Question

If the value of Kp for N2+3H2...2NH3 is 4/27, then for which of following reaction the value of Kp is reciprocal of above reaction: A. 2SO2+O2...2SO3 B. N2O4...2NO2 C. 2O3...3O2 D. PCI5...PCL3+CL2

Answer 1

C. 2O₃ ⇌ 3O₂

Step-by-step explanation:

Kp is the equilibrium constant calculated from the partial pressures of a gas-phase reaction equation.

For a general gas-phase reaction aA + bB ⇌ nC + xD

the expression for the Kp = (pC)ⁿ(pD)ˣ / (pA)ᵃ(pB)ᵇ

where pA = partial pressure of A; pB = partial pressure of B; pC = partial pressure of C; pD = partial pressure of D

From the given reaction in equilibrium; N₂ + 3H₂ ⇌ 2NH₃

Kp = (pNH₃)² / (pN₂)¹ * (pH₂)³ = 4/7

(pNH₃)² / (pN₂)¹ * (pH₂)³ = (2)²/ (1)¹ * (3)³

Therefore, number of mole of reactants and products is equivalent to partial pressure.

A. 2SO₂ ⇌ O₂ + 2SO₃

pSO₂ = 2, pO₂ = 1, pSO₃ = 2,

Kp = 2²/ (2² * 1²) = 4/4 = 1

B. N₂O₄ ⇌ 2NO₂

pN₂O₄ = 1, pNO₂ = 2

Kp = 2²/1² = 4

C. 2O₃ ⇌ 3O₂

pO₃ = 2, pO₂ = 3

Kp = 3³/2² = 27/4

D. PCl₅ ⇌ PCl₃ + Cl₂

pPCl₅ = 1, pPCl₃ = 1, pCl₂ = 1

Kp = (1¹ * 1¹) / 1¹ = 1