practica7 - Calculando concentraciones molares para Na2S2O3...

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Calculando concentraciones molares para Na2S2O3 Cm 2 = Cm 1 V 1 V 2 Para el ensayo 1 tenemos que: Cm1= 0.15M V1=25 V2=30 Cm 2 = 0.1525 30 = 0125 De manera similar para el Na2S2O3 tenemos que: La velocidad la calculamos con v = 1 / seg , así para el evento 1 con t=22.5 v = 1 22.5 = 0.044 Orden de reacción: 0
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-1.7 -1.6 -1.5 -1.4 -1.3 -1.2 -1.1 -1 -0.9 -0.8 -2.5 -2 -1.5 -1 -0.5 0 f(x) = 1.08x - 0.36 LogV-log[Na2S2O3] Calculando concentraciones molares para HCl Cm 2 = Cm 1 V 1 V 2 Para el ensayo 1 tenemos que: Cm1= 0.3M V1=25 V2=30 Cm 2 = ( 0.3 ) 25 30 = 0.25 De forma similar para el HCl tenemos que: Como v = 1 / seg , así para el evento 1 con t=22.5 v = 1 12.26 = 0.08156
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Orden de reacción: 0 -1.4 -1.3 -1.2 -1.1 -1 -0.9 -0.8
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Unformatted text preview: -0.8-0.7-0.6-0.5-1.6-1.4-1.2-1-0.8-0.6-0.4-0.2 f(x) = 0.42x - 0.87 Log V- log[HCL] Para cada graFca ¿Qué representa el valor de la pendiente y la ordenada al origen? En ambas graFcas la pendiente representa el orden de la reacción mientras que la ordenada al origen es el logaritmo de la constante de velocidad. Comprobación Primer graFca Tomando dos eventos arbitrariamente-1.352182518=logk +n(-0.903089987)+mlog0.3…. .(1)-1.447158031=logk+n(-1)+mlog0.3…. .(2) Restando 1-2 0.09=0.09n n=1 Que concuerda con la ecuación que obtuvimos en la que y=1.08x-0.3637...
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  • Summer '14
  • ANTonio
  • Gráfica, Centímetro, Módulo de mando y servicio, V2=30

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