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Chemical Equilibrium. Reaction Types So far this year we have been writing chemical formulas as completion reaction. So far this year we have been writing.

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Presentation on theme: "Chemical Equilibrium. Reaction Types So far this year we have been writing chemical formulas as completion reaction. So far this year we have been writing."— Presentation transcript:

1 Chemical Equilibrium

2 Reaction Types So far this year we have been writing chemical formulas as completion reaction. So far this year we have been writing chemical formulas as completion reaction. In other words, the reaction completely goes to products. In other words, the reaction completely goes to products. We represented this by using  in the chemical equation. We represented this by using  in the chemical equation. Example: Example: S 8 + 8O 2  8SO 2

3 Reversible Reactions However, many of the reactions we study are actually reversible. However, many of the reactions we study are actually reversible. Reversible reactions are chemical reactions in which the products re-form the original reactants. Reversible reactions are chemical reactions in which the products re-form the original reactants. Eventually, a chemical equilibrium is formed or a state of balance where the rate forward reaction equals the rate in reverse. Eventually, a chemical equilibrium is formed or a state of balance where the rate forward reaction equals the rate in reverse.

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5 Chemical equilibria are dynamic In other words, even though it looks like the reaction has stopped, the substances are still reacting. In other words, even though it looks like the reaction has stopped, the substances are still reacting. However, for every new product that was made, an old product turned back into reactant. However, for every new product that was made, an old product turned back into reactant. Note that this does not mean that products and reactants need to be equal in concentration. Note that this does not mean that products and reactants need to be equal in concentration.

6 That is why we have the Equilibrium Constant A number that relates the concentrations of starting materials to products of a reversible reaction at a given temperature. A number that relates the concentrations of starting materials to products of a reversible reaction at a given temperature. Symbolized by K eq Symbolized by K eq They do not have units They do not have units Remember, concentrations are expressed in molarity (mol/L) Remember, concentrations are expressed in molarity (mol/L)

7 K eq can show whether a reaction is favorable If K eq equals a large number, then there will be more products than reactants. (bigger arrow pointing to the right) If K eq equals a large number, then there will be more products than reactants. (bigger arrow pointing to the right) Reactions that produce more products than reactants are known as favorable. Reactions that produce more products than reactants are known as favorable. If K eq equals a small number, then there will be more reactants than products. (bigger arrow pointing to the left) If K eq equals a small number, then there will be more reactants than products. (bigger arrow pointing to the left) If the value is 1, then it is 50:50 If the value is 1, then it is 50:50

8 How do we determine if K eq is small or large? We need to write the K eq formula (expression) for the reaction We need to write the K eq formula (expression) for the reaction Remember, these constants only apply to equations in equilibrium Remember, these constants only apply to equations in equilibrium Step 1 – Write the balanced equation for the reaction. Step 1 – Write the balanced equation for the reaction.

9 Step 2 – Write the equilibrium expression. Place the product concentrations in the numerator and the reactant concentration in the denominator. Place the product concentrations in the numerator and the reactant concentration in the denominator. All will be multiplied All will be multiplied This is done by writing each ion, compound, etc, in brackets This is done by writing each ion, compound, etc, in brackets Leave out any solids or pure liquids Leave out any solids or pure liquids This includes water if dealing with an aqueous solution This includes water if dealing with an aqueous solution

10 Step 3 Complete the equilibrium expression Complete the equilibrium expression This is done by raising each concentration by its coefficient from the balanced equation. This is done by raising each concentration by its coefficient from the balanced equation. Examples Examples CaCO 3 (s) + 2H 3 O + (aq) Ca 2+ (aq) + CO 2 (g) + 3H 2 O(l) N 2 (g) + O 2 (g) 2NO(g)

11 We can then use these expressions to solve problems. Let’s look at page 506; problems 1 and 2.

12 Solubility Product Constant, K sp A special equilibrium constant A special equilibrium constant These are equilibrium constants for the dissolution of slightly soluble salts These are equilibrium constants for the dissolution of slightly soluble salts It is for describing the saturation of a salt in solution. It is for describing the saturation of a salt in solution. The expression is written in the same way as the other equilibrium constants. The expression is written in the same way as the other equilibrium constants. Remember, solids and pure liquids are not included in the expression. Remember, solids and pure liquids are not included in the expression.

13 Examples The K sp for silver carbonate is 8.4x10 -12 at 298. The concentration of carbonate ions in a saturated solution is 1.28 x 10 -4 M. What is the concentration of silver ions? The K sp for silver carbonate is 8.4x10 -12 at 298. The concentration of carbonate ions in a saturated solution is 1.28 x 10 -4 M. What is the concentration of silver ions? The ionic substance T 3 U 2 ionizes to form T 2+ and U 3- ions. The solubility of T 3 U 2 is 3.77 x 10 -20 mol/L. What is the value of the solubility-product constant? The ionic substance T 3 U 2 ionizes to form T 2+ and U 3- ions. The solubility of T 3 U 2 is 3.77 x 10 -20 mol/L. What is the value of the solubility-product constant?

14 Before we work on some problems in Groups, all groups to a board Part 1) Methane, CH 4 and water, H 2 O, react to make carbon monoxide, CO, and hydrogen, H 2. All are gases. Write the balanced equation. Part 2) Write the equilibrium expression. Part 3) Suppose you were given the concentrations: H 2 = 1.56 M, CH 4 = 3.70 x 10 -2 M, H 2 O = 0.827, and the K eq = 3.112 x 10 2. What is the concentration of CO?

15 Homework Page 522-523: 17 (pick 4 of the eight to complete), 27, 28, 31, 35 Page 522-523: 17 (pick 4 of the eight to complete), 27, 28, 31, 35


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