lecture+8sf

lecture+8sf - H 2(g I 2(g ⇄ 2HI(g K p = 1.00 x 10 2 A...

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Unformatted text preview: H 2 (g) + I 2 (g) ⇄ 2HI(g) K p = 1.00 x 10 2 A container holds 0.0100 atm H 2 (g), 0.00500 atm I 2 (g), and 0.500 atm HI(g). What are all equilibrium pressures? First do a Q test to predict the direction of the reaction. Q = 2 2 I H 2 HI p p p = ) 0050 . )( 0100 . ( 500 . 2 = 5000 > 1.00 x 10 2 Since Q > K, reaction goes to the left H 2 (g) + I 2 (g) ⇄ 2HI(g) Initial 0.0100 atm 0.0050 atm 0.500 atm Change +x +x -2x Equil 0.0100+x 0.0050+x 0.500-2x ) x 0050 . )( x 0100 . ( ) x 2 500 . ( 2 + +- = 1.00 x 10 2 Solving: x = 0.0355 atm At equilibrium: p HI = 0.500-2x = 0.500-2(0.0355) = 0.429 atm p H 2 = 0.0100+x = 0.0100+0.0355 = 0.0455 atm p I 2 = 0.0050+x = 0.00500 + 0.0355 = 0.0405 atm Check: ) 0405 )(. 0455 . ( 429 . 2 = 99.9 (close enough to 1.00 x 10 2 ) NH 4 CO 2 NH 2 (s) ⇄ 2NH 3 (g) + CO 2 (g) Solid NH 4 CO 2 NH 2 is introduced into a sealed container. When equilibrium is reached, the total pressure in the container is 0.363 atm. Calculate K p for the reaction. NH 4 CO 2 NH 2 (s) ⇄ 2NH 3 (g) + CO 2 (g) Initial solid (enough) 0 Change -x +2x +x Equil excess solid 2x x K p = p 2 NH 3 p HCl = (2x) 2 x = 4x 3 To get the value of x, use Dalton’s Law of Partial Pressures: The sum of the partial pressures equals the total pressure. 2x + x = 0.363 atm 3x = 0.363 atm x = 0.121 atm K p = (0.242) 2 (0.121) = 0.00709 2NO 2 (g) ⇄ 2NO(g) + O 2 (g) NO 2 (g) is placed in a container having an initial pressure of 1.000 atm. When equilibrium is reached, the total pressure is 1.463 atm. Calculate K p for this reaction. 2NO 2 (g) ⇄ 2NO(g) + O 2 (g) Initial 1.000 Change -2x +2x x Equil 1.000-2x 2x x Sum of partial pressures = total pressure (1.000-2x) + 2x + x = 1.463 1+ x = 1.463 x = 0.463 P NO 2 = 1.000-2x = 1.000-2(0.463) = 0.074 atm P NO = 2x = 2(0.463) = 0.926 atm P O 2 = x = 0.463 K p = 2 NO O 2 NO 2 2 p p p = 2 2 074 . ) 463 . )( 926 . ( = 72.5 2SO 3 (g) ⇄ 2SO 2 (g) + O 2 (g) K p = 1.21 x 10-5 SO 3 (g) is placed in a container, exerting 0.500 atm before any reaction. When equilibrium is reached, what are all partial pressures 2SO 3 (g) ⇄ 2SO 2 (g) + O 2 (g) Initial 0.500 Change -2x +2x +x Equil 0.500-2x 2x x 2 2 ) x 2 500 . ( x ) x 2 (- = 1.21 x 10-5 This is a cubic equation, difficult to solve numerically. There is a useful approximation technique that we will use fairly often. This approximation occurs when K is small. When K is small, the extent of reaction, measured by x (or in this case 2x) is small. Specifically, we can 2x is small compared with 0.500 or: (0.500 - 2x) is approximately equal to 0.500 Using this approximation, our equation becomes 2 2 ) 500 . ( x ) x 2 ( = 1.21 x 10-5 This is easy to solve. Collecting terms: x 3 = 7.56 x 10-7 Using the cube root function: x = 0.00911 How good is our approximation?...
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This note was uploaded on 07/17/2010 for the course 160 162 taught by Professor Kimmel during the Spring '10 term at Rutgers.

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lecture+8sf - H 2(g I 2(g ⇄ 2HI(g K p = 1.00 x 10 2 A...

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