Guest funkygg Posted June 19, 2005 Posted June 19, 2005 I tried to do it, but please check if it's right or not, The equilibrium expression is: 2ICl -> I2 + Cl2 Kc is 4.8x10-6 Find out the equilibrium concentration of Cl2, if the initial concentration of ICl is 1.33 mol/L. There is no I2 or C2 initially present. so... Kc = [iCl] / ([i2] [Cl2]) Substituting gives: 4.8x10-6 = 1.33 - x / (x times x) Next we arrive at this quadratic equation in standard form: 4.8x10-6x2 + x - 1.33 = 0 Using the quadratic formula, which is x = (- b ± square root[b2 - 4ac]) / 2a, we obtain: x = (- 1 + square root[(1)2 - (4) (4.8x10-6) (-1.33)]) / 9.6x10-6 After suitable calculations, i got x = 1.33 strange tho, thanx for looking over.
EwenM Posted June 19, 2005 Posted June 19, 2005 No, you have the wrong expression for Kc: Kc = [i2] [Cl2] / [iCl]^2 (That's divided by the concentration of ICl squared - sorry about the formatting). Because [i2] will be equal to [Cl2] you can simplify this by taking the square root: [Cl2]/[iCl] = sqrt(Kc) [Cl2] turns out to be 2.91 x 10^-3
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