test2solution_pdf

# 027 32points the equilibrium constant k for the

This preview shows pages 8–10. Sign up to view the full content.

027 3.2points The equilibrium constant K for the dissocia- tion of N 2 O 4 (g) to NO 2 (g) is 1700 at 500 K. Predict its value at 300 K. For this reaction, Δ H is 56 . 8 kJ · mol 1 . 1. 1 . 11 × 10 4 2. 1 . 54 × 10 7 3. 1 . 32 × 10 6 4. 0.188 correct 5. 15.5 Explanation: 028 3.2points At 990 C, K c = 1 . 6 for the reaction H 2 (g) + CO 2 (g) H 2 O(g) + CO(g) How many moles of H 2 O(g) are present in an equilibrium mixture resulting from the addition of 1.00 mole of H 2 , 2.00 moles of CO 2 , 0.75 moles of H 2 O, and 1.00 mole of CO to a 5.00 liter container at 990 C? 1. 0.80 mol 2. 1.1 mol correct 3. 1.0 mol 4. 1.4 mol 5. 0.60 mol 6. 1.7 mol Explanation: K c = 1 . 6

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Version 324 – EXAM 2 – holcombe – (51395) 9 [H 2 ] = 1 . 00 mol 5 L [CO 2 ] = 2 . 00 mol 5 L [H 2 O] = 0 . 75 mol 5 L [CO] = 1 . 00 mol 5 L Q = [H 2 O][CO] [H 2 ][CO 2 ] = parenleftbigg 0 . 75 mol 5 L parenrightbigg parenleftbigg 1 mol 5 L parenrightbigg parenleftbigg 1 mol 5 L parenrightbigg parenleftbigg 2 mol 5 L parenrightbigg = 0 . 375 <K c = 1 . 6 Therefore equilibrium moves to the right. H 2 (g) + CO 2 (g) H 2 O(g) + CO(g) 0.2 0.4 0.15 0.2 - x - x x x 0 . 2 - x 0 . 4 - x 0 . 15 + x 0 . 2 + x (0 . 15 + x )(0 . 2 + x ) (0 . 2 - x )(0 . 4 - x ) = 1 . 6 x 2 + 0 . 35 x + 0 . 03 x 2 - 0 . 6 x + 0 . 08 = 1 . 6 x 2 + 0 . 35 x + 0 . 03 = 1 . 6 x 2 - 0 . 96 x + 0 . 128 0 . 6 x 2 - 1 . 31 x + 0 . 098 = 0 x = 1 . 31 ± radicalbig (1 . 31) 2 - 4(0 . 6)(0 . 098) 1 . 2 = 7 . 756 × 10 2 M mol H 2 O = 5 . 00 L × ( 0 . 15 + 7 . 75 × 10 2 ) mol L = 1 . 1 mol H 2 O 029 3.2points Given K p = 4 . 6 × 10 14 and Δ H 0 = 115 kJ/mol for the reaction 2 Cl 2 (g) + 2 H 2 O(g) 4 HCl(g) + O 2 (g) at 25 C, what is K p at 400 C? 1. 1 . 3 × 10 2 2. 1 . 4 × 10 5 3. 7 . 7 × 10 3 correct 4. 3 . 9 × 10 4 5. 7 . 9 × 10 2 Explanation: T 1 = 298.15 K T 2 = 673.15 K Use the van’t Hoff equation: ln parenleftbigg K 2 K 1 parenrightbigg = Δ H 0 rxn R parenleftbigg 1 T 1 - 1 T 2 parenrightbigg ln K 2 - ln K 1 = Δ H 0 rxn R parenleftbigg 1 T 1 - 1 T 2 parenrightbigg ln K 2 = Δ H 0 rxn R parenleftbigg 1 T 1 - 1 T 2 parenrightbigg + ln K 1 = parenleftbigg 115000 J / mol 8 . 314 J / mol · K parenrightbigg × parenleftbigg 1 298 . 15 K - 1 673 . 15 K parenrightbigg + ln ( 4 . 6 × 10 14 ) = - 4 . 86538 K 2 = e 4 . 86538 = 0 . 0077089 030 3.2points Which of the following equilibrium reactions is NOT affected by changes in pressure? 1. H 2 (g) + Br 2 ( ) 2 HBr(g) 2. H 2 (g) + I 2 (s) 2 HI(g) 3. 2 H 2 O 2 ( ) 2 H 2 O( ) + O 2 (g) 4. 2 BrCl(g) Br 2 (g) + Cl 2 (g) correct 5. 2 CO 2 (g) 2 CO(g) + O 2 (g) Explanation: 031 3.2points Consider the reaction A + B C + D
Version 324 – EXAM 2 – holcombe – (51395) 10 at equilibrium at 25 C with Δ S rxn = 12 cal/(K · mol) and Δ H rxn = - 22 kcal/mol. In- creasing the temperature by 10 C will have what effect on the equilibrium constant? 1. increase 2. decrease correct 3. no change Explanation: Increasing temperature adds heat to the exothermic reaction and favors the reverse reaction which absorbs heat and decreases K which reflects the ratio of product to reactant concentrations. 032 3.2points The equilibrium N 2 (g) + O 2 (g) ←→ 2 NO(g) exists in a closed container: If the partial pressures of N 2 , O 2 and NO at equilibrium are 1.5 atm, 2.5 atm and 0.75 atm, respectively, what is the equilbrium constant?
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern