Chapter 14 TAVSS

Chapter 14 TAVSS - CHAPTER 14 NON ACID/BASE CHEMICAL...

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Unformatted text preview: CHAPTER 14 NON ACID/BASE CHEMICAL EQUILIBRIUM Problems to prepare students for hourly exam II. • Non acid-base equilibrium concepts • Non acid-base equilibrium calculations • Le Chatelier’s Principle E. Tavss 4/06 1 NON ACID-BASE EQUILIBRIUM CONCEPTS CHEM162-2003 2 ND WEEK RECITATION CHAPTER 13 - NON ACID-BASE CHEMICAL EQUILIBRIA NON-ACID-BASE EQUILIBRIUM CONCEPTS 13.21 At a given temperature, K = 1.3 x 10-2 for the reaction N 2 (g) + 3H 2 (g) 2NH 3 (g) Calculate values of K for the following reactions at this temperature. a. 1/2N 2 (g) + 3/2H 2 (g) NH 3 (g) Half of the equation = square root of K = √(1.3 x 10-2 ) = 0.114 b. 2NH 3 (g) N 2 (g) + 3H 2 (g) Reverse the equation = inverse of K = 1/(1.3 x 10-2 ) = 76.9 c. NH 3 (g) ½ N 2 (g) + 3/2H 2 (g) Reverse and ½ the equation = the inverse square root of K = √(1/(1.3 x 10-2 )) = 8.77 d. 2N 2 (g) + 6H 2 (g) 4NH 3 (g) Double the equation = square the K = (1.3 x 10-2 ) 2 = 1.69 x 10-4 CHEM162-2003 2 ND WEEK RECITATION CHAPTER 13 - NON ACID-BASE CHEMICAL EQUILIBRIA NON-ACID-BASE EQUILIBRIUM CONCEPTS 31 Write expressions for K for the following reactions. a. P 4 (s) + 5O 2 (g) P 4 O 10 (s) 2 K = 1/[O 2 ] 5 ) = [O 2 ]-5 b. NH 4 NO 3 (s) N 2 O(g) + 2H 2 O(g) K = ([N 2 O][H 2 O] 2 ) c. CO 2 (g) + NaOH(s) NaHCO 3 (s) K = 1/([CO 2 ] d. S 8 (s) + 8O 2 (g) 8SO 2 (g) K = [SO 2 ] 8 /[O 2 ] 8 CHEM162-2003 3 RD WEEK RECITATION CHAPTER 13 - NON ACID-BASE CHEMICAL EQUILIBRIA NON-ACID-BASE EQUILIBRIUM CONCEPTS 39 At 900 o C, Kp = 1.04 for the reaction CaCO 3 (s) CaO(s) + CO 2 (g) At a low temperature, dry ice (solid CO 2 ), calcium oxide, and calcium carbonate are introduced into a 50.0-L reaction chamber. The temperature is raised to 900 o C, resulting in the dry ice converting to gaseous CO 2 . For the following mixtures, will the initial amount of calcium oxide increase, decrease, or remain the same as the system moves toward equilibrium at 900 o C? a. 655 g CaCO 3 , 95.0 g CaO, P CO2 = 2.55 atm Kp = [CO 2 ] Qp = 2.55 CaO will decrease. d. 715 g CaCO 3 , 813 g CaO, P CO2 = 0.211 atm Kp = [CO 2 ] 3 Qp = 0.211 CaO will increase. 5. CHEM162-2000 HOURLY EXAM II CHAPTER 13 - NON ACID-BASE CHEMICAL EQUILIBRIA NON-ACID-BASE EQUILIBRIUM CONCEPTS For a certain chemical reaction: K p = 6.0 x 10-5 at 298 K K p = 1.0 x 10-5 at 400 K These data imply that the reaction A . is exothermic. B. is endothermic. C. has fewer moles of gas on the right side. D. has fewer moles of gas on the left side. E. has molecules which move slower at higher temperatures. Let’s say that the reaction is A + B ← → C + D The difference between 298 K and 400 K is the heat applied. Heat is added to the reaction to bring it up from 298 to 400 K. Let’s use the delta sign to represent heat, and arbitrarily put it on the left side of the equilibrium equation. Putting the heat sign on the left side means that this is an endothermic reaction, i.e., the reaction absorbs heat....
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This note was uploaded on 09/23/2011 for the course CHEMISTRY 101 taught by Professor Rabeony during the Fall '08 term at Rutgers.

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Chapter 14 TAVSS - CHAPTER 14 NON ACID/BASE CHEMICAL...

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