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# chapter 16 - 22 16.24 H 2 O(g CO(g H 2(g CO 2(g K = 1.6...

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Chapter 16 Homework 16.10 (a) K = [H 2 O][CO] [H 2 ][CO 2 ] = (0.11)(0.11) (0.087)(0.087) = 1.6 (b) H 2 + CO 2 ' H 2 O + CO H 2 CO 2 H 2 O CO Initial (M) 0.050mol/2.0L 0.050mol/2.0L 0 0 Change (M) -x -x +x +x Equilibrium (M) 0.025 – x 0.025 – x +x +x ( )( ) ( )( ) ( ) ( ) mol L M CO mol O H mol M x x x x x x x x x K 028 . 0 2 014 . 0 014 . 0 025 . 0 6 . 1 025 . 0 6 . 1 025 . 0 025 . 0 6 . 1 2 2 2 2 2 = × = = = = = = = 16.16 [N 2 O 4 ] = 15.6 g 5.00 L 1 mol N 2 O 4 92.01 g = 0.0339 M N 2 O 4 ' 2 NO 2 N 2 O 4 2 NO 2 Initial (M) 0.0339 0 Change (M) -x +2x Equilibrium (M) 0.0339 – x 2x

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[ ] [ ] ( ) ( ) 00783 . 0 00636 . 0 ) 4 ( 2 ) 10 99 . 1 )( 4 ( 4 ) 10 88 . 5 ( ) 10 88 . 5 ( 2 4 0 10 99 . 1 10 88 . 5 4 0339 . 0 2 10 88 . 5 10 88 . 5 4 2 3 3 2 4 3 2 2 3 3 4 2 2 2 = × × ± × = ± = = × × + = × × = = or a ac b b x x x x x O N NO K (a) mol L M m equilibriu at pressent NO of Amount 0636 . 0 ) 00 . 5 ))( 00636 . 0 ( 2 ( 2 = = = (b) % 8 . 18 100 0339 . 0 00636 . 0 % 4 2 = = M M d dissociate O N 16.20 The second equation has been reversed and multiplied by 1 / 2 K 2 = 1/( K 1 ) 1/2 16.22 The second equation has been reversed and multiplied by 2 K new = 1/ K 2 = 1/(6.66 × 10 –12 ) 2 = 2.25 × 10 22 16.24 H
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Unformatted text preview: 22 16.24 H 2 O(g) + CO(g) ' H 2 (g) + CO 2 (g) K = 1.6 Fe(s) + CO 2 (g) ' FeO(s) + CO(g) K = 1/(0.67) Fe(s) + H 2 O(g) ' FeO(s) + H 2 (g) K net = (1.6)/(0.67) = 2.4 16.46 (a) 2 CH 3 CO 2 H ' (CH 3 CO 2 H) 2 CH 3 CO 2 H (CH 3 CO 2 H) 2 Initial (M) 5.4 × 10 –4 0 Change (M) –2 x + x Equilibrium (M) 5.4 × 10 –4 – 2 x x ( ) ( ) ( ) % 84 100 10 4 . 5 10 3 . 2 2 % 10 3 . 2 10 2 . 3 10 3 . 9 70 10 3 . 1 2 10 4 . 5 10 2 . 3 4 4 4 4 3 2 5 2 4 4 = × × = × × = × + − × = − × = × = − − − − − − M M or x x x x x K (b) Increasing the temperature will shift the equilibrium to the left....
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