CHEM 113 1nd Edition Lecture 13 Outline of Last Lecture I. K and Q for heterogeneous equilibriumII. Expressing equilibria with pressure termsIII. Determining the direction of a reactionOutline of Current Lecture II. Solving Equilibrium ProblemsIII. Practice problemCurrent Lecture- Here is a plan of action for solving equilibrium problemso When we know equilibrium quantities. Then you simply substitute those values into the K of concentration expression and calculate.o When we know some of the quantities. Then we use a reaction table to calculate quantities and find K of concentration by doing the following.- 1). Balance the equation- 2). Write the expression for K and Q- 3). Set up a Reaction Table by entering the followingo Initial quantities of reactants and products o Changes in the previous quantities during the reactiono Equilibrium quantities These amounts are calculated by adding the changeto the initial amount- 4). For most problems, values we put into the Reaction Table will be in terms of x. So then we use other information given in the problem to solve for x. - Sample problem 17.7These notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute.o Step 1 is to balance the equation but this has already been done.o Step 2 is to identify the expression for K, which is the product of concentrations of H2 and I2 gas divided by the concentration of HI squared. The reason why the concentration of HI is squared is because of the coefficient 2, which was found by balancing the equation.o We are given the concentration HI at equilibrium therefore we need to find the concentration of the products at equilibrium.o Now we set up a reaction tableConcentration 2HI H2 I2Initial .100 mol/L 0 0Change -2x X XEquilibrium .100-2x X X The initial concentration of HI was found by using the given information .200 mol of Hi gas and 2.00-L. The book gives a good explanation on how to fill in the rest of the table.o Solving for x Now we can set 0.078 M= .100-2x. This will gives us an equilibrium concentration for H2 and I2 of .011. We now plug the that value into our established expression for K This will give us a final K concentration value of
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