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lecture+8sf

# lecture+8sf - H2(g I2(g 2HI(g Kp = 1.00 x 102 A container...

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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 . 0 )( 0100 . 0 ( 500 . 0 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 . 0 )( x 0100 . 0 ( ) x 2 500 . 0 ( 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 . 0 ( 429 . 0 2 = 99.9 (close enough to 1.00 x 10 2 )

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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 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 0 0 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 . 0 ) 463 . 0 )( 926 . 0 ( = 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 0 0 Change -2x +2x +x Equil 0.500-2x 2x x 2 2 ) x 2 500 . 0 ( 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 . 0 ( 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? We assumed 0.500 - 2x = 0.500 With x = 0.00911 0.500 - 2(0.00911) = 0.482 The difference between 0.500 and 0.482 is 0.0182 (equal to 2(0.00911). Taken as a percentage of original pressure:

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x 100 = 3.6% This is fairly small. As a rule of thumb, if our approximation is good to within 5%, it’s considered reasonable. We could also refine our approximation by substituting our first answer (0.00911) into the denominator of our original equation. This is called the method of successive approximations. 2 2 ) x 2 500 . 0 ( x ) x 2 ( - = 1.21 x 10 -5 2 2 )) 00911 . 0 ( 2 500 . 0 ( x ) x 2 ( - = 1.21 x 10 -5 Solving now: x = 0.00889 This is the correct answer to three significant figures. 2SO 3 (g) 2SO 2 (g) + O 2 (g) K p = 1.21 x 10 -5 SO 3 (g) and SO 2 (g) are placed in a container.
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lecture+8sf - H2(g I2(g 2HI(g Kp = 1.00 x 102 A container...

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