Problem Set #2 Sol

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

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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

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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
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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.

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

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