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chapter_2015 - CHAPTER 15 CHEMICAL EQUILIBRIUM 15.7 (a) Kc...

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CHAPTER 15 CHEMICAL EQUILIBRIUM 15.7 (a) 2 2 c 2 2 [CO] [O ] [CO ] = K 2 2 2 CO O 2 CO = P PP K P (b) 2 3 c 3 2 [O ] [O ] = K 3 2 2 O 3 O = P P K P (c) 2 c 2 [COCl ] [CO][Cl ] = K 2 2 COCl CO Cl = P P K (d) 2 c 2 [CO][H ] [H O] = K 2 2 CO H HO = P K P (e) c [H ][HCOO ] [HCOOH] +− = K (f) c2 [O ] = K 2 O = P K P 15.8 (a) 22 CO H O = P KP P (b) 2 2 2 O SO = P K 15.9 (a) 3 2 2 NH 3 c 27 2 7 NO H [NH ] [NO ] [H ] == P P KK (b) 2 2 2 2 SO 2 c 33 2 O [SO ] [O ] P P P (c) 2 2 2 CO c 2C O [CO] [CO ] P P P (d) 65 c [C H COO ][H ] [C H COOH] −+ = K 15.11 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.
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CHAPTER 15: CHEMICAL EQUILIBRIUM 315 (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 . 15.12 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. 15.13 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 15.14 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 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?
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CHAPTER 15: CHEMICAL EQUILIBRIUM 316 15.15 Using Equation (15.5) 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 15.16 Strategy: The relationship between K c and K P is given by Equation (15.5) 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?
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This note was uploaded on 05/22/2008 for the course CHEM 122 taught by Professor Bellew during the Spring '07 term at New Mexico.

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chapter_2015 - CHAPTER 15 CHEMICAL EQUILIBRIUM 15.7 (a) Kc...

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