Question

Consider the following reaction: SO2Cl2(g)⇌SO2(g)+Cl2(g) Kc=2.99×10−7 at 227 ∘C If a reaction mixture initially contains 0.150 MSO2Cl2, what is the equilibrium concentration of Cl2 at 227 ∘C?

Answer 1

SOLUTION The given reaction is on SoCl₂ (9) - SO, (g) + Cl₂ lg) we know that at equilibrium, Ke= [so] [Cl₂ [s0,4] 1 where ke is the equilibrium constant. [So,] is the concentration of so, at equilibriun [Ca] and [50, C₂] are the concentration of Cl, and soul respectively at equilibrium uiven - K for reaction = 2.99 x 10-7 Initial [S0₂ C₂ = 0.150 m So Cl 19) - So, (g) + Cl₂ (9) 150M 0 0 Initial Concentration final +71 0.150-2 Concentration Using relation as stated above, 2.99x 10-7 = n. 0.150-2

0x 2.99 xlo? - ? 0.150 - as n is. neglecting an in the denominator very small we get, 2.99x107- 0.150 On solving, na 2.12 & 10-4 M So Equilibrium concentration of cle = 2.12x1024 M
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