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Homework 6.) 1.9 x 10 -2 7.) 1.2 x 10 2 8.) 1.3 x 10 -2 9.) 0.046 mol/L 10.) 0.15.

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Presentation on theme: "Homework 6.) 1.9 x 10 -2 7.) 1.2 x 10 2 8.) 1.3 x 10 -2 9.) 0.046 mol/L 10.) 0.15."— Presentation transcript:

1 Homework 6.) 1.9 x ) 1.2 x ) 1.3 x ) mol/L 10.) 0.15

2 True or False Game 1.At equilibrium, changes in macroscopic properties of a system can be observed. 2.Equilibrium involves dynamic change at the molecular level 3.Equilibrium can be reached in closed systems only. 4.At equilibrium, the concentration of reactants does not have to equal the concentration of products. 5.Equilibrium constants,K c, have the same units as molar concentration

3 Review of Equilibrium Constants If K c < 1 The reactants are favored If K c > 1 The products are favored Every reaction has a unique K c value for every temperature. Pure solids and Pure liquids are not included in the K c expression because their concentrations remain constant.

4 Review of Equilibrium Constants H 2(g) + I 2(g) 2HI (g) We can calculate the equilibrium constant, K c if we know the equilibrium concentrations of each of the reactants and products. What if we only know the initial concentrations? How could we find our equilibrium concentrations?

5 ICE Tables H2H2 I2I2 2HI I C E ICE Table YouTube Video :

6 What are ICE Tables used for? Used to organize given and calculated data Table includes: Initial Concentration, Change in Concentration, and Equilibrium Concentration

7 Steps to Follow 1.Make sure reaction is balanced. Draw ICE table and enter data given in question. 2.Use given data and your stoichiometry skills to determine change in concentration for each reactant and product. 3.Calculate the equilibrium concentrations of each reactant and product

8 Example of an ICE Table Question mol of hydrogen gas and mol of iodine gas were mixed in a 1.00 L reaction vessel and heated to 450 °C. At equilibrium, concentration of iodine was mol/L. Calculate the equilibrium concentration of hydrogen gas and hydrogen iodide gas.

9 Example #2 A mixture of iron(III) nitrate and potassium thiocyanate, in aqueous solution, react to form the iron(III) thiocyanate ion. Fe 3+ (aq) + SCN - (aq) Fe(SCN) 2+ (aq) The initial concentration of Fe 3+ is M, and SCN - is M. The final concentration of Fe(SCN) 2+ is measured to be 4.5x Calculate the value for K c. Are the products or reactants favored?

10 Work on Practice Problem Handout. See pages in purple book for further explanation or theory if needed


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