Problem Set #2 Sol - Chem 102 Klobukowski Winter 2009...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
Chem 102 Klobukowski Winter 2009 Problem Set 2 page 1 Problem Set 2 (Unit 2) Solutions to Additional Problems 1. At a certain temperature, the reaction, NO (g) + O 3 (g) NO 2 (g) + O 2 (g), has an equilibrium constant K = 1.32 × 10 10 . Calculate the equilibrium position (partial pressures of all species) when 0.28 atm of nitric oxide (NO) are mixed with 0.58 atm of ozone (O 3 ). K is large therefore let the reaction go to completion and work with the reverse reaction: Set up a reaction table and let the reaction go to completion: NO (g) + O 3 (g) NO 2 (g) + O 2 (g) i 0.28 0.58 0 0 c -0.28 -0.28 +x +x e 0 0.30 0.28 0.28 Set up i.c.e. table for the reverse reaction (K = 1/1.32 × 10 10 = 7.58 × 10 -11 ) NO 2 (g) + O 2 (g) NO (g) + O 3 (g) i 0.28 0.28 0 0.30 c -x -x +x +x e 0.28-x 0.28-x x 0.30+x (you don’t have to write the reaction in this reverse way, you can keep the reaction written as above, only you’ll have –x on the right and +x on the left, but the result of the values you put into K is the same. ) Substitute these values into the expression for K: K = P NO P O 3 P NO 2 P O 2 = x (0.30 + x ) (0.28 ! x )(0.28 ! x ) = 7.58 " 10 ! 11 Since K is small, make the approximations (0.28 ! x ) " 0.28 and (0.30 + x ) ! 0.30 Therefore, x (0.30) (0.28)(0.28) = 7.58 ! 10 " 11 x = 2.0 × 10 -11 atm The equilibrium pressures of all species are P NO 2 = 0.28 atm , P O 2 = 0.28 atm , P O 3 = 0.30 atm , and P NO = 2.0 ! 10 " 11 atm . 2. 1.00 atm each of H 2 (g) and P 2 (g) were placed in a reaction vessel at 800 K. When the reaction 3 H 2 (g) + P 2 (g) 2 PH 3 (g) reached equilibrium, the total pressure in the vessel was 1.91 atm. Calculate the equilibrium constant. Set up an i.c.e. table 3H 2 (g) + P 2 (g) 2PH 3 (g) i 1.00 1.00 0 c -3x -x +2x e 1.00-3x 1.00-x 2x We know that at equilibrium P tot = P H 2 + P P 2 + P PH 3 = 1.91 atm
Background image of page 1

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

View Full DocumentRight Arrow Icon
Chem 102 Klobukowski Winter 2009 Problem Set 2 page 2 From the i.c.e. table, we know the equilibrium pressures in terms of x, i.e., (1.00 ! 3 x ) + (1.00 ! x ) + 2 x = 1.91 2.00 ! 2 x = 1.91 2 x = 0.09 x = 0.045 atm Therefore, at equilibrium, P H 2 = 1.00 ! 3(0.045) = 0.865 atm , P P 2 = 1.00 ! 0.045 = 0.955 atm , and P PH 3 = 2(0.045) = 0.090 atm . Solving for the equilibrium constant K, we obtain K = ( P PH 3 ) 2 ( P H 2 ) 3 P P 2 = (0.090) 2 (0.865) 3 (0.955) = 1.31 ! 10 " 2 3. The following reaction at 200 o C has an equilibrium constant value of 4.65 x 10 -12 : CO (g) + 3 H 2 (g) CH 4 (g) + H 2 O (g) What are the equilibrium partial pressures of all components when the initial partial pressures are: P CH4 = 1.0 atm and P H2O = 2.0 atm Q ~ ; Q > K; Reaction goes to the left are there are no reactants initially present. For reverse reaction K = 2.15 x 10
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 03/25/2009 for the course CHEM 102 taught by Professor Drk during the Spring '09 term at University of Alberta.

Page1 / 11

Problem Set #2 Sol - Chem 102 Klobukowski Winter 2009...

This preview shows document pages 1 - 3. Sign up to view the full document.

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