1 94 c 2 151 c 3 77 c 4 31 c 5 50 c 6 135 c correct 7

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1. 94 C 2. 151 C 3. 77 C 4. 31 C 5. 50 C 6. 135 C correct 7. 65 C Explanation: K 2 = 3 K 1 Use the van’t Hoff equation. For the K ’s just put in 3 for K 2 and 1 for K 1 , and use the temperatures as appropriate ( T 1 is room tem- perature, T 2 is the unknown temperature). Remember to use Kelvin for temperature, not Celsius degrees. 014 3.3points The combustion of octane has a standard change in enthalpy Δ H = 5500 kJ · mol 1 . If the reaction were performed at T 1 = 200 K and T 2 = 300 K, which pair of equilibrium constants might be correct? 1. K 1 = 100 ,K 2 = 10 2. K 1 = 10 300 ,K 2 = 1 correct 3. K 1 = 10 ,K 2 = 100 4. K 1 = 1 ,K 2 = 10 300 Explanation: Since the reaction is exothermic, the value of K should be inversely proportional to T , so the higher temperature must have the smaller value of K . This eliminates two of the an- swer choices. Since Δ H = 5500 kJ · mol 1 is a very large number, it also means that dramatic changes in K are expected in re- sponse to changes in T , eliminating two an- swer choices, and leaving only one that satis- fies both criteria. 015 3.3points Consider the reaction CO(g) + 2 H 2 (g) CH 3 OH(g) . At room temperature, K is approximately 2 × 10 4 , but at a higher temperature K is substantially smaller. Which of the following is true? 1. The reaction is endothermic. 2. The reaction becomes spontaneous at higher temperatures. 3. At the higher temperature, more CH 3 OH(g) is produced. 4. The value of K c for this reaction is smaller at all temperatures. 5. The reaction is exothermic. correct Explanation: 016 3.3points A certain reaction has an equilibrium con- stant of 300 at 25 C and 276 at 35 C. What is Δ H 0 per mole for this reaction? Correct answer: 6 . 36393 kJ / mol rxn. Explanation: T 1 = 25 C+273 = 298 K K 1 = 300 T 2 = 35 C+273 = 308 K K 2 = 276 ln parenleftbigg K 2 K 1 parenrightbigg = Δ H 0 R parenleftbigg 1 T 1 1 T 2 parenrightbigg Δ H 0 = R ln parenleftbigg K 2 K 1 parenrightbigg parenleftbigg 1 T 1 1 T 2 parenrightbigg 1 = 8 . 314 J / mol · K
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casey (rmc2555) – Homework 6 – holcombe – (51395) 5 × ln parenleftbigg 276 300 parenrightbigg parenleftbigg 1 298 K 1 308 K parenrightbigg 1 = 6 . 36393 kJ / mol rxn 017 3.3points A particular small molecule drug works by binding to the active site in a given enzyme Drug ( aq ) + Enzyme ( aq ) BoundComplex ( aq ) If the equilibrium constant for this reaction is 10 10 , at what concentration of free drug is there 1000 times more bound enzyme (com- plex) than unbound enzyme? 1. 10 10 M 2. 10 3 M 3. 10 13 M 4. 10 7 M correct 5. 1 M Explanation: K = [complex]/[drug][enzyme] [drug] = [complex]/[enzyme] x 1/K = 1000/10 10 = 10 7 M 018 3.3points The reaction N 2 (g) + 3 H 2 (g) 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?
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