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SIDLkUkGIA ¥ ACLkIA
2011
ÞkC8LLMA Nº 2
Una mezc|a de S0 Ʒ Co
2
2SƷ Co
2
y
2SƷ n
2
en vo|umen se a||menta en
un a|to horno a 900 º C encontrar |a
compos|c|ón en equ|||br|o sab|endo
que |a pres|ón tota| es 1 atmŦ


DA1CSť

1Ŧ keacc|onesť
C +
1
2
O
2
CO Aͩ = -2ó7ûû -2û. 95Ͱ
C + O
2
CO
2
Aͩ = -942ûû - û. 2Ͱ
ͪ
2
+
1
2
O
2
ͪ
2
O Aͩ = -589ûû + 13. 1ûͰ
2Ŧ keacc|ón de| agua
C +
1
2
O
2
CO Aͩ = -2ó7ûû - 2û. 95Ͱ
CO
2
C +O
2
Aͩ = 942ûû + û. 2Ͱ
ͪ+
1
2
O
2
ͪ
2
O Aͩ = -589ûû +13. 1ûͰ
ͪ
2
+ CO
2
CO+ ͪ
2
O Aͩ = 8óûû - 7. ó5Ͱ
3Ŧ Ca|cu|o de |a energ|a ||bre
fť 1 ƹ 900 ¨ C ƹ 900+273 ƹ1173k
Aͩ = 8óûû -7. ó5 Ͱ
Aͩ = 8óûû -7. ó5(1173)
Aͩ = 373. 45
Aͩ = -RͰͮ;ͻ
Aͩ = -2. 3û25 - 1. 987ͰͮͿgͻ

SIDLkUkGIA ¥ ACLkIA
2011
Luego despe[amos k
ͮͿgͻ = -

4. 575 Ͱ

ͻ = ͷ;t|ͼͿg(-

4. 575 Ͱ
)
ͻ = ͷ;t|ͼͿg(-
8óûû -7. ó5 Ͱ
4. 575 Ͱ
)
ͻ = ͷ;t|ͼͿg(-
8óûû -7. ó5(1173)
4. 575 Ͱ
)
ͻ = 1. 175

Luegoť
ͻ =
P
CO
. P
ͪ
2
O
P
ͪ
P
CO
2

P
CO
. P
ͪ
2
O
P
ͪ
P
CO
2
= 1. 175

De La reacc|ón y de| enunc|ado tenemosť
ͪ
2
+ CO
2
CO + ͪ
2
O
û. 25 -ʹ û. 25 -ʹ ʹ +û. 5 ʹ


Luegoť
;Ͱ = P
ͪ
2
+ P
CO
2

+P
CO
+ P
ͪ
2
O
= 1ͷtͽ
;Ͱ = û. 25 -ʹ +û. 25 - ʹ + ʹ +û. 5 + ʹ = 1ͷtͽ

(ʹ + û. 5)(ʹ)
(û. 25 - ʹ)(û. 25 - ʹ)
= 1. 175

LCUACICN CUADkÁ1ICA
û. 174ʹ
2
-1. û87ʹ+û. û73
k
1
ţkƗƹ
-͸_¸͸`-4ͷ͹


C° lf ecufcló° de bf¾ff be°em¾ť
k
1
ƹ0Ŧ06828
k
2
ƹ6Ŧ14601

SIDLkUkGIA ¥ ACLkIA
2011


LULGCť
LA CCMÞCSICICN LN LÇUILI8kIC
P
ͪ
2
= û. 25 - ʹ = û. 182 P
ͪ
2
= 18. 2%
P
CO
2

= û. 25 - ʹ = û. 182 P
CO
2

= 18. 2%
P
CO
= ʹ + û. 5 = û. 5ó8 P
CO
= 5ó. 8%
P
ͪ
2
O
= ʹ = û. ûó8 P
ͪ
2
O
= ó. 8%
P
Ͱ
= 1ûû%









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