2 temperatures below 1015 k 3 temperatures above 985

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2. temperatures below 1015 K. 3. temperatures above 985 K. 4. all temperatures. correct 5. temperatures below 985 K. Explanation: Δ G = Δ H - T Δ S is used to predict spon- taneity. G is negative for a spontaneous reaction.) T is always positive; for the reverse reaction, we reverse the sign of Δ H and Δ S . We thus have Δ G = ( - ) - T (+) for the re- verse reaction, so Δ G will be negative for any physically possible value of T . 005 10.0 points At a certain elevated temperature and pres- sure, diamond and graphite are in equilib- rium. When graphite changes to diamond under these conditions 1. the molar Gibbs free energy for diamond is zero and so is that of graphite. 2. the change in molar Gibbs free energy is a minimum. 3. the change in molar Gibbs free energy is zero. correct 4. the change in standard molar Gibbs free
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estrada (gse77) – Homework 2 – Sutcliffe – (52410) 3 energy is zero. 5. the change in molar Gibbs free energy is a maximum. Explanation: At equilibrium, Δ G for the process is zero. 006 10.0 points If the reaction quotient Q for some hypothet- ical reaction is 100, how does Δ G compare to Δ G 0 ? 1. Δ G > Δ G 0 correct 2. Δ G = Δ G 0 3. There is insufficient information given to determine the answer. 4. Δ G < Δ G 0 Explanation: Δ G = Δ G 0 + RT ln Q If Q = 100, then the natural log term will be positive. This means that the non standard value of Δ G will be bigger (more positive) than the value of Δ G 0 . 007 10.0 points Which of the following is TRUE? 1. A large value of K means that the equi- librium concentrations of products are large compared to the equilibrium concentrations of the reactants. correct 2. A small value of K means that the equilib- rium concentrations of the reactants are small compared to the equilibrium concentrations of the products. 3. When the value of K is small, the equi- librium lies on the product side of the equilib- rium reaction. 4. When the value of Q is large, the equilib- rium lies on the product side of the equilib- rium reaction. 5. When the value of K is large, the equilib- rium lies on the reactant side of the equilib- rium reaction. Explanation: 008 10.0 points Consider the reaction NO(g) + 1 2 O 2 (g) NO 2 (g) Find the equilibrium constant for the re- action at 298 K if Δ H = - 56 . 52 kJ and Δ S = - 72 . 60 J · K 1 at 298 K. 1. 8 . 08 × 10 9 2. 1 . 30 × 10 6 correct 3. 7 . 67 × 10 7 4. 3 . 23 × 10 4 5. 1 . 65 × 10 4 Explanation: Δ G = Δ H - T Δ S = - 56520 - (298)( - 72 . 60) = - 34885 . 2 J K = e Δ G/ ( RT ) = e 34885 . 2 / (8 . 314 · 298) = 1 . 3033 × 10 6 009 10.0 points Given the reaction 2 ICl(s) I 2 (s) + Cl 2 (g) and the thermodynamic data Species ΔH f S 0 kJ/mol J/mol · K ICl(s) 17 . 78 242 . 4 I 2 (s) 0 . 0 116 . 1 Cl 2 (g) 0 . 0 223 . 0
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estrada (gse77) – Homework 2 – Sutcliffe – (52410) 4 calculate K p at 100 C . 1. 7.562 2. 0.0023 correct 3. 0.023 4. 0.57 5. 0.75 Explanation: T = 100 C + 273 K = 373 K Δ H 0 = 0 + 0 + 0 - (2)(17 . 78) = - 35 . 56 kJ / mol Δ S 0 = 116 . 1 + 223 . 0 - (2)(242 . 4) = - 145 . 7 J / K Δ G 0 = Δ H 0 - T Δ S 0 = - 35 . 56 - (373)( - 0 . 1457) = 18 . 7861 kJ / mol Δ G 0 = - RT ln K K = e Δ G 0 / ( RT ) = exp bracketleftbigg - 18786 . 1 J / mol (8 . 314 J / mol · K) (373 K) bracketrightbigg = 0 . 0023 010 10.0 points For the decomposition of ammonia to nitro- gen and hydrogen, the equilibrium constant is 1 . 47 × 10 6 at 298 K. Calculate the temper- ature at which K = 1 . 00.
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