# Balance de energia

Páginas: 3 (676 palabras) Publicado: 15 de noviembre de 2011
Balance de Energía
Calor de materia prima
QMP=i=1nmiCpiTR-TA
TA = 22°C = 295.15K
TR = 110°C =383.15 K
TS = 26°C =299.15 K
| | A | B | C | D | CP liq. |
C2H4O2 | acetic acid | -18.944 |1.0971E+00 | -2.8921E-03 | 2.9275E-06 | 128.194627 |
C8H18O | 1-octanol | 93.554 | 1.6251E+00 | -4.1118E-03 | 4.2771E-06 | 324.97978 |
HCl | hydrogen chloride | 73.993 | -1.2946E-01 | -7.8980E-05 |2.6409E-06 | 96.8043037 |
H2O | water | 92.053 | -3.9953E-02 | -2.1103E-04 | 5.3469E-07 | 75.6250174 |
| | | | T | 298.15 | |
CP esta en J/mol k asi que es necesario pasarlo a J/g k = KJ/Kgk, para esto dividimos entre sus pesos moleculares
C2H4O2 | acetic acid | 2.1347 |
C8H18O | 1-octanol | 2.4998 |
HCl | hydrogen chloride | 2.7099 |
H2O | water | 4.2014 |QMP=(mh2oCph2o+mHClCpHCl+mC8H18OCpC8H18O+mCH3COOHCpCH3COOH)TR-TA
QMP = [(1.3162*4.2014)+(0.6626*2.7099)+(22.67*2.4998)+(10.46*2.1347)]( 383.15-295.15)
QMP = 7596.5903 J = 1815.6288 calorias
Calor de reacción∆HRT=∆HR°+298.15TR∆Cp dT
Para calcular el Hf° C8H17OH y Hf°C10H20O2 es necesario calcularlo por contribución de pesos moleculares
C2H6O + C6H17OH C8H17OH
46 84130
0.35 0.65

Hf° C8H17OH= (0.35*-277690)+(0.65*-156230) = -198741 J/mol
Hf°CH3COOH= -484500 J/mol
Hf°H2O= -285830 J/mol
C8H17OH + CH3COOHC10H20O2 + H2O
130 60 172 18
0.68 0.32
Hf°C10H20O2= (0.68*-198741) + (0.32*-484500) = -290183.88 J/mol
ΔHR= [-290183.88 -285830] -[-484500 -198741] = 104227.12 J/mol = 24910 cal/mol
| | A | B | C | D | CP liq. |
C2H4O2 | acetic acid | -18.944 | 1.0971E+00 | -2.8921E-03 | 2.9275E-06 | 128.194627 |
C8H18O | 1-octanol |93.554 | 1.6251E+00 | -4.1118E-03 | 4.2771E-06 | 324.97978 |
H2O | water | 92.053 | -3.9953E-02 | -2.1103E-04 | 5.3469E-07 | 75.6250174 |
C10H20O2 | 2-ethylhexyl acetate | 196.577 | 1.0336E+00...

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