CH 3 COOH C 2 H 5 OH CH 3 COO C 2 H 5 H 2 O La concentración con un volumen L

Ch 3 cooh c 2 h 5 oh ch 3 coo c 2 h 5 h 2 o la

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CH 3 COOH + C 2 H 5 OH ↔CH 3 COO C 2 H 5 + H 2 O La concentración con un volumen L: [ CH 3 COOH ¿ = 1 3 1 L = 1 3 L [ C 2 H 5 OH ]= 1 3 1 L = 1 3 L [ CH 3 COOC 2 H 5 ]= 2 3 1 L = 1 3 L [ H 2 O ]= 2 3 1 L = 1 3 L kc = [ CH 3 COOC 2 H 5 ] [ H 2 O ] [ C 2 H 5 OH ][ CH 3 COOH ] = 2 mol 3 L 2 mol 3 L 1 mol 3 L 1 mol 3 L = 4 mol L 30. A las 1100k, la constante de equilibrio para la reacción: 2 SO ( g ) 2 SO 2 ( g )+ O 2 ( g ) Es 0.0271 mol/L ¿Cuál es el valor de Kp a esta temperatura? ∆n = 3 2 = 1 RT ¿ ¿ kp = kc ¿ kp = 2.45 atm
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31. PARA LA REACCION: H 2 ( g ) + CO 2 ( g ) ↔H 2 O ( g ) + CO (g) KC ES 0.771 A 750ºC .Si 0.01 moles de H2 y 0.01 moles de CO2 se mezclan en un recipiente de 1 litro de 750 ºC ¿Cuáles son las concentraciones de todas las sustancias presentes en equilibrio? H 2 ( g ) + CO 2 ( g ) ↔H 2 O ( g ) + CO ( g ) 0.010.0100 x x x x 0.01 x 0.01 x x x kc = [ H 2 O ] [ CO ] [ H 2 ] [ CO 2 ] = x 2 ( 0.01 x ) 2 = 0,771 x ( 0.01 x ) = 0.878 x = 0.00468 Por consiguiente en el equilibrio: [H2]=[CO2]=0.01-0.00468=0.0053mol/L [H2O]= [CO] =0.00468 mol/L 32. Una muestra de 2 moles de HI se introduce en un recipiente de 5 litros. Cuando se calienta el sistema hasta una temperatura de 900 K, el HI se disocia según la reacción: 2 HI ÁH 2 + I 2 , cuya constante es: K C = 3,8·10 -2 . Determina el grado de disociación del HI. a) Equilibrio: N 2 O 4 (g) Á 2 NO 2 (g) n 0 (mol) n 0 0 n equil (mol) n 0 (1– ) 2 n 0 n TOTAL = n 0 (1– ) + 2 n 0 = n 0 (1+ ) m p·V = n TOTAL ·R·T = n 0 (1 + )·R·T = –––––– · (1+ )·R·T M(N 2 O 4 )
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p·M(N 2 O 4 ) 1,0 · 92 = ––––––––– – 1 = –––––––––––––– – 1 = 0,5 ya que la masa se conserva aunque se disocie. (m/V) ·R·T 2,24 · 0,082 · 333 b) Primero calcularemos K P a partir de los datos anteriores, para lo cual necesitamos conocer las presiones parciales de cada gas: n 0 (1– ) 1– 0,5 p(N 2 O 4 ) = ––––––– · p = –––– · p = ––– · 1 atm = 0,33 atm n 0 (1+ ) 1+ 1,5 2 n 0 2 1 p(NO 2 ) = ––––––– · p = –––– · p = ––– · 1 atm = 0,67 atm n 0 (1+ ) 1+ 1,5 p(NO 2 ) 2 (0,67 atm) 2 K P = ––––––– = ––––––––– = 1,33 atm p(N 2 O 4 ) 0,33 atm p(NO 2 ) 2 [(2 /1+ )·p] 2 4 2 K P = ––––––– = ––––––––––– = –––– · 10 atm = 1,33 atm p(N 2 O 4 ) (1– /1+ )·p 1– 2 Despejando se obtiene que: = 0,18 BIBLIOGRAFIA “Química General” – Raymond Chang. “Ciencia química” –Ing. Carlos E.Armas “Quimica General” 5º edición – McMurry- Fay
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  • Fall '19
  • Oxígeno, Cloro, Hidrógeno, Dióxido de azufre, Dióxido de nitrógeno

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