Chem 162-2011 review session exam II

Chem 162-2011 review session exam II - CHEM 162-2011 EXAM...

Info iconThis preview shows pages 1–5. Sign up to view the full content.

View Full Document Right Arrow Icon
CHEM 162-2011 EXAM II REVIEW SESSION from 1
Background image of page 1

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

View Full DocumentRight Arrow Icon
14-1CHEM 162-2010 EXAM II Chapter 14 - Non Acid/Base Chemical Equilibrium Non acid-base equilibrium concepts ET: Emphasize direction of Q 25. SO 2 Cl 2 (g) SO 2 (g) + Cl 2 (g) K = 0.070 Flask X: 0.50 mol SO 2 Cl 2 , 0.10 mol SO 2 , 0.20 mol Cl 2 in 0.50L Flask Y: 0.50 mol SO 2 Cl 2 , 0.10 mol SO 2 , 0.20 mol Cl 2 in 1.00L Flask Z: 0.50 mol SO 2 Cl 2 , 0.10 mol SO 2 , 0.20 mol Cl 2 in 2.00L Which flask will result in a reaction forming more Cl 2 ? A. Flask X only B. Flask Y only C. Flask Z only D. Flasks X and Y only E . Flasks Y and Z only Flask X: SO 2 Cl 2 (g) SO 2 (g) + Cl 2 (g) Initial 2 x 0.50 = 1.00M 2 x 0.10 = 0.20M 2 x 0.20 = 0.40M Change Equilibrium Q = ([SO 2 ][Cl 2 ])/[SO 2 Cl 2 ] Q = ([0.20][0.40])/[1.00] = 0.080. Q > K. Reaction will go to the left; [Cl 2 ] will be decreased. Flask Y: SO 2 Cl 2 (g) SO 2 (g) + Cl 2 (g) Initial 0.50 0.10 0.20 Change Equilibrium Q = ([SO 2 ][Cl 2 ])/[SO 2 Cl 2 ] Q = ([0.10][0.20])/[0.50] = 0.040. Q < K. Reaction will go to the right; [Cl 2 ] will be increased. Flask Z: SO 2 Cl 2 (g) SO 2 (g) + Cl 2 (g) Initial 0.50/2 = 0.25 0.10/2 = 0.05 0.20/2 = 0.10 Change Equilibrium Q = ([SO 2 ][Cl 2 ])/[SO 2 Cl 2 ] Q = ([0.05][0.10])/[0.25] = 0.020. Q < K. Reaction will go to the right; [Cl 2 ] will be increased. 2
Background image of page 2
14-1CHEM 162-2010 EXAM II Chapter 14 - Non Acid/Base Chemical Equilibrium Non acid-base equilibrium concepts 16. Given: C(s) + CO 2 (g) 2CO K p = 1.3 x 10 14 COCl 2 (g) CO(g) + Cl 2 (g) K p = 167 Calculate K p for: C(s) + CO 2 (g) + 2Cl 2 (g) 2COCl 2 A. 2.2 x 10 16 B. 7.8 x 10 11 C. 1.3 x 10 12 D. 2.6 x 10 11 E . 4.7 x 10 9 Use Hess’s Law. For the 1 st equation, leave as is. C(s) + CO 2 (g) 2CO K p = 1.3 x 10 14 For the 2 nd equation, reverse it and double it; correspondingly, invert and square the K p . 2CO(g) + 2Cl 2 (g) 2COCl 2 (g) K p = (1/167) 2 Add the equations and multiply the equilibrium constants. C(s) + CO 2 (g) + 2Cl 2 (g) 2COCl 2 K = 4.66 x 10 9 3
Background image of page 3

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

View Full DocumentRight Arrow Icon
Chapter 14 - Non Acid/Base Chemical Equilibrium Non acid-base equilibrium concepts 6. Which of the following is true about a chemical reaction at equilibrium? ET: Skip X. The rate of the forward reaction equals the rate of the reverse reaction. Y. There is no observable change in any property of the system. Z. The molecules have less kinetic energy than before equilibrium was reached. A . X and Y only B. X only C. Y only D. X and Z only E. X, Y, and Z At equilibrium, the forward and reverse reactions proceed at equal rates. At equilibrium, the concentrations of reactants and products remain constant. The kinetic energy is a function of temperature, and not dependent on
Background image of page 4
Image of page 5
This is the end of the preview. Sign up to access the rest of the document.

This document was uploaded on 11/02/2011 for the course GEN CHEM 162 at Rutgers.

Page1 / 33

Chem 162-2011 review session exam II - CHEM 162-2011 EXAM...

This preview shows document pages 1 - 5. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online