There will be two values possible for the unknown x, and for a quadratic equation with the general formula ax 2 + bx + c = 0 (where a, b, and c are the coefficients of the quadratic equation), the two possible values are as follows: One of the more common equations has an x 2 term in it and is called a quadratic equation. Sometimes when an ICE chart is set up and the K eq expression is constructed, a more complex algebraic equation will result. ![]() The + sign is included explicitly in the change row of the ICE chart to avoid any confusion. The ICE chart is a more formalized way to do these types of problems. Which, of course, is the same expression we have already solved and yields the same answers for the equilibrium concentrations. Substituting the last row into the expression for the K eq yields The ICE chart for the above example would look like this: The expressions in the equilibrium row are substituted into the K eq expression, which yields an algebraic equation that we try to solve. Finally, the equilibrium expressions in the last row are a combination of the initial value and the change value for each species. However, the change values must be in the proper stoichiometric ratio as indicated by the balanced chemical equation. The change values, usually algebraic expressions because we do not yet know their exact numerical values, go in the next row. The initial values go in the first row of the chart. We formalize this process by introducing the ICE chart, where ICE stands for initial, change, and equilibrium. To check, we simply substitute these concentrations and verify that we get the numerical value of the K eq, in this case 4.0: To determine the equilibrium concentrations, we need to go back and evaluate the expressions 1 − x and 2 x to get the equilibrium concentrations of our species: Now we have to remind ourselves what x is - the amount of H 2 and Cl 2 that reacted - and 2 x is the equilibrium. ![]() Now we rearrange and solve (be sure you can follow each step): So we can take the square root of both sides: The fraction is a perfect square, as is the 4.0 on the right. This expression may look formidable, but first we can simplify the denominator and write it as a perfect square as well: This is an equation in one variable, so we should be able to solve for the unknown value. We can substitute these concentrations into the K eq expression for this reaction and combine it with the known value of K eq: So now we know the equilibrium concentrations of our species: How much HCl is made? We start with zero, but we also see that 2 mol of HCl are made for every mole of H 2 (or Cl 2) that reacts (from the coefficients in the balanced chemical equation), so if we lose x M H 2, we gain 2 x M HCl. If we start with 1.0 M Cl 2 at the beginning and we react x M, we have (1.0 − x) M Cl 2 left at equilibrium. This means that if x M H 2 reacts, x M Cl 2 reacts as well. How do we know that? The coefficients of these two species in the balanced chemical equation are 1 (unwritten, of course). This means that at equilibrium, we have (1.0 − x) M H 2 left over.Īccording to the balanced chemical equation, H 2 and Cl 2 react in a 1:1 ratio. Let us assume that x M H 2 reacts as the reaction goes to equilibrium. But by how much will it proceed? We don’t know, so let us assign it a variable. Its associated K eq is 4.0, and the initial concentration of each reactant is 1.0 M:īecause we have concentrations for the reactants but not the products, we presume that the reaction will proceed in the forward direction to make products. Such calculations are not difficult to do, especially if a consistent approach is applied. There are some circumstances in which, given some initial amounts and the K eq, you will have to determine the concentrations of all species when equilibrium is achieved. ![]() Calculate equilibrium concentrations from the values of the initial amounts and the K eq.Square brackets: means the concentration of x. Square brackets mean the concentration of the compound written within the The equation for the equilibrium constant uses the concentrations of the reactants and
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