H03 Solutions

# H03 Solutions - bleiweiss(emb853 H03 Equilibrium...

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bleiweiss (emb853) – H03: Equilibrium – Mccord – (90540) 1 This print-out should have 21 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 10.0 points An equilibrium in which processes occur con- tinuously, with NO NET change, is called 1. homogeneous equilibrium. 2. static equilibrium. 3. heterogeneous equilibrium. 4. dynamic equilibrium. correct Explanation: ±or a system at dynamic equilibrium, al- though the concentrations oF the components do not change, the processes continue to oc- cur in the Foward and reverse directions at the same rate. 002 10.0 points The expression For K c For the reaction at equi- librium is 4 NH 3 (g) + 5 O 2 (g) ± ² 4 NO(g) + 6 H 2 O(g) 1. [NO] 4 [H 2 O] 6 [NH 3 ] 4 [O 2 ] 5 correct 2. [NH 3 ] 4 [O 2 ] 5 3. [NO] 4 [H 2 O] 6 4. [NH 3 ] 4 [O 2 ] 5 [NO] 4 [H 2 O] 6 Explanation: The equation must be written with the ap- propriate Formula and correctly balanced. K c is the equilibrium constant For species in so- lution and equals the mathematical product oF the concentrations oF the chemical prod- ucts, divided by the mathematical product oF the concentrations oF the chemical reactants. In this mathematical expression, each concen- tration is raised to the power oF its coe²cient in the balanced equation. ±or K c the molar concentrations are used For the activities oF the components. 003 10.0 points At 600 C, the equilibrium constant For the reaction 2 HgO(s) 2 Hg( ³ ) + O 2 (g) is 2.8. Calculate the equilibrium constant For the reaction 1 2 O 2 (g) + Hg( ³ ) HgO(s) . 1. 0.36 2. 0.60 correct 3. 1.7 4. - 1 . 7 5. 1.1 Explanation: 004 10.0 points Calculate the equilibrium constant at 25 C For a reaction For which Δ G 0 = - 5 . 64 kcal / mol. 1. - 13697 . 9 2. 13697 . 9 correct 3. 6848 . 96 4. 27395 . 9 5. 1 . 36979 × 10 5 Explanation: T = 25 C + 273 = 298 K Δ G 0 = - 5640 cal / mol At equilibrium Δ G 0 = - RT ln K - 5640 = ( - 1 . 987 cal / mol · K) × (298 K)(ln K ) K = 13697 . 9 005 10.0 points

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bleiweiss (emb853) – H03: Equilibrium – Mccord – (90540) 2 Given the reversible reaction equation 2 CO(g) + O 2 (g) ± ² 2 CO 2 (g) which is the relationship between K c and K p ? 1. K p = K c ( RT ) - 1 correct 2. K p = K c 3. K p = K c 4. K p = K c ( ) 2 5. K p = K c ( ) - 2 Explanation: The ideal gas law can be used to derive the link between K c and K p : P = n V = M RT M = P K c = [CO 2 ] 2 [CO] 2 [O 2 ] = P 2 CO 2 ( ) 2 P 2 CO ( ) 2 · P O 2 = P 2 CO 2 P 2 CO · P O 2 = K p K p = K c ( ) - 1 006 10.0 points The reaction N 2 (g) + 3 H 2 (g) ± ² 2 NH 3 (g) , has an equilibrium constant of 4 . 0 × 10 8 at 25 C. What will eventually happen if 44.0 moles of NH 3 , 0.452 moles of N 2 , and 0.108 moles of H 2 are put in a 10.0 liter container at 25 C? 1. Nothing; the system is at equilibrium. 2. More N 2 and H 2 will be formed. 3. More NH 3 will be formed. correct Explanation: K =4 . 0 × 10 8 [NH 3 ]= 44 . 0 mol 10 L [N 2 0 . 452 mol 10 L [H 2 0 . 108 mol 10 L Q = [NH 3 ] 2 [N 2 ] [H 2 ] 3 = (4 . 40 M) 2 (0 . 0452 M) (0 . 0108 M) 3 =3 . 4 × 10 8 Since Q < K equilibrium will shift to the right, forming more NH 3 .
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## This note was uploaded on 09/01/2010 for the course CH 302 taught by Professor Holcombe during the Summer '07 term at University of Texas.

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H03 Solutions - bleiweiss(emb853 H03 Equilibrium...

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