Kno3 Delta H
Páginas: 2 (435 palabras)
Publicado: 2 de octubre de 2011
Tabla 1 Masa de KNO3 =4 gramos
n KNO3(mol) | Vol. de agua (mL) | Vol. Total desolución (mL) | Temperatura(°C) |Temperatura(K) |
0.0397 | 3 | 5.5 | 77.3 | 350.45 |
0.0397 | 1 | 6.5 | 70.8 | 343.95 |
0.0397 | 1 | 7.5 | 65.6 | 338.75 |
0.0397 | 1 | 8.4 | 50.8 | 323.95 |
0.0397 | 1 | 9 | 45.2 | 318.35 |0.0397 | 1 | 10 | 41.4 | 314.55 |
0.0397 | 1 | 11 | 35.3 | 308.45 |
0.0397 | 1 | 12 | 33.4 | 306.55 |
1 mol KNO3101.11 g x 4 g=0.0397 mol Concentración = mol L= 0.0397 mol0.0055L=7.218 M =s
EVENTO 1. Temperatura = 350.45 K
a) Calcular la concentración de los iones para cada evento
[K+]=[NO3-] = s
s = 7.218 M = [K+] = [NO3-]
b) Calcular la constante de equilibrioK
K= [K+][NO3-] = s2 = (7.218)2 = 52.099
c) Calcular ΔG a partir de la constante de equilibrio para cada temperatura
ΔG= -RTlnK = -(8.314 J/molK)(350.45 K)ln(52.099) = -11 518.048 J/mold) Calcular ΔS a partir de los valores de ΔG obtenidos para cada evento
Vol. Total desolución (mL) | Temperatura(K) | 1/T(K-1) | Solubilidad(mol/L) | K | ln K | ΔG (J/mol) | ΔS (J/molK) |
5.5| 350.45 | 0.002853 | 7.218 | 52.099 | 3.953 | -11 518.048 | 112.118 |
6.5 | 343.95 | 0.002907 | 6.108 | 37.308 | 3.619 | -11 304.417 | 113.616 |
7.5 | 338.75 | 0.002952 | 5.294 | 28.026 | 3.333 |-11 133.511 | 114.855 |
8.4 | 323.95 | 0.003087 | 4.726 | 22.335 | 3.106 | -10 647.088 | 118.601 |
9 | 318.35 | 0.003141 | 4.412 | 19.466 | 2.969 | -10 463.035 | 120.109 |
10 | 314.55 |0.003179 | 3.97 | 15.761 | 2.757 | -10 338.143 | 121.163 |
11 | 308.45 | 0.003242 | 3.609 | 13.025 | 2.567 | -10 137.658 | 122.909 |
12 | 306.55 | 0.003262 | 3.308 | 10.943 | 2.393 | -10 075.211 |123.467 |
ΔS=-ΔG+ ΔHT = --11 518.048 J/mol+ 27 773.748 J/mol 350.45 K = 112.118 J/molK
Obtención del ΔH para KNO3
1/T (K-1) | ln K |
0.002853 | 3.953 |
0.002907 | 3.619 |
0.002952 | 3.333...
n KNO3(mol) | Vol. de agua (mL) | Vol. Total desolución (mL) | Temperatura(°C) |Temperatura(K) |
0.0397 | 3 | 5.5 | 77.3 | 350.45 |
0.0397 | 1 | 6.5 | 70.8 | 343.95 |
0.0397 | 1 | 7.5 | 65.6 | 338.75 |
0.0397 | 1 | 8.4 | 50.8 | 323.95 |
0.0397 | 1 | 9 | 45.2 | 318.35 |0.0397 | 1 | 10 | 41.4 | 314.55 |
0.0397 | 1 | 11 | 35.3 | 308.45 |
0.0397 | 1 | 12 | 33.4 | 306.55 |
1 mol KNO3101.11 g x 4 g=0.0397 mol Concentración = mol L= 0.0397 mol0.0055L=7.218 M =s
EVENTO 1. Temperatura = 350.45 K
a) Calcular la concentración de los iones para cada evento
[K+]=[NO3-] = s
s = 7.218 M = [K+] = [NO3-]
b) Calcular la constante de equilibrioK
K= [K+][NO3-] = s2 = (7.218)2 = 52.099
c) Calcular ΔG a partir de la constante de equilibrio para cada temperatura
ΔG= -RTlnK = -(8.314 J/molK)(350.45 K)ln(52.099) = -11 518.048 J/mold) Calcular ΔS a partir de los valores de ΔG obtenidos para cada evento
Vol. Total desolución (mL) | Temperatura(K) | 1/T(K-1) | Solubilidad(mol/L) | K | ln K | ΔG (J/mol) | ΔS (J/molK) |
5.5| 350.45 | 0.002853 | 7.218 | 52.099 | 3.953 | -11 518.048 | 112.118 |
6.5 | 343.95 | 0.002907 | 6.108 | 37.308 | 3.619 | -11 304.417 | 113.616 |
7.5 | 338.75 | 0.002952 | 5.294 | 28.026 | 3.333 |-11 133.511 | 114.855 |
8.4 | 323.95 | 0.003087 | 4.726 | 22.335 | 3.106 | -10 647.088 | 118.601 |
9 | 318.35 | 0.003141 | 4.412 | 19.466 | 2.969 | -10 463.035 | 120.109 |
10 | 314.55 |0.003179 | 3.97 | 15.761 | 2.757 | -10 338.143 | 121.163 |
11 | 308.45 | 0.003242 | 3.609 | 13.025 | 2.567 | -10 137.658 | 122.909 |
12 | 306.55 | 0.003262 | 3.308 | 10.943 | 2.393 | -10 075.211 |123.467 |
ΔS=-ΔG+ ΔHT = --11 518.048 J/mol+ 27 773.748 J/mol 350.45 K = 112.118 J/molK
Obtención del ΔH para KNO3
1/T (K-1) | ln K |
0.002853 | 3.953 |
0.002907 | 3.619 |
0.002952 | 3.333...
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