Prueba de Entrada TErmo

8/17/2019 Prueba de Entrada TErmo

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PRUEBA DE ENTRADA CICLO 2016-1

CURSO: TERMODINAMICA QUIMICA 1 PQ-321

ALUMNO: CONDORI PACO, MARCOS JHAYR20144090A

P1.- En un a!"#$ %an&$ '( )(!(%a)u%a, "a (n(%&*a "#+%( IS/ n$%a" '($%a#n '(" a!$% '( a&ua a "a )(!(%a)u%a )(%$'#n#a T5/, )#(n( "a6#&u#(n)( (ua#n:

ΔGf o = (-240 000+6.9 T+12.9 T !o"(10#T # $ %o!-1

Ca"u"a%: a/ 7a (n)a"!#a n$%a" 8 "a (n)%$!*a n$%a" '( $%a#n '(" a!$%'( a&ua a 1000.

+/ E" '( '#6$#a#n '(" a!$% '( a&ua a 10005 8 !%(6#n '( 20a).

RESOLUCION:

(a (n(%&*a "#+%( n$%a" '( $%a#n '(" a!$% '( a&ua a 1000 5$ (6:

ΔG °f ,1000 K =(−240000+(6.95)(1000)+(12.9)(1000)( log1000)) J

mol

ΔG °f ,1000 K =−194350 J

mol

Ha""a$6 "a En)%$!*a n$%a" '( $%a#n: Δ S°f =

−d ( ΔG °f )dT

Δ S °f =−d (−240000+6.95T +12.9T logT )

dT

J

K mol

Δ S°f =−(6.95+ 12.9

2.303+12.9 logT )

J

K mol

Δ S°f ,1000 K =−(6.95+ 12.9

2.303

+12.9log1000) J

K mol

Δ S°f ,1000 K =−51.25 J

K mol

R($%'an'$ "a %("a#n:

ΔG °f = Δ H °f −TΔS°f

Δ H 0

f ,1000 K =−194350 J

mol

+(1000 K )(−51.25 J

K mol

)


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Δ H °f ,1000 K =−245600 J

mol

('#D#6$#a#n '(" a!$% '( a&ua, '$n'( p (6 "a !%(6#n )$)a" 8 α ("

&%a'$ '( '#6$#a#n.

H 2O →H 2+1

2O

2

(1−α ) p

(1+1

2 α )

αp

(1+1

2 α )

1

2 αp

(1+1

2 α )

O+)(n($6 "a $n6)an)( '( (;u#"#+%#$ !:

p

p H 2 x(¿¿O2)

1

2

p H 2

O

Kp=¿

Kp=(

αp

(1+1

2α ))(

12

αp

(1+1

2α ))

1

2

(1−α ) p

(1+1

2α )

Da'$ ;u( α ≪1 "a (ua#n !u('( 6#!"#


3/7


7a a%#a#n '( "a (n(%&*a "#+%( n$%a" '( '#6$#a#n '(" a!$% '(a&ua a 1000 5 (6>24? @00 J$" 8 (6) %("a#$na'a $n ! 6(&Bn:

ΔG °=− RT ln Kp

R((!"aan'$ a"$%(6 24? @00 J$", R.314 J$".5, T1000 /$+)(n($6:

Kp=1.481525 x10−13 atm1/2,además p=20atm

R((!"aan'$ (n:

α =(2 Kp2

p )1

3 →α =1.3 x 10−9

conclusion :elvapor de aua, a1000 K ! 20atmestádisociado1.3 x 10−7


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P2.-7a '(6$!$6##n '(" A$n#a$ 6$+%( F$"%a#$, "a #'a ('#a '(%(a#n a%*a $n "a !%(6#n:

P)*,/o 26 10 163&

M)3&%/o*

G.@ 3.G 1.G 1.0

Ca"u"a%:

a) ¿Cuál es el orden de la reacción?

S(a "a %(a#n '( $%'(n n:

2 "H 3 ()→ " 2 ()+3 H 2( )

Eua#n '( ("$#'a':

"H

[¿¿ 3]n

r=−12

d [ "H 3]dt

=# ¿

Eua#n #n)(&%a'a: [ "H 3]1−n=[ "H 3]o+2# (n−1)t

T#(!$ '( #'a ('#a: t 1 /2= 2

n−1−1

2# (n−1)[ "H 3]on−1

S( $+6(%a ;u( "a #'a ('#a '( "a %(a#n (6) %("a#$na'a $n ("

$%'(n n '( "a %(a#n 8 $n "a $n(n)%a#n #n##a" [ "H 3]o

t 1 /2 α

1

[ "H 3]on−1

A!"#an'$ "$&a%#)$6

t

log (¿¿1/2)=constante−(n−1) log [ "H 3]o¿

T%a+a=an'$ $n "a6 !%(6#$n(6 #n##a"(6:

t

"H p(¿¿3)o

log (¿¿ 1/2)=constante−(n−1) log¿¿


5/7


%a


6/7


(%$, (n u8$ a6$ "a %(a#n 6(%*a '( 1(% $%'(n P$% )an)$ 6( !u('('(#% ;u( (" $%'(n S DEPENDE '( "a !%(6#n #n##a", "u(&$: (" $%'(nNO (6 #n'(!(n'#(n)(

c) ¿Cuál es el signicado del orden de la reacción respecto del

!ecanis!o de la reacción?

Cuan'$ 6( )#(n(n +a=a6 !%(6#$n(6, "a %a#n '( 6u!(%


7/7


?G; - 4GG0.91@>@?

.@?G2'

O 20.21 A - 2a > ' > 2N 20.G9 A - 20.G9A

N$6 '#(n ;u( a8 un ((6$ '( a#%( '( 30 (n a6a:

Rn1: CH4 > 2O2 V CO2 > 2H2OP$% a'a $" '( CH4 6( $n6u(n 2 $"(6 '( O2

Rn2: C2H@ > G2/O2 V 2CO2 > 3H2OP$% a'a $" '( C2H@ 6( $n6u(n G2/ $"(6 '( O2

⇒O2 )($%#$ 2GG0.91@ > G2/?.@?G/.32/⇒ O2 '( A 0.21A

((6$ '( a#%( '( 30 (n a6a

0 .3= [0.21 (−(

2 x770.916+ 72 x 85.657)](32)

[2 x770.916+ 72 x85.657 ](32)

A 11400.?G?9? $"(6 '( O2

C$n "a an)#'a' '( A, a""a$6 "a $!$6##n '( :

a 942.23 $"(6 '( CO2

??2.49 $"(6 '( O2' 1G9.03 $"(6 '( H2O0.G9A [email protected]?? $"(6 '( N2

7u(&$ n)$)a"(6 : 1229.9G $"(6 '( "$6 $!$n(n)(6 '( "$6 &a6(6 '($+u6)#n.

Pa%a Ha""a% (" $"u(n ('#an)( "a (ua#n '( (6)a'$ !a%a &a6(6#'(a"(6:

& =nRT

% =

(1299.978mol)(0.082 atm'l

mol'K )(403.15 K )1atm

=406616.3627 l

& =406.6163627 m3

Conclusión: el volumen de los gases de combustión que se generanes de 4!.!" m3

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