MSJ2e_Ch14_ISM2_June26_p640

MSJ2e_Ch14_ISM2_June26_p640 - Chapter 14 Chemical...

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Chapter 14: Chemical Equilibrium 52. Use a method similar to that described in the answer to Question 45(b). (conc. N 2 ) = (conc. H 2 ) = 1.00 mol N 2 O 4 /5.00 L = 0.200 M N 2 (g) + 3 H 2 (g) 2 NH 3 (g) conc. initial (M) 0.200 0.200 0 change conc. (M) – x – 3x + 2x equilibrium conc. (M) 0.200 – x 0.200 – 3x 2x At equilibrium Κ χ = ΝΗ 3 [ ] 2 Ν 2 [ ] Η 2 [ ] 3 = (2ξ29 2 (0.200- ξ29( 0.200- 3ξ29 3 = 5.97 × 10 –2 The reactant-favored reaction will have larger reactant concentrations than product concentrations, so let’s assume x is small, such that subtraction from the product concentrations is negligible: 0.200 – x 2245 0.200, and 0.200– 3x 2245 0.200. (2ξ29 2 (0.200 29( 0.200 29 3 = 5.97 × 10 –2 x = 5.97 10 -2 0.200 ( 29 0.200 ( 29 3 4 = 0.00489 This value of x is not very small, so let’s iteratively plug it back into the equation in the places where we ignored it, to obtain a more accurate value. (This method is described in Appendix A.7.) Repeat the procedure until the value of x stops changing: x = 5.97 10 -2 0.200- 0.00489 ( 29 0.200- 3(0.00489 29 ( 29 3 4 = 0.00431 x = 5.97 10 -2 0.200- 0.00431 ( 29 0.200 - 3(0.00431 29 ( 29 3 4 = 0.00437 x = 5.97 10 -2 0.200- 0.00437 ( 29 0.200 - 3(0.00437 29 ( 29 3 4 = 0.00437 Percentage N 2 converted = αμουντοφ Ν 2 χονωερτεδ ινιτιαλαμουντοφ Ν 2 100 % = 0.00437 Μ 0.200 100 % = 2.19 % Check your answers : The equilibrium concentrations should combine to reproduce K c : [N 2 ] = 0.200 – x = 0.196 M, [H 2 ] = 0.100 – 3x = 0.187 M, [NH 3 ] = 2x = 0.00874 M K c = (0.00874 29 2 (0.196 29( 0.187 29 3 = 5.96 × 10 –2 . 104
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Chapter 14: Chemical Equilibrium Shifting a Chemical Equilibrium: Le Chatlier’s Principle 53. As described in Section 14.6, the equilibrium is not affected by the addition of any solid, so adding more BaSO 4 (s) to an equilibrium mixture will not affect the equilibrium state; therefore the answer is (c). 54 . As described in Section 14.6, the positive H indicates that heat is a reactant in this reaction. Increasing the temperature will drive the reaction toward products and away from N 2 O 4 . That means the N 2 O 4 concentration will decrease, and the answer is (b). 55. Use the methods described in Section 14.6 to answer this question. Changing concentrations or pressures of reactants and products or changing the available energy in the system will take the system out of equilibrium. The response to that change will be a shift away from the increase. Changing the amounts of solids or liquids will not affect the equilibrium position. Only temperature can affect the value of the equilibrium constant. Because this reaction is exothermic, energy is a product. H 2 (g) + Br 2 (g) 2 HBr(g) + 103.7 kJ Change [Br 2 ] [HBr] K c K p Some H 2 (a reactant) is added to the container decrease increase no change no change The temperature of the gases in the container is increased. (increase in energy, a product ) increase decrease decrease decrease The pressure of HBr ( a product ) is increased. increase increase (since extra is added) no change no change 56 . Use the methods described in Section 14.6 to answer this question. Changing concentrations or pressures of reactants and products or changing the available energy in the system will take the system out of equilibrium. The response to that change will be to shift away from an increase (or shift towards a decrease). Changing the amounts of solids or liquids will not affect the equilibrium position.
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