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# solution_pdf - parvaresh(sep782 – Homework 1 –...

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parvaresh (sep782) – Homework 1 – Sutcliffe – (52440) 1 This print-out should have 18 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 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 . 002 10.0 points If Δ G = 27 . 1 kJ at 25 C for the reaction CH 3 COOH(aq) + H 2 O( ) CH 3 COO (aq) + H 3 O + (aq) , calculate K a for this reaction at 298 K. 1. 1 . 15 × 10 11 2. 1.01 3. 5 . 63 × 10 4 4. 9 . 89 × 10 1 5. 1 . 78 × 10 5 correct Explanation: 003 10.0 points Calculate the equilibrium constant at 305 K for the reaction HgO(s) Hg( ) + 1 2 O 2 (g) , given the data S Δ H f ( J K · mol ) ( kJ mol ) HgO(s) 70 . 29 - 90 . 83 Hg( ) 76 . 02 0 1 2 O 2 (g) 205 . 14 0 Correct answer: 1 . 26185 × 10 10 . Explanation: HgO(s) Hg( ) + 1 2 O 2 (g) , Δ H r = - H f , HgO(s) ] = - ( - 90 . 83) = 90 . 83 kJ / mol Δ S r = S Hg( ) + 1 2 S O 2 (g) - S HgO(s) = 76 . 02 J K · mol + 1 2 parenleftbigg 205 . 14 J K · mol parenrightbigg - 70 . 29 J K · mol = 108 . 3 J K · mol At 305 K, Δ G r(305 K) = 90 . 83 K - (305 K) × parenleftbigg 108 . 3 J K · mol parenrightbigg parenleftbigg 1 kJ 1000 J parenrightbigg = 57 . 7985 kJ / mol Δ G r(305 K) = - RT ln K ln K = - Δ G r(305 K) RT = - 57 . 7985 kJ (8 . 314 J / K)(305 K) × 1000 J 1 kJ = - 22 . 7933 K = e 22 . 7933 = 1 . 26185 × 10 10 . 004 10.0 points The standard state for phosphorus is white phosphorus. (There’s a red form too, but thats not the standard).

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parvaresh (sep782) – Homework 1 – Sutcliffe – (52440) 2 What is the numerical value of the equi- librium constant K p at 25 C for the de- composition of phosgine (PH 3 ) gas into the elements white phosphorus and hydrogen? Δ G 0 f = 13 . 4 kJ / mol for PH 3 (g). 1. None of these 2. 1 . 00 × 10 1 3. 9 . 95 × 10 1 4. 2 . 23 × 10 2 correct 5. 4 . 48 × 10 3 Explanation: Δ G = - 13 . 4 kJ / mol = - 13400 J / mol T = 25 C = 298 K R = 8 . 314 J mol · K K p = e Δ G/ ( RT ) = exp bracketleftBigg 13400 J / mol (8 . 314 J mol · K ) (298 K) bracketrightBigg = 223 . 301 005 10.0 points A group who investigated the formation of ammonia from its elements determined that K c is 90 at 773 K and 3 at 873 K. Is this reaction endothermic or exothermic?
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solution_pdf - parvaresh(sep782 – Homework 1 –...

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