Student Doctor Network Forums > Pre-Medical Forums > MCAT Discussions > MCAT Study Question Q&A > Keq coefficients problem (reward = cookies) PDA View Full Version : Keq coefficients problem (reward = cookies) BoneMental02-21-2011, 05:55 PMI am having trouble figure out when you should multiply by the coefficients in an equilibrium constant expression. For example, take the reaction: 3A + 2B --> 4C Assume the reaction is only this one step, and thus, the coefficients can be used as the exponents. Keq = (A)^3 * (B)^2 ...............(C)^4 I have noticed that when you are given the molar solubility (let's say, 5M) and asked to solve for Keq, you would use the coefficients when evaluating: Keq = (3x)^3 * (2x)^2 where x = 5M ...............(4x)^4 However, if given the molarities of each and asked to solve for Keq, you do NOT use the coefficients. For example, if you are told you have 4M of A, 2M of B, and 5M of C: Keq = (4)^3 * (2)^2 ...............(5)^4 Can someone please explain why this is? I can't figure out when I should use the coefficients or not!!!! AHHHHH!!! Rabolisk02-21-2011, 08:59 PMAssume the reaction is only this one step, and thus, the coefficients can be used as the exponents. It is not necessary to assume anything about the mechanism of a reaction. Keq = (A)^3 * (B)^2 ...............(C)^4 You reversed this. The answer should be the inverse of this. I have noticed that when you are given the molar solubility (let's say, 5M) and asked to solve for Keq, you would use the coefficients when evaluating: Keq = (3x)^3 * (2x)^2 where x = 5M ...............(4x)^4 Molar solubility means the moles of solute that will dissolve in a liter of solution. In a reaction A --> B + 2C, every mole of A that dissolves produces a mole of B and 2 moles of C. Ksp = [B][C]^2 = (x)(2x)^2, where x is the molar solubility of A. At equilibrium, there should be no net dissolution or precipitation, so the maximum amount of A that can dissolve in a given amount of solvent (molar solubility) will dissolve. The reason you use coefficients is because [C] has to be 2[B], by stoichiometry, assuming that you initially start out with only A in a solvent. Essentially you're using ICE box to solve for K. However, if given the molarities of each and asked to solve for Keq, you do NOT use the coefficients. Because you are already given the concentration of each species, stoichiometry does not matter. Just because a reaction produces 1 mole of B and 2 moles of C does not mean that the concentration of C is necessarily twice that of B, because either one of those can be selectively removed. When you are given the equilibrium concentrations, you don't know how you arrived at this state, only that the reaction is at equilibrium.