Chemistry
gabbimay
17

What is the answer? How to do ? Thx

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(1) Answers
Cesiacaylen

C - concentration of PCl3 and Cl2 C = n/V = 1mol/5dm3 = 0,2 mol/dm3 ............... START: PCl3 + Cl2 <----> PCl5 0,2........0,2...............0 EQUILIBRIUM: PCl3 + Cl2 <----> PCl5 0,2-x..0,2-x..............x K = [PCl5]/([Cl2][PCl3]) 25 = x/(0,2-x)² 25 = x/(0,04-0,4x+x²) 1-10x+25x² = x 25x²-11x+1 = 0 Δ = b² - 4ac Δ = (-11²) - (4×25×1) Δ = 121 - 100 = 21 √Δ = 4,583 x₁ = (-b+√Δ)/2a = 11+4,583/50 = 0,31166 0,31166 > 0,2 x₂ = (-b-√Δ)/2a = 11-4,583/50 = 0,12834 ≈ 0,128M _____________________________ [PCl5] = x = 0,128M >>> ( answer D ) :) ;)

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