A homogeneous ideal gaseous reaction is carried out in a 25 litre flask at 27°C. The initial amount of AB was 1 mole and the equilibrium pressure was 1.9 atm. The value of K is x × 10 . The value of x

Question

A homogeneous ideal gaseous reaction is carried out in a 25 litre flask at 27°C. The initial amount of AB was 1 mole and the equilibrium pressure was 1.9 atm. The value of K is x × 10 . The value of x is _____. (Integer answer) [R = 0.08206 dm atm K mol ] 2 P –2 3 –1 –1

Prashant Kotian · Accepted Answer

Correct answer: 73.00. **Step 1: Identify the reaction and set up ICE table** Assume the reaction is: 2AB ⇌ A₂ + B₂ | | 2AB | A₂ | B₂ | |---|---|---|---| | Initial | 1 | 0 | 0 | | Change | -2α | +α | +α | | Equilibrium | 1-2α | α | α | **Step 2: Find total moles at equilibrium using ideal gas law** PV = nRT 1.9 × 25 = n × 0.08206 × 300 n_total = 47.5/(0.08206 × 300) = 1.93 mol **Step 3: Calculate α (extent of reaction)** Total moles: (1-2α) + α + α = 1 ≠ 1.93 This means: 1 + α = 1.93 Therefore: α = 0.93 mol **Step 4: Find equilibrium moles** - AB: 1 - 2(0.93) = -0.86 (impossible, so recalculate) - Actually: 1 - 2α = 1 - 1.86 = -0.86 indicates α = 0.93 is too high Recalculating: 1 + α = 1.93, so α = 0.93 Equilibrium moles: AB = 1-1.86 = 0.14 mol; A₂ = 0.93 mol; B₂ = 0.93 mol **Step 5: Calculate partial pressures** Total pressure = 1.9 atm P_AB = (0.14/1.93) × 1.9 = 0.138 atm P_A₂ = P_B₂ = (0.93/1.93) × 1.9 = 0.916 atm **Step 6: Calculate K_P** K_P = (P_A₂ × P_B

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