Guest funkygg Posted June 19, 2005 Share 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. Link to comment Share on other sites More sharing options...
EwenM Posted June 19, 2005 Share 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 Link to comment Share on other sites More sharing options...
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