![]() ![]() However, the calculations must be done in molarity. Notice that moles are given and volume of the container is given. So the root of 1.92 is rejected in favor of the 0.26 value and the three equilibrium concentrations can be calculated.Ĭalculate all three equilibrium concentrations when 0.500 mole each of H 2 and Br 2 are mixed in a 2.00 L container and K c = 36.0. It is the term (0.5 − x) which must be positive. The answer lies in the fact that x is not the final answer, whereas (0.5 − x) is. The question then becomes how to determine which root is the correct one to use.Ĭalculate all three equilibrium concentrations when K c = 0.680 with o = 0.500 and o = 1.00 M.ģ) After some manipulation (left to the student), we arrive at this quadratic equation, in standard form:Ĥ) Using a quadratic equation solver, we wind up with this:ĥ) Both roots yield positive values, so how do we pick the correct one? The third example will be one in which both roots give positive answers. The answer you get will not be exactly 16, due to errors introduced by rounding. An amount of a substance with physical reality cannot be represented with negative numbers.Ħ) Determination of the equilibrium amounts and checking for correctness by inserting back into the equilibrium expression is left to the student. Amounts of substances are always represented with positive numbers. Why?īecause we are dealing with the amount of a physical substance in mol / L. The answer obtained in this type of problem CANNOT be negative. In the second example, the quadratic formula will be used.Ĭalculate all three equilibrium concentrations when K c = 16.0 and o = 1.00 M.ģ) After suitable manipulation (which you can perform yourself), we arrive at this quadratic equation in standard form:ĥ) Please notice that the negative root was dropped, because −b turned out to be −1. This mistake happens a LOT!!Ĩ) Both sides are perfect squares (done so on purpose), so we square root both sides to get:įrom there, the solution should be easy and results in x = 0.160 M.ĩ) This is not the end of the solution since the question asked for the equilibrium concentrations, so:ġ0) You can check for correctness by plugging back into the equilibrium expression: For convenience, here is the equation again:Ħ) Plugging values into the expression gives:ħ) Two points need to be made before going on:ġ) Where did the 64.0 value come from? It was given in the problem.Ģ) Make sure to write (2x) 2 and not 2x 2. It is simply the initial conditions with the change applied to it:ĥ) We are now ready to put values into the equilibrium expression. Since we have only one equation (the equilibrium expression) we cannot have two unknowns.Ĥ) The equilibrium row should be easy. In problems such as this one, never use more than one unknown. In fact, always use the coefficients of the balanced equation as coefficients on the "x" terms. HI is being made twice as fast as either H 2 or I 2 are being used up. The positive signifies that more HI is being made as the reaction proceeds on its way to equilibrium. For every one H 2 used up, one I 2 is used up also. We know this from the coefficients of the equation. What we do know is that an EQUAL amount of each will be used up. X signifies that we know some H 2 and I 2 get used up, but we don't know how much. The minus sign comes from the fact that the H 2 and I 2 amounts are going to go down as the reaction proceeds. This is the one that causes the most difficulty in understanding: ![]() The first two values were specified in the problem and the last value ( = 0) come from the fact that the reaction has not yet started, so no HI could have been produced yet.ģ) Now for the change row. The ChemTeam hopes you notice that I, C, E are the first initials of Initial, Change, and Equilibrium.Ģ) Now, let's fill in the initial row. Just in case you are not sure, the subscripted zero, as in o, means the initial concentration.Ĭalculate all three equilibrium concentrations when o = o = 0.200 M and K c = 64.0.ġ) The solution technique involves the use of what is most often called an ICEbox. In this type of problem, the K c value will be given You just plug into the equilibrium expression and solve for K c.Ĭalculating equilibrium concentrations from a set of initial concentrations takes more calculation steps. ChemTeam: Calculating Equilibrium Concentrations from Initial ConcentrationsĬalculating Equilibrium Concentrations from Initial ConcentrationsĬalculating K c from a known set of equilibrium concentrations seems pretty clear. ![]()
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