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Lec 11 Jan 28 Wed - 2 O 4 and NO 2 at 25 ° C given 2 NO...

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2.7 3. Equilibrium Calculations (15-7) There are two types of calculations: A. Given equilibrium concentrations or pressures, find K B. Given K, find equilibrium concentrations or pressures. Example : A sample of 1.0 atm of NO 2 placed in an evacuated flask dimerizes to form N 2 O 4 (g). At equilibrium, the total pressure of the system is 0.61 atm at 25 ° C. Calculate K. Step 1 . Write balanced reaction and calculate Q 2NO 2 (g) N 2 O 4 (g) Q = 0/1 2 = 0 (so Q < K even if K turns out to be very small: rxn A. Given [] eq solve for K Step 2 . Make up reaction (i.c.e.) table. P N 2 O 4 = 0.39 atm P NO 2 = 1.0 ! 2(0.39) = 0.22 atm K = 0.39 0.22 ( ) 2 = 8.1 Step 3 . Put equilibrium values into equation for K 2 NO 2 (g) N 2 O 4 (g) Initial 1.0 0 Change - 2x +x (rxn goes right, ) Equil 1.0-2x x according to Q) We know the total pressure at equilibrium: 0.61 atm = P NO2 + P N2O4 0.61 atm = (1.0 - 2x) + x x = 0.39 atm
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2.8 (1) Intermediate K (K 10 3 - 10 2 ): at equilibrium, significant amounts of both reactants and products. A flask contains 3.0 atm N 2 O 4 and 0.50 atm NO 2 . Calculate the equilibrium pressures for N
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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. +2x-x (+ 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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