1 1 k 2 therefore the equilibrium moves to the right

This preview shows page 7 - 10 out of 10 pages.

1 . 0 = 1 . 0 <K = 2 Therefore, the equilibrium moves to the right. A B(g) ini, M 1 . 0 1 . 0 Δ, M x x eq, M 1 . 0 x 1 . 0 + x K = [B] [A] = 2 1 . 0 + x 1 . 0 x = 2 1 . 0 + x = 2 . 0 2 x x = 0 . 33 [A] = 1 . 0 x = 0 . 67 M [B] = 1 . 0 + x = 1 . 33 M 024 3.3points If 5 L of HCl(g) at 1 bar and 273 K and 26 g of I 2 (s) are transferred to a 8 L reaction vessel and heated at 25 C, what will be the equilibrium concentration of HCl? K c = 1 . 6 × 10 34 at 25 C for 2 HCl(g) + I 2 (s) 2 HI(g) + Cl 2 (g) . Correct answer: 0 . 0275364 mol / L. Explanation: We use the ideal gas relationship to find the initial concentration of HCl(g) at 25 C: n = PV RT = (1 bar) (5 L) ( 0 . 08314 L · bar K · mol ) (273 K) = 0 . 220291 mol [HCl] = 0 . 220291 mol 8 L = 0 . 0275364 mol / L Analyzing the reaction using concentra- tions, 2 HCl(g) + I 2 (s) 2 HI(g) + Cl 2 (g) 0 . 0275364 0 0 2 x +2 x + x 0 . 0275364 2 x 2 x x K c = [HI] 2 [Cl 2 ] [HCl] 2 1 . 6 × 10 34 = (2 x ) 2 ( x ) (0 . 0275364 2 x ) 2 Since the equilibrium constant is very small, we can assume that x 0 . 0275364, so 1 . 6 × 10 34 = 4 x 3 (0 . 0275364) 2 x 3 = (1 . 6 × 10 34 ) (0 . 0275364) 2 4 x = 3 radicalbigg (1 . 6 × 10 34 ) (0 . 0275364) 2 4 = 3 . 11859 × 10 13 At equilibrium, [HCl] = 0 . 0275364 2 x = 0 . 0275364 2 (3 . 11859 × 10 13 ) = 0 . 0275364 mol / L . 025 3.3points 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. 1.7 mol
Image of page 7

Subscribe to view the full document.

casey (rmc2555) – Homework 6 – holcombe – (51395) 8 2. 0.80 mol 3. 0.60 mol 4. 1.0 mol 5. 1.1 mol correct 6. 1.4 mol Explanation: K c = 1 . 6 [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 026 3.3points 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. 0.096 atm 2. None of these 3. 0.067 atm 4. 0.012 atm correct 5. 0.168 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 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 027(part1of2)3.3points Calculate the reaction free energy of H 2 (g) + I 2 (g) 2 HI(g) at 700 K when the concentrations of H 2 , I 2 , and HI are 0 . 16 mol / L, 0 . 44 mol / L, and
Image of page 8
casey (rmc2555) – Homework 6 – holcombe – (51395) 9 2 . 14 mol / L, respectively. For this reaction, K c = 54 at 700 K.
Image of page 9

Subscribe to view the full document.

Image of page 10
You've reached the end of this preview.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    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.

    Student Picture

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

  • Left Quote Icon

    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.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    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.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern