I₂ + Br₂, P_Br₂ = 4.0 atm, so P_I₂ = 4.0 atm. The pressure of IBr reacted = 2 × 4.0 = 8.0 atm. Equilibrium P_IBr = 10 - 8.0 = 2.0 atm. K_p = (P_I₂ × P_Br₂) / (P_IBr)² = (4.0 × 4.0) / (2.0)² = 16 / 4 = 4. Therefore, correct answer is 4.

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