Presentation is loading. Please wait.

Presentation is loading. Please wait.

A.P. Chemistry Chapter 13 Equilibrium. 13.1 Equilibrium is not static, but is a highly dynamic state. At the macro level everything appears to have stopped.

Similar presentations


Presentation on theme: "A.P. Chemistry Chapter 13 Equilibrium. 13.1 Equilibrium is not static, but is a highly dynamic state. At the macro level everything appears to have stopped."— Presentation transcript:

1 A.P. Chemistry Chapter 13 Equilibrium

2 13.1 Equilibrium is not static, but is a highly dynamic state. At the macro level everything appears to have stopped but at the molecular level, there is frantic activity. The concentration of products builds as the reaction proceeds. There will come a time, in some reactions, where the products will collide and thus react and reform the reactants. When both the forward and reverse reaction occur at the same rate, there is no change in concentration of reactants and products, and the reaction is said to be at equilibrium A double arrow (  ) is used to show that a reaction can occur in either direction.

3 13.2 For the generic equation jA + kB  mC + nD(p. 615) A, B, C, and D represent chemical species, and j, k, m, and n are their coefficients in the balanced equation. Since the concentration of the products and reactants remains constant at equilibrium, we can set up a mass action expression: which is the concentration of the products (raised to the power of their coefficients) divided by the concentration of the reactants (raised to the power of their coefficients), K c = [C] m [D] n [A] j [B] k (p. 615) The mass action expression can then be used to find a constant for the reaction at a given temperature. This constant is known as the equilibrium expression (K c or K eq ). (p. 617)

4 The equilibrium expression for a reaction is the reciprocal of that for the reaction written in reverse. When the balanced equation for a reaction is multiplied by a factor n, the equilibrium expression for the new reaction is the original expression raised to the nth power. (K new = (K original ) n ) The units for K are determined by the powers of the various concentration terms, and depend on the reaction being considered. What does it mean when the value of K eq is very large? (products are favored heavily) What does it mean when the value of K eq is very small? (reactants are favored heavily) Equilibrium position- set of equilibrium concentrations. There is only one equilibrium constant for a particular system at a particular temperature, but there are an infinite number of equilibrium positions. This can tell us in what direction the reaction will proceed. (p. 618)

5 Example: For the Haber process 3H 2 (g) + N 2 (g)  2NH 3 (g), K c = 8.7 at a given temperature, T. Write the equilibrium expression for the equation. K = [NH 3 ] 2 /([H 2 ] 3 [N 2 ]) Calculate the equilibrium constant if the reaction is written as: 3/2 H 2 + ½ N 2  NH 3 K’ c = (8.7) ½ 2NH 3  3H 2 + N 2 K’’ c = 1/K = (8.7) -1

6 13.3 Equilibrium Positions Involving Pressures Rearrange the Ideal Gas Law to include molarity (use C for concentration) PV = nRT or P = (n/V)RT = CRT (p. 619) Where C = n/V, or the number of moles n of gas per unit volume V. Thus C represents the molar concentration of the gas. For the Haber process: The equilibrium expression would look like: K = [NH 3 ] 2 = C NH3 2 [H 2 ] 3 [N 2 ] C H2 3 C N2 Or, in terms of the equilibrium partial pressures of the gases: (p. 620) Kp = P NH3 2 P H2 3 P N2 Since we learned there is a relationship which exists between concentration and pressure, so there must be a relationship between K c and K p : (p. 621) K p = K(RT) /\n Where /\n is the sum of the coefficients of the gaseous products minus the sum of the coefficients of the gaseous reactants (see a derivation of this equation on p. 621)

7 13.4 Heterogeneous Equilibria Homogeneous equilibria- where all reactants and products are all in the same phase (p. 622) (Haber process is an example of a homogeneous equilibrium; all substances are in gas phase) Heterogeneous equilibria- where all reactants and products are not in the same phase. These equilibria do not depend on the amounts of pure solids or liquids (Therefore, pure solids and pure liquids are not included in the equilibrium expression)(p. 622) Example: CaCO 3 (s)  CaO(s) + CO 2 (g) K = [CO 2 ]

8 13.5 Applications of the Equilibrium Constant (p. 627) Reaction Quotient (Q) – obtained by applying the law of mass action using initial concentrations instead of equilibrium concentrations. This will help us to determine which direction the reaction will proceed in to reach equilibrium. For example, for the Haber process again, the expression for the reaction quotient would be Q = [NH 3 ] 2 /([H 2 ] 3 [N 2 ] From this equation, 3 conditions may occur: Q = K. The system is at equilibrium. No shift will occur. Q>K In this case, the ratio of initial concentrations of products to initial concentrations of reactants is too large. To reach equilibrium, a net change of products to reactants must occur. The system shifts to the left, consuming products and forming reactants, until equilibrium is reached. Q { "@context": "http://schema.org", "@type": "ImageObject", "contentUrl": "http://images.slideplayer.com/14/4217051/slides/slide_8.jpg", "name": "13.5 Applications of the Equilibrium Constant (p.", "description": "627) Reaction Quotient (Q) – obtained by applying the law of mass action using initial concentrations instead of equilibrium concentrations. This will help us to determine which direction the reaction will proceed in to reach equilibrium. For example, for the Haber process again, the expression for the reaction quotient would be Q = [NH 3 ] 2 /([H 2 ] 3 [N 2 ] From this equation, 3 conditions may occur: Q = K. The system is at equilibrium. No shift will occur. Q>K In this case, the ratio of initial concentrations of products to initial concentrations of reactants is too large. To reach equilibrium, a net change of products to reactants must occur. The system shifts to the left, consuming products and forming reactants, until equilibrium is reached. Q

9 13.6 Solving Equilibrium Problems (p. 635) 1. write the balanced equation for the reaction. 2. write the equilibrium expression using the law of mass action. 3. list the initial concentrations. 4. calculate Q, and determine the direction of the shift to equilibrium. 5. define the change needed to reach equilibrium, and define the equilibrium concentrations by applying the change to the initial concentrations. 6. substitute the equilibrium concentrations into the equilibrium expression; solve for the unknown. 7. check your calculated equilibrium concentrations by making sure that they give the correct value of K. Small values of K and the resulting small shift to the right to reach equilibrium allow for simplification. This is okay as long as the resulting change in value is less than 5% of the original value. If not, then you must use the quadratic formula.

10 Example: 3.0 mol of iodine and 4.0 mol of bromine are placed in a 2.0 L reaction vessel at 150 o C. The reaction comes to equilibrium at which point 3.2 mol of iodine bromide are present. Determine the equilibrium constant for the reaction: I 2 (g) + Br 2 (g)  2IBr(g) Set up a rice!! K c = 3.0

11 Example: Consider the equilibrium: H 2 (g) + I 2 (g)  2HI(g) If 2.0 mol of hydrogen gas and 2.0 mol of iodine gas are introduced into a 2.0 L reaction vessel at 400 o C, determine the concentrations of all species at equilibrium. (K c = 64 @ 400 o C) [H 2 ] = 1.0M - x = 0.20 M [I 2 ] = 1.0M - x = 0.20 M [HI] = 2x = 1.60M

12 Example: Using above reaction, Calculate K p for this reaction @ 400 o C. Determine the equilibrium partial pressure of HI @ 400 o C if initially the partial pressures of H 2 and I 2 are 2.0 and 4.0 atm respectively, at 400 o C. K p = 64(K p = K c whenever the # of moles of gas on each side of the equation is equal) P HI = 2x = 2(1.9) = 3.8 atm

13 13.7 LeChatelier’s Principle- if a change is imposed on a system in equilibrium, the position of the equilibrium will shift in a direction that tends to reduce that change (or, in other words, it will shift in a direction to alleviate the stress applied to the system). (p. 641) Effect of change in concentration If a reactant or product is added to a system at equilibrium, the system will shift away from the added component to decrease the amount of the added component. If a reactant or product is removed, the system will shift toward the removed component, to make more of the removed component. (p. 641)

14 Effect of a change in pressure(p. 643) The addition of an inert gas increases the total pressure but has no effect on the concentrations or partial pressures of the reactants or products. When the volume of the container holding a gaseous system is reduced, the system responds by reducing its own volume. This is done by decreasing the total number of gaseous molecules in the system, so the reaction will shift in the direction which has the least number of molecules. Adding or removing a gas product or reactant (see Effect in Change of Concentration)

15 Effect of a Change in Temperature It is important to realize that although the changes discussed so far may alter the equilibrium position, they do not alter the equilibrium constant. The value of K is altered with a change in temperature. If energy is added to an exothermic system at equilibrium by heating it, the shift will occur in the direction which will consume energy, that is, to the left. Opposite would be true for an endothermic reaction. (In other words, this acts like concentration, the equilibrium will shift away if there is an increase on the side where the energy term is written.)(p. 646)

16 Example: Consider the equilibrium CaCO 3 (s)  CaO(s) + CO 2 (g) A 1.0 L reaction vessel contains 10.0 g of CaO at 1000 o C. At this temperature, 0.20 atm of carbon dioxide gas is introduced into the reaction vessel. Will any calcium carbonate form? Justify your answer. At 1000 o C, K c = 0.0370 for this reaction. K p = K c (RT) /\n = (0.0370)[(0.0821)(1273)] 1 = 3.87 Q = 0.20 Q { "@context": "http://schema.org", "@type": "ImageObject", "contentUrl": "http://images.slideplayer.com/14/4217051/slides/slide_16.jpg", "name": "Example: Consider the equilibrium CaCO 3 (s)  CaO(s) + CO 2 (g) A 1.0 L reaction vessel contains 10.0 g of CaO at 1000 o C.", "description": "At this temperature, 0.20 atm of carbon dioxide gas is introduced into the reaction vessel. Will any calcium carbonate form. Justify your answer. At 1000 o C, K c = 0.0370 for this reaction. K p = K c (RT) /\n = (0.0370)[(0.0821)(1273)] 1 = 3.87 Q = 0.20 Q

17 For the above reaction, /\H o = 175 kJ/mol How will the equilibrium be affected by each of the following? Adding CaCO 3 (s) (no effect; solids never appear in the equilibrium expression) Removing CO 2 (g) (a net shift to the right; system acts to replace the carbon dioxide) Halving the volume of the reaction vessel (a net shift to the left; halving the volume doubles all of the partial pressures; the reaction shifts to the side w/ fewer gas moles) Adding helium gas to the reaction vessel (no effect; even though it increases the total pressure, it does not change the partial pressure of carbon dioxide) Heating the reaction vessel (a net shift to the right; endothermic reaction. Think of heat as a reactant term. You will consume it and send the reaction to the other side) Adding a catalyst to the reaction vessel (no effect; changes the rate, equilibrium is attained faster, but it does not change equilibrium positions)


Download ppt "A.P. Chemistry Chapter 13 Equilibrium. 13.1 Equilibrium is not static, but is a highly dynamic state. At the macro level everything appears to have stopped."

Similar presentations


Ads by Google