1.0 L container NO 2 = 0.12 mole. time

Transcription

1 CHEM 1105 GAS EQUILIBRIA 1. Equilibrium Reactions - a Dynamic Equilibrium Initial amounts: = mole = 0 mole 1.0 L container = 0.12 mole moles = 0.04 mole 0 time (a) 2 In a 1.0 L container was placed 4.00 moles of NO(g) and the system was allowed to come to equilibrium. At equilibrium, 1.82 moles of N 2 (g) were present. Calculate the moles of the other gases at equilibrium. The equation is give below. (Ans moles O 2 ; 0.36 mole NO) 2NO(g) k N 2 (g) + O 2 (g) (b) In a 3.0 L container was placed 0.60 mole of NH 3 (g) and the system was allowed to come to equilibrium. At equilibrium, 0.18 mole of H 2 (g) was present. Calculate the moles of the other gases at equilibrium. The equation is give below. (Ans mole N 2 ; 0.48 mole NH 3 ) 2NH 3 (g) k N 2 (g) + 3H 2 (g) [see also Ebbing, 14.1 CO(g) + 3H 2 (g) k CH 4 (g) + H 2 O(g)] 2. Equilibrium Constant At 100 C Experiment 1 initial moles/l 0.00 equilibrium moles/l [ ] 2 [ ] 0.36 Experiment Experiment ([A] = moles A/L) 0.37

2 -2- Although the equilibrium concentrations of and vary with their initial concentrations, the ratio [ ] 2 /[ ] is constant and is called the Equilibrium Constant, K c. For a given equilibrium system, K c is independent of the initial concentrations, the volume and the total pressure but depends on the temperature. For the equation aa + bb k cc + dd, K c = [C]c [D] d [A] a [B] b [See also Ebbing, pp 601, 602 CO(g) + 3H 2 (g) k CH 4 (g) + H 2 O(g)] (a) Calculate K c for 1(a). (Ans. 26) (b) Calculate K c for 1(b). (Ans. 1.7 x 10-4 ) (c) (d) Consider the equilibrium 2SO 2 (g) + O 2 (g) k 2SO 3 (g) Into a 1.0 L container were placed mole SO 2 and mole O 2. At equilibrium, there was mole of SO 3. Calculate K c. (Ans. 2.8 x 10 2 ) Consider the equilibrium 2NH 3 (g) + 4H 2 O(g) k 2 (g) + 7H 2 (g) Into a 2.0 L container were placed mole NH 3, mole H 2 O, mole and mole H 2. When the system had reached equilibrium, there was mole of H 2 O present. Calculate the equilibrium concentrations of all of the gases and hence K c. (Ans. [NH 3 ] = 0.517; [H 2 O] = 0.433; [ ] = 0.254; [H 2 ] = 0.429; K c = ) (1) Reaction reversed: K reversed = 1 K forward (2) Reaction multiplied by n: K new = (K old ) n (3) Reaction 3 = Reaction 1 + Reaction 2: K 3 = K 1 x K 2 3. Heterogeneous Equilibria The concentration of pure liquids and solids are constant and hence do not appear in the expression for K c. (a) Consider the equilibrium 3Fe(s) + 4H 2 O(g) K Fe 3 O 4 (s) + 4H 2 (g) In a 3.0 L container is placed 5.0 g of H 2 O and 3.0 g of Fe. At equilibrium, there was only 2.0 g of Fe present. Calculate K c for the equilibrium. (Ans. 7.8 x 10-5 ) 4. Qualitative Significance of an Equilibrium Constant (1) K c is very small: At 25 C, K c = 1 x for N 2 (g) + O 2 (g) k 2NO(g) Very little NO present; negligible Forward Reaction, Backward Reaction goes to completion. ([NO] is so small that it cannot be measured; it must be calculated) (2) K c is very large: At 25 C, K c = 1 x for 2Cl(g) k Cl 2 (g) Very little Cl present; negligible Backward Reaction, Forward Reaction goes to completion. (3) K c is neither large nor small: Both "reactants" (LHS) and "products" (RHS) present in significant amounts. These are the reactions we are considering.

3 5. Using K c to Predict the Direction of Reaction for a System not at Equilibrium Consider the equilibrium 2SO 2 (g) + O 2 (g) k 2SO 3 (g) at a temperature at which K c = 280. In a 3.0 L container were placed 1.0 mole SO 3, 0.20 mole SO 2 and 0.90 mole O 2. Is the system at equilibrium, will the reaction go to the right (form more SO 3 ) or will the reaction go to the left (form more SO 2 and O 2 ) to attain equilibrium? To answer the question, one must compare the reaction quotient, Q c, with K c. The expression for Q c has the same form as that for K c except that the concentrations are not necessarily the equilibrium concentrations. 2 [SO 3 ] i = 3.0 ([A] i = moles/l at a particular instant) [SO 2 ] 2 = 83 i [O 2 ] i Q c is less than K c ; hence the reaction will go to the right (more SO 3 will form) until Q c = K c (i.e., equilibrium is attained). Obviously, if Q c > K c, the reaction will go to the left (to produce more "reactants" and less "products"). (a) Consider the equilibrium 2NO(g) k N 2 (g) + O 2 (g) at a temperature at which K c = 26. In a 2.0 L container were placed 0.20 mole NO, 1.0 mole O 2 and 1.4 mole N 2. Will the reaction go to the left (more NO produced) or to the right (more O 2 and N 2 produced) as the system attains equilibrium? (Ans. To the left, since Q c = 35) (See also Example 14.5, Exercise 14.8 and Problems and 14.48) 6. Using K c to Calculate Equilibrium Concentrations In the trivial case, the value of K c and all but one of the equilibrium concentrations are given and the unknown equilibrium concentration is to be found. -3- (a) For the equilibrium H 2 (g) + I 2 (g) K 2HI(g), K c = 55. Decomposing HI at 55 C gave [HI] at equilibrium = 0.50 M. Calculate [H 2 ] and [I 2 ] at equilibrium. (Ans. [H 2 ] = [I 2 ] = M) (See also Example 14.6, Exercise 14.9 and Problems and 14.52) More commonly, K c and the starting amount (or concentration) for one or more of the substances is given and all the equilibrium amounts (or concentrations) are to be calculated. (See Example 14.7) (b) Consider the equilibrium 2HI(g) K H 2 (g) + I 2 (g) at the same temperature as the equilibrium in 6(a). In a 1.0 L container is placed 0.80 mole of HI. Calculate the moles of each substance at equilibrium. (Ans mole HI, mole I 2 and mole H 2 ) (See also Exercise and Problems and 14.54) (c) To the equilibrium system in 6(b) [i.e., with the molar amounts given in the answer to 6(b)] is added mole H 2 and mole of I 2. Calculate the moles of each substance when equilibrium has been re-established. (Ans mole HI, mole I 2 and mole H 2 ) The problems in Example 14.7 and in 6(b) and 6(c) were easily solved since they involved perfect squares. If perfect squares are not present, a quadratic (or higher order) equation will have to be solved (See Example 14.8). Because this takes too much time or is too difficult, if such a problem is given only setting up the expression for K c will be required. (d) Consider the system 2NH 3 (g) k N 2 (g) + 3H 2 (g) (K c = 1.8 x 10 4 ) Into a 4.0 L container was placed 0.80 mole of NH 3. What are the equilibrium concentrations?

4 7. Effect of Changing the Conditions of an Equilibrium System (Le Chatelier s Principle). The most important, and common, reason to change the conditions of an equiluibrium system is to increase the amount of a desired product. The conditions whose change can affect the equilibrium composition and hence increase the amount of a desired product are concentration (by adding or removing any of the gaseous components of the equilibrium system), volume (resulting in a change of pressure ) and temperature. Le Chatelier s Principle can be used to predict how the equilibrium composition will change on changing any of the conditions above. This principle states that when a system in chemical equilibrium is disturbed by changing concentration, pressure or temperature, the equilibrium composition changes in a way that tends to offset, or counteract, the change in the condition(s). In considering changes in concentration and pressure, the equilibrium expression can also be used. -4- (1) Changes in Concentration Increasing a concentration by adding more substance is easy. Decreasing a concentration by removing a substance is more difficult. Examples are cooling to remove H 2 O, adding acid to remove NH 3 and NaOH to remove CO 2. (2) Changes in Volume ( Change in Pressure) When the pressure is increased by decreasing the volume, the system shifts so as to decrease the number of moles of gas (i.e., in the direction of fewer gas molecules). Conversely, when the pressure is decreased by increasing the volume, the system shifts so as to increase the number of moles of gas (i.e., in the direction of more gas molecules). It follows that if there are the same number of gas molecules on each side of the equation, such a system is not affected by changes in volume (and hence changes in pressure). Also, equilibrium systems are not affected by changes in pressure caused by adding an inert gas (a gas that cannot react with any of the equilibrium components). Consider the equilibrium N 2 (g) + 3H 2 (g) k 2NH 3 (g) n [NH 3 ] 2 NH n x V V NH 3 K c = = = [N 2 ] [H 2 ] 3 nn n 2 H 2 V V 3 n n 3 x N 2 H 2 (n = moles V = volume in L) If V is decreased (P is increased) the numerator will get smaller and hence Q c K c. The system will then shift to make more NH 3 by the reaction of N 2 with H 2 until Q c = K c (the system is shifting to the right (in the direction of fewer gas molecules). For equilibria with the same number of gas molecules on each side, V will not appear in the K c expression using moles and hence changing volume will not affect such equilibria. (3) Changes in Temperature Increase in temperature favours the endothermic reaction. Decrease in temperature favours the exothermic reaction. If the forward reaction is endothermic, K c increases as T increases. If the forward reaction is exothermic, K c decreases as T increases. Consider the equilibrium: (g) k 2 (g); H = kj If the temperature is increased, the system will counteract this by trying to get rid of the heat supplied. To do this, the system will shift to the right (i.e., the forward reaction will proceed since it is endothermic and will absorb heat).

5 3 (a) Consider the equilibrium 3Fe(s) + 4H 2 O(g) K Fe 3 O 4 (s) + 4H 2 (g) In a 3.0 L container is placed 5.0 g of H 2 O and 3.0 g of Fe. At equilibrium, there was only 2.0 g of Fe present. Calculate K c for the equilibrium. (Ans. 7.8 x 10-5 ) L Start: 3Fe(s) + 4H 2 O(g) K Fe 3 O 4 (s) + 4H 2 (g) 3.0 g 5.0 g mol Change: -1.0 g mol mol mol Equilibrium: 2.0 g mol mol = mol mol Moles/L mol/l mol/l K c = / = 7.8 x (b) Consider the equilibrium C(s) + CO 2 (g) K 2CO(g). When this system is at equilibrium in a 2.0 L container at 700 C, there are mol CO, 0.20 mol CO 2 and 4.8 g C. When cooled to 600 C, an additional 0.48 g of C forms. Calculate K c at 700 C and at 600 C. (Ans. 2.5 x 10-2 ; 8.3 x 10-4 )