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Unformatted text preview: 2 O 4 and NO 2 at 25 ° C given: 2 NO 2 (g) ⇔ N 2 O 4 (g) K = 8.1 at 25 ° C B. Given K solve for eq Step 1 . Calculate Q at initial conditions and compare to K to determine which way the reaction will go to reach equilibrium. Q = P N 2 O 4 initial P NO 2 initial ( ) 2 = 3.0 0.50 ( ) 2 = 12 Q > K so reaction goes ! Step 2 . Make up i.c.e. table. 2NO 2 (g) ⇔ N 2 O 4 (g) i. 0.50 atm 3.0atm c. +2xx (+ sign on left, as e. 0.50 + 2x 3.0  x determined by Q) Step 3 . Substitute equilibrium values into equation for K and solve for x K = 8.1 = 3.0 ! x 0.50 + 2x ( ) 2 " # $ $ % & ’ ’ = 3.0 ! x 0.25 + 2.0x + 4x 2 Rearrange to get : 32.4 x 2 + 17.2 x − 1.0 = 0 Solve quadratic equation : x = ! 17.2 ± 17.2 ( ) 2 ! 4 32.4 ( ) ! 1.0 ( ) 2 32.4 ( ) x = 0.053,  0.58 Reject solution that does not make physical sense. So, x = 0.053 atm 2.9 P atm 0.61 P 4 2 2 O N NO = = Therefore, the equilibrium pressures are :...
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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.
 Spring '09
 DrK
 Equilibrium

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