Chang Chapter 14 solutions

Chang Chapter 14 solutions - CHAPTER 14 CHEMICAL...

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CHAPTER 14 CHEMICAL EQUILIBRIUM 14.13 c [B] [A] = K (1) With K c = 10, products are favored at equilibrium. Because the coefficients for both A and B are one, we expect the concentration of B to be 10 times that of A at equilibrium. Choice (a) is the best choice with 10 B molecules and 1 A molecule. (2) With K c = 0.10, reactants are favored at equilibrium. Because the coefficients for both A and B are one, we expect the concentration of A to be 10 times that of B at equilibrium. Choice (d) is the best choice with 10 A molecules and 1 B molecule. You can calculate K c in each case without knowing the volume of the container because the mole ratio between A and B is the same. Volume will cancel from the K c expression. Only moles of each component are needed to calculate K c . 14.14 Note that we are comparing similar reactions at equilibrium – two reactants producing one product, all with coefficients of one in the balanced equation. (a) The reaction, A + C U AC has the largest equilibrium constant. Of the three diagrams, there is the most product present at equilibrium. (b) The reaction, A + D U AD has the smallest equilibrium constant. Of the three diagrams, there is the least amount of product present at equilibrium. 14.15 When the equation for a reversible reaction is written in the opposite direction, the equilibrium constant becomes the reciprocal of the original equilibrium constant. 34 11 4.17 10 == = × × 33 '2 . 4 0 1 0 K K 14.16 The problem states that the system is at equilibrium, so we simply substitute the equilibrium concentrations into the equilibrium constant expression to calculate K c . Step 1: Calculate the concentrations of the components in units of mol/L. The molarities can be calculated by simply dividing the number of moles by the volume of the flask. 2 2.50 mol [H ] 0.208 12.0 L M 5 6 2 1.35 10 mol [S ] 1.13 10 12.0 L × × M 2 8.70 mol [H S] 0.725 12.0 L M
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CHAPTER 14: CHEMICAL EQUILIBRIUM 377 Step 2: Once the molarities are known, K c can be found by substituting the molarities into the equilibrium constant expression. 2 2 2 22 6 [H S] (0.725) [H ] [S ] (0.208) (1.13 10 ) == = × 7 c 1.08 10 K × If you forget to convert moles to moles/liter, will you get a different answer? Under what circumstances will the two answers be the same? 14.17 Using Equation (14.5) of the text: K P = K c (0.0821 T ) Δ n where, Δ n = 2 3 = 1 and T = (1273 + 273) K = 1546 K K P = (2.24 × 10 22 )(0.0821 × 1546) 1 = 1.76 × 10 20 14.18 Strategy: The relationship between K c and K P is given by Equation (14.5) of the text. What is the change in the number of moles of gases from reactant to product? Recall that Δ n = moles of gaseous products moles of gaseous reactants What unit of temperature should we use? Solution: The relationship between K c and K P is given by Equation (14.5) of the text. K P = K c (0.0821 T ) Δ n Rearrange the equation relating K P and K c , solving for K c . c (0.0821 ) Δ = P n K K T Because T = 623 K and Δ n = 3 2 = 1, we have: 5 1.8 10 (0.0821)(623 K) (0.0821 ) Δ × = 7 c 3.5 10 P n K T × K 14.19 We can write the equilibrium constant expression from the balanced equation and substitute in the pressures.
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Chang Chapter 14 solutions - CHAPTER 14 CHEMICAL...

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