bracketrightbigg 6 88395 006 10 points A 0 835 mole sample of HI is placed in a

Bracketrightbigg 6 88395 006 10 points a 0 835 mole

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bracketrightbigg = 6 . 88395 006 1.0 points A 0 . 835 mole sample of HI is placed in a 5.000 liter container and held at a constant temperature until the gas-phase reaction 2 HI H 2 + I 2 comes to equilibrium. At equilibrium, it is found that the concentration of I 2 is 2 . 1 × 10 2 mol/L. What is the numerical value of the equilibrium constant K c at this temperature? Correct answer: 0 . 028224. Explanation: [HI] ini = 0 . 835 mol 5 L = 0 . 167 M 2 HI (g) H 2 (g) + I 2 (g) ini, M 0 . 167 0 0 Δ, M - 2 x x x eq, M 0 . 167 - 2 x x x [I 2 ] eq = x = 2 . 1 × 10 2 M [H 2 ] eq = x = 2 . 1 × 10 2 M [HI] eq = (0 . 167 - 2 x ) = 0 . 125 M
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To (st22362) – Homework 3 – Fakhreddine – (52420) 3 K c = [H 2 ] [I 2 ] [HI] 2 = (2 . 1 × 10 2 M) 2 (0 . 125 M) 2 = 0 . 028224 007 1.0 points Consider the equilibrium A(g) 2 B(g) + 3 C(g) at 25 C. When A is loaded into a cylinder at 9 . 63 atm and the system is allowed to come to equilibrium, the final pressure is found to be 16 . 17 atm. What is Δ G r for this reaction? Correct answer: - 12 . 5403 kJ / mol. Explanation: Considering the pressures (in atm), A 2 B + 3 C ini 9 . 63 0 0 Δ - x +2 x +3 x fin 9 . 63 - x +2 x +3 x The total pressure is P total = 9 . 63 - x + 2 x + 3 x 16 . 17 atm = 9 . 63 + 4 x x = 1 . 635 atm , so P A = 9 . 63 - x = 7 . 995 atm P B = 2 x = 3 . 27 atm P C = 3 x = 4 . 905 atm The equilibrium expression is K = P 2 B · P 3 C P A = (3 . 27) 2 (4 . 905) 3 7 . 995 = 157 . 832 and the energy is Δ G = - RT ln K = - parenleftbigg 8 . 314 J K · mol parenrightbigg (298 K) × ln (157 . 832) = - 12540 . 3 J / mol = - 12 . 5403 kJ / mol . 008 1.0 points At 25 C, K c = 1 . 58 × 10 8 for the reaction NH 4 (NH 2 CO 2 )(s) 2 NH 3 (g) + CO 2 (g) . Calculate K at 25 C for this reaction. 1. 3 . 87 × 10 7 2. 5 . 69 × 10 3 3. 9 . 45 × 10 5 4. 2 . 31 × 10 4 correct 5. 1 . 36 × 10 7 Explanation: 009 1.0 points The equilibrium constant K p for the reaction I 2 (g) + Br 2 (g) 2 IBr(g) + 11 . 7 kJ is 280 at 150 C. Suppose that a quantity of IBr is placed in a closed reaction vessel and the system is allowed to come to equilibrium at 150 C. When equilibrium is established, the pressure of IBr is 0.200 atm. What is the pressure of I 2 at equilibrium? 1. 0.168 atm 2. 0.067 atm 3. 0.012 atm correct 4. None of these 5. 0.096 atm Explanation: At equilibrium, P IBr = 0 . 200 atm K p = 280 I 2 (g) + Br 2 (g) 2 IBr(g) + 11.7 kJ 0 0 y x x - 2 x x x y - 2 x
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To (st22362) – Homework 3 – Fakhreddine – (52420) 4 y - 2 x = 0 . 2 K p = P 2 IBr P I 2 · P Br 2 = 280 0 . 2 2 x 2 = 280 x = 0 . 2 280 = 0 . 0119523 P I 2 = 0 . 0119523 atm 010 1.0 points A 1.000 L vessel is filled with 2.000 moles of N 2 , 1.000 mole of H 2 , and 2.000 moles of NH 3 . When the reaction N 2 (g) + 3 H 2 (g) 2 NH 3 (g) comes to equilibrium, it is observed that the concentration of H 2 is 2 . 25 moles/L. What is the numerical value of the equilibrium con- stant K c ? Correct answer: 0 . 0494458. Explanation: [N 2 ] ini = 2 mol 1 L = 2 M [NH 3 ] ini = 2 mol 1 L = 2 M [H 2 ] ini = 1 mol 1 L = 1 M Therefore equilibrium moves to the left. N 2 (g) + 3 H 2 (g) 2 NH 3 (g) ini, M 2 1 2 Δ, M + x +3 x - 2 x eq, M 2+ x 1+3 x 2 - 2 x [H 2 ] eq = 2 . 25 M , so [H 2 ] eq = 1 . 0 + 3 x = 2 . 25 M x = 2 . 25 M - 1 M 3 = 0 . 416667 M [N 2 ] eq = (2 + x ) M = 2 . 41667 M [NH 3 ] eq = (2 - 2 x ) M = 1 . 16667 M K c = [NH 3 ] 2 [N 2 ] [H 2 ] 3 = (1 . 16667) 2 M (2 . 41667 M) (2 . 25 M) 3 = 0 . 0494458 011 1.0 points The vapor pressure of benzene at 298 K is 94.4 mm of Hg. The standard molar Gibbs free energy of formation of liquid benzene at 298 K is 124.5 kJ/mol. What is the stan-
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