WS7_key - Chem 142 Name Worksheet 7 Chemical Equilibrium...

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Chem 142 Summer 2011 Page 1 of 7 Name ________________________________________ Section ________________ Worksheet 7: Chemical Equilibrium and Acids/Bases 1. For each of the following reactions, write the expressions for K c and K p . What is the relationship between K c and K p ? Does the reaction represent a homogeneous or heterogeneous equilibrium? a. H 2 ( g ) + I 2 ( g ) ! 2HI ( g ) K c = HI [ ] 2 H 2 [ ] I 2 [ ] K p = P HI ( ) 2 P H 2 P I 2 K c = K p homogeneous b. CH 3 OH ( g ) ! CO ( g ) + H 2 ( g ) K c = H 2 [ ] CO [ ] CH 3 OH [ ] K p = P H 2 P CO P CH 3 OH K c RT ( ) = K p homogeneous c. 2NH 3 ( g ) + CO 2 ( g ) ! N 2 CH 4 O ( s ) + H 2 O ( g ) K c = H 2 O [ ] NH 3 [ ] 2 CO 2 [ ] K p = P H 2 O P NH 3 ( ) 2 P CO 2 K c RT ( ) ! 2 = K p heterogeneous d. 2NBr 3 ( s ) ! N 2 ( g ) + 3Br 2 ( g ) K c = Br 2 [ ] 3 N 2 [ ] K p = P Br 2 ( ) 3 P N 2 K c RT ( ) 4 = K p heterogeneous e. 2KClO 3 ( s ) ! 2KCl ( s ) + 3O 2 ( g ) K c = O 2 [ ] 3 K p = P O 2 ( ) 3 K c RT ( ) 3 = K p heterogeneous f. CuO ( s ) + H 2 ( g ) ! Cu ( l ) + H 2 O ( g ) K c = H 2 O [ ] H 2 [ ] K p = P H 2 O P H 2 K c = K p heterogeneous 2. The equilibrium constant for the reaction A ( g ) + B ( g ) ! C ( g ) + D ( g ) is K. What is the equilibrium constant for each of the following reactions? a. C ( g ) + D ( g ) ! A ( g ) + B ( g ) K -1 b. 2A ( g ) + 2B ( g ) ! 2C ( g ) + 2D ( g ) K 2 c. ½ A ( g ) + ½ B ( g ) ! ½ C ( g ) + ½ D ( g ) K 1/2 d. 2C ( g ) + 2D ( g ) ! 2A ( g ) + 2B ( g ) K -2 e. ¼ C ( g ) + ¼ D ( g ) ! ¼ A ( g ) + ¼ B ( g ) K -1/4
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Chem 142 Summer 2011 Page 2 of 7 3. For the following reactions calculate K using the given equilibrium conditions. Do the second set of conditions represent a system at equilibrium? If not, which way will the reaction shift in order to reach equilibrium? a. 2N 2 ( g ) + O 2 ( g ) ! 2N 2 O ( g ) i. 2.80 x 10 -4 mol N 2 , 2.50 x 10 -5 mol O 2 , 2.00 x 10 -2 mol N 2 O in a 2.00 L flask N 2 O [ ] = 2.00 ! 10 " 2 mol 2.00 L = 1.00 ! 10 " 2 M N 2 [ ] = 2.80 ! 10 " 4 mol 2.00 L = 1.40 ! 10 " 4 M O 2 [ ] = 2.50 ! 10 " 5 mol 2.00 L = 1.25 ! 10 " 5 M K c = N 2 O [ ] 2 N 2 [ ] 2 O 2 [ ] = 1.00
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