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casey (rmc2555) – Homework 4 – holcombe – (51395) 1 This print-out should have 23 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 4.5points Consider the reaction CaCO 3 (s) CaO(s) + CO 2 (g) , at equilibrium at a certain temperature. Now a small amount of Ca 14 CO 3 (s) ( 14 C la- beled calcium carbonate) is added. What species will be present after equilibrium is re-established? 1. CaO(s), 14 CO 2 (g), CO 2 (g) 2. CaCO 3 (s), CaO(s), 14 CO 2 (g), CO 2 (g) 3. CaCO 3 (s), Ca 14 CO 3 (s), CaO(s), CO 2 (g) 4. Ca 14 CO 3 (s), CaO(s), 14 CO 2 (g), CO 2 (g) 5. CaCO 3 (s), Ca 14 CO 3 (s), CaO(s), 14 CO 2 (g), CO 2 (g) correct Explanation: The addition of extra reactant Ca 14 CO 3 will favor the production of more products (CaO(s) and 14 CO 2 (g)). However, since this is a dynamic equilibrium, both forward and backward reactions occur. Therefore, an equilibrium will be established where 14 C is present in both products and reactants con- taining C. 002 4.5points What is the main overall driving force for any spontaneous reaction or change? Consider only the reaction system, not the surround- ings. 1. To maximize electrostatic interactions. 2. To obey the laws of gravity. 3. To release heat energy. 4. To maximize entropy. 5. To lower the available free energy. cor- rect Explanation: 003 4.5points Pure liquids or solids do NOT appear in the equilibrium constant expression. 1. False 2. True correct Explanation: 004 4.5points Consider the chemical equation below: N 2 O 4 (g) + 4H 2 (g) ←→ 4H 2 O(g) + N 2 (g) What would K eq be for this reaction? 1. K eq = [H 2 O] 4 [N 2 O 4 ][H 2 ] 4 2. K eq = [N 2 O 4 ][H 2 ] 4 [H 2 O] 4 3. K eq = [H 2 O] 4 [N 2 ] [N 2 O 4 ][H 2 ] 4 correct 4. K eq = [N 2 O 4 ][H 2 ] 4 [H 2 O] 4 [N 2 ] Explanation: For a reaction of the form: aA(s) + bB(aq) cC( ) + dD(g) The equilibrium constant will take the form: K eq = [D] d [B] b Only species that have a changeable con- centration, gases and aqueous species, have activities that change. Solids and liquids have activities which are constants, so they don’t appear in mass action expressions. 005 4.5points Today, ammonia is synthesized through a se- ries of reactions (called the Haber-Bosch pro- cess) that take place between methane, air (which is four parts N 2 , one part O 2 ), and potassium carbonate:
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casey (rmc2555) – Homework 4 – holcombe – (51395) 2 7 CH 4 (g) + 8 N 2 (g) +2 O 2 (g) + 17 H 2 O(g) + 7 K 2 CO 3 (s) 16 NH 3 (g) + 14 KHCO 3 (s) What is the equilibrium expression for the Harber-Bosch process?
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