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Chemical Equilibrium A Balancing Act. During a game, players enter and leave... But there are always the same number of players on field.

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Presentation on theme: "Chemical Equilibrium A Balancing Act. During a game, players enter and leave... But there are always the same number of players on field."— Presentation transcript:

1 Chemical Equilibrium A Balancing Act

2 During a game, players enter and leave... But there are always the same number of players on field.

3 H 2 O(l) H 2 O(g)

4 4 Equilibrium Remember: We’ve already used the phrase “equilibrium” when talking about rxns. In principle, every chemical reaction is reversible... capable of moving in the forward or backward direction. 2 H 2 + O 2 2 H 2 O Some reactions are easily reversible... Some not so easy...

5 5 The Concept of Equilibrium - Consider colorless frozen N 2 O 4. At room temperature, it decomposes to brown NO 2 : N 2 O 4 (g)  2NO 2 (g). At some time, the color stops changing and we have a MIXTURE of N 2 O 4 and NO 2. Chemical equilibrium is the point at which the rate of the forward reaction is equal to the rate of the reverse reaction. At that point, the concentrations of all species are constant.

6 Using the collision model: –as the amount of NO 2 builds up, there is a chance that two NO 2 molecules will collide to form N 2 O 4. –At the beginning of the reaction, there is no NO 2 so the reverse reaction (2NO 2 (g)  N 2 O 4 (g)) does not occur.

7 7 The Concept of Equilibrium As the substance warms it begins to decompose: N 2 O 4 (g)  2NO 2 (g) When enough NO 2 is formed, it can react to form N 2 O 4 : 2NO 2 (g)  N 2 O 4 (g) At equilibrium, as much N 2 O 4 reacts to form NO 2 as NO 2 reacts to re-form N 2 O 4 The double arrow implies the process is dynamic.

8 8 The Concept of Equilibrium As the reaction progresses –[A] decreases to a constant, –[B] increases from zero to a constant. –When [A] and [B] are constant, equilibrium is achieved.

9 9 Equilibrium CONSTANT For a general reaction the equilibrium constant expression is where K c is the equilibrium constant (or Kp if they are all gases) But we can use abbreviation K for all equilibrium constant … for now

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11 11 The Equilibrium Expression Write the equilibrium expression for the following reaction:

12 12 The Equilibrium Constant Heterogeneous Equilibria When all reactants and products are in one phase, the equilibrium is homogeneous. If one or more reactants or products are in a different phase, the equilibrium is heterogeneous. Consider: –experimentally, the amount of CO 2 does not seem to depend on the amounts of CaO and CaCO 3. Why? *Ignore (s) & ( l ) from equilibrium expression

13 13 The Equilibrium Constant Heterogeneous Equilibria

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16 K eq > 1: Products are equilibrium K eq < 1: Reactants are equilibrium Value of Equilibrium Constant

17 Calculate the value of K eq if [N 2 ] = 0.20 mol/L, [O 2 ] = 0.15 mol/L, and [NO] = mol/L. What does the value of K tell you about the equilibrium?

18 ICE ICE BABY

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21 21 Calculating Equilibrium Constants Steps to Solving Problems: 1.Write an equilibrium expression for the balanced reaction. 2.Write an “ICE table”. Fill in the given amounts. 3.Use stoichiometry (mole ratios) on the change in concentration line. 4.Deduce the equilibrium concentrations of all species. Usually, the initial concentration of products is zero. (But of course this is not always the case.)

22 Equilibrium example 1: The equilibrium constant is 7.6X for the rxn of decomposition of water. Write the balanced equation for the process. Suppose you start with initial concentration of water 1.27 M. Calculate the concentrations of all other species at equilibrium.

23 Equilibrium example 2: The equilibrium constant for a reaction between nitrogen and oxygen gas to produce nitrogen monoxide is 4.8X Write the balanced equation. Initial pressure of N 2 is and the initial pressure of O 2 is Calculate the pressures for all species at equilibrium.

24 Equilibrium Example 3: The equilibrium constant is 3.2 X for a reaction of decomposition of HCl into hydrogen and chlorine gas. Suppose you start with M solution of HCl. Find equilibrium concentrations of all species.

25 Equilibrium example 4: The Kc for reaction for formation of ammonia is 4.51X If we start with 1 atm of N 2 and 1.4 atm of H 2, find the equilibrium pressures.

26 26 Applications of Equilibrium Constants Predicting the Direction of Reaction We define Q, the reaction quotient, for a reaction at conditions NOT at equilibrium as where [A], [B], [P], and [Q] are molarities at any time. Q = K only at equilibrium

27 27 Applications of Equilibrium Constants Predicting the Direction of Reaction If Q > K then the reverse reaction must occur to reach equilibrium (go left) If Q < K then the forward reaction must occur to reach equilibrium (go right)

28 28 Le Chatelier’s Principle: if you disturb an equilibrium, it will shift to undo the disturbance. Remember, in a system at equilibrium, come what may, the concentrations will always arrange themselves to multiply and divide in the Keq equation to give the same number (at constant temperature). Le Châelier’s Principle

29 29 Change in Reactant or Product Concentrations Adding a reactant or product shifts the equilibrium away from the increase. Removing a reactant or product shifts the equilibrium towards the decrease. To optimize the amount of product at equilibrium, we need to flood the reaction vessel with reactant and continuously remove product (Le Châtelier). We illustrate the concept with the industrial preparation of ammonia

30 30 Change in Reactant or Product Concentrations Consider the Haber process If H 2 is added while the system is at equilibrium, the system must respond to counteract the added H 2 (by Le Châtelier). That is, the system must consume the H 2 and produce products until a new equilibrium is established. Therefore, [H 2 ] and [N 2 ] will decrease and [NH 3 ] increases.

31 31 Change in Reactant or Product Concentrations The unreacted nitrogen and hydrogen are recycled with the new N 2 and H 2 feed gas. The equilibrium amount of ammonia is optimized because the product (NH 3 ) is continually removed and the reactants (N 2 and H 2 ) are continually being added. Effects of Volume and Pressure As volume is decreased pressure increases. Le Châtelier’s Principle: if pressure is increased the system will shift to counteract the increase.

32 32 Consider the production of ammonia As the pressure increases, the amount of ammonia present at equilibrium increases. As the temperature decreases, the amount of ammonia at equilibrium increases. Le Châtelier’s Principle: if a system at equilibrium is disturbed, the system will move in such a way as to counteract the disturbance.

33 33 Change in Reactant or Product Concentrations

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36 36 Example

37 37 Effects of Volume and Pressure The system shifts to remove gases and decrease pressure. An increase in pressure favors the direction that has fewer moles of gas. In a reaction with the same number of product and reactant moles of gas, pressure has no effect. Consider

38 38 Effects of Volume and Pressure An increase in pressure (by decreasing the volume) favors the formation of colorless N 2 O 4. The instant the pressure increases, the system is not at equilibrium and the concentration of both gases has increased. The system moves to reduce the number moles of gas (i.e. the forward reaction is favored). A new equilibrium is established in which the mixture is lighter because colorless N 2 O 4 is favored.

39 39 Effect of Temperature Changes The equilibrium constant is temperature dependent. For an endothermic reaction,  H > 0 and heat can be considered as a reactant. For an exothermic reaction,  H < 0 and heat can be considered as a product. Adding heat (i.e. heating the vessel) favors away from the increase: –if  H > 0, adding heat favors the forward reaction, –if  H < 0, adding heat favors the reverse reaction.

40 40 Effect of Temperature Changes Removing heat (i.e. cooling the vessel), favors towards the decrease: –if  H > 0, cooling favors the reverse reaction, –if  H < 0, cooling favors the forward reaction. Consider for which  H > 0. –Co(H 2 O) 6 2+ is pale pink and CoCl 4 2- is blue.


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