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    INSTITUTO POLITCNICO NACIONAL

    UNIDAD PROFESIONAL INTERDISCIPLINARIA DEBIOTECNOLOGA

    DEPARTAMENTO DE BIOINGENIERA

    LABORATORIO DE BIOINGENIERA

    PRCTICA NO. 4

    DETERMINACION DE CALOR DE REACCIN

    EQUIPO 4

    INTEGRANTES: ALVARADO DANIEL

    CRUZ ENRIQUEZ ANAYANCY

    IDIAQUEZ RIOS SHANDIRA ITZELLLANOS BARRANCO VICTOR SAL

    ROJAS FALCON NANCY

    PROFESORES(AS):CARTAS ROMERO ADA LUZ

    LOBATO AZUCENO LINALOE

    ABRICA GONZLEZ PAULINA

    ANGELES GUADALUPE RIVERA

    Fecha de en !e"a:#$%&e'%#4

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    DETERMINACION DE CALOR DE REACCIN

    INTRODUCCIN

    Se e!"#$!% &%'"( e (e%&&$)! % '% &%!*$ % e e!e(+,% &%'"(,-$&% /e e' 0$0*e#%1% e &e e( " %20"(2e( 3%(% /e '% *e#3e(%*/(% 3e(#%!e &% &"!0*%!*e /(%!*e *" " e'3("&e0" e '% (e%&&$)! /,#$&%.

    S$ e' #e $" e5*e($"( (e&$2e e!e(+,% '% (e%&&$)! 0e e!"#$!% e5"*6(#$&% 7 0$ e'0$0*e#% %20"(2e e!e(+,% 0e ''%#% e! "*6(#$&%. E! '%0 (e%&&$"!e0 e5"*6(#$&%08 0!e&e0$*% #e!"0 &%'"( 3%(% ("#3e( '"0 e!'%&e0 e' /e 0e e03(e! e( %' -"(#%( '"0!/e9"08 7 e! '%0 (e%&&$"!e0 e! "*6(#$&%08 "&/((e %' &"!*(%($"8 0e !e&e0$*% #:0 &%'"( 3%("#3e( '"0 e!'%&e0 e' /e 0e e03(e! e %' -"(#%( '"0 !/e9"0.

    De %&/e( " &"! e' &($*e($" *e(#" $!:#$&" e 0$+!"08 3%(% /!% (e%&&$)! e5"*6(#$&%8 e'&%'"( e (e%&&$)! 0e(: !e+%*$9" 7 3%(% /!% e! "*6(#$&% 0e(: 3"0$*$9".

    L% Le7 e He00 $&e /e &/%! " e! /!% (e%&&$)! /,#$&% '"0 (e%&*$9"0 0e &"!9$e(*e! e!3(" /&*"08 '% 9%($%&$)! e '% e!*%'3,% e0 e' #$0#" $! e3e! $e!*e#e!*e e /e 0e 9e($-$ /e'% (e%&&$)! e! /! 3%0" " e! /!% 0e($e e 3%0"0. E0*" 0$+!$-$&% /e8 0$ 0e 0e3%(% /!%(e%&&$)! e! /!% 0e($e e (e%&&$"!e0 &"! 9%'"(e0 e e!*%'3,% e (e%&&$)! &"!"&$ "08 0e3/e e &%'&/'%( ;H

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    MATERIALES

    MATERIAL EQUIPO REACTI OS>? &%'"(,#e*("0>? *e(#)#e*("0>@ %+$*% "( e 9$ ($">? 3("2e*%0 e @ #L>? 9%0"0 e 3(e&$3$*% "0 e ?#L

    >@ 3%(($''% e'6&*($&% e H?

    Le!*e=%0 e N%OHS"'/&$)! e N%OH . MS"'/&$)! e N%OH .? MS"'/&$)! e HC' .? MS"'/&$)! e HC' . M

    DESARROLLO E*PERIMENTAL

    A) DETERMINACION DEL CALOR DE DISOLUCIN DEL NaO+

    @. C"'"&%( @ #' e H?O*e#3e(%*/(% %#2$e!*e e! e'

    &%'"(,#e*(".?. Pe0%( @+ e N%OH 7 %+(e+%('"%' &%'"(,#e*("8 %+$*%( 1%0*% /e0e $0/e'9% &"#3'e*%#e!*e e'N%OH.

    B) DETERMINACION DEL CALORDE NEUTRALI,ACION DE UN

    ACIDO FUERTE POR UNA BASE FUERTE

    @. C"'"&%( @ #' e 0"'/&$)! e HC' .? M e! e' &%'"(,#e*("?. Me $( @+ e N%OH8 %+(e+%( %' &%'"(,#e*("8 %+$*%( 1%0*% /e 0e $0/e'9%

    &"#3'e*%#e!*e e' N%OH.

    C) DETERMINACION DEL CALOR DE NEUTRALI,ACION DE UN ACIDO FUERTEPOR UNA BASE FUERTE

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    @. C"'"&%( @ #' e 0"'/&$)! e N%OH .? M e! e' &%'"(,#e*(".?. Me $( @ #' e HC' .? M8 % $&$"!%( %' &%'"(,#e*(" /e &"!*$e!e N%OH.

    RESULTADOS

    De e!-.nac./n de 0a ca'ac.dad ca01!23.ca de0 ca01!2-e !1

    Ta 0a #5 Da 1& 1 en.d1& e6'e!.-en a0e& 'a!a de e!-.na! 0a ca'ac.dad ca01!23.ca de0ca01!2-e !1

    V"'/#e!e %+/%

    %T a (m

    V"'/#e!e %+/%&%'$e!*e

    #L

    M%0%e

    %+/%-(,% +

    m1

    M%0% e%+/%

    &%'$e!*e

    m2

    Te#3e(%*/( % e %+/%

    -(,%T@

    Te#3e(%*/( % e %+/%&%'$e!*e

    T?

    T final

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    Ta 0a 45 Da 1& 1 en.d1& 'a!a 0a de e!-.nac./n de0 ca01! de ne8 !a0.9ac./n en !e e0 NaO+(&) +C0(ac)

    N" ee53e($#e!*"

    Volumen de

    HCl0.25 M

    mL

    Cantidadde NaOH

    g

    No demoles

    deHCl

    moles

    No demoles

    deNaOHmoles

    Temperatura delagua

    ambienteT 1

    Temperatura fnal de

    ladisolucin

    T final

    C ! C "# #00 #.052$ 0.025 0.02%&2 2&

    2'%.#5 2(

    &0#.#5

    2 #00 #.0$ 0.025 0.025$' 2#2').#

    5 2$&00.#

    5

    Ta 0a ;5 Da 1& 1 en.d1& 'a!a 0a de e!-.nac./n de0 ca01! de ne8 !a0.9ac./n en !e e0 NaO+ (ac) +C0 (ac)

    N" ee53e($#e!*"

    Volumen de

    HCl0.25

    MmL

    Volumende

    NaOH0.25

    g

    No de

    molesde HClmoles

    No demoles

    deNaOHmoles

    Temperaturadel agua

    ambienteT 1

    Temperaturafnal de la

    disolucinT final

    C ! C "

    # #00 #00 0.025 0.025 22 2'5.#5 2)2'$.#5

    2 #00 #00 0.025 0.025 22 2'5.#5 2)2'$.#5

    Ta 0a 4? . KJ mol

    HC'%& >@ .4

    KJ mol

    N%C'%& > 411.119

    KJ mol

    H 2 o ' >? .

    KJ mol

    OH%&>?? . 4 KJ mol

    N%OH%&>4 .N%%& 0

    KJ mol

    N%

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    KJ mol

    Ta 0a >5 a01!e& e/!.c1& de 0a& en a0'.a& de d.&108c./n ne8 !a0.9ac./n

    N" ee53e($#e!*

    "

    H $0"'/&$)! eN%OH 0J

    H !e/*(%'$ %&$)! eN%OH 0 HC' %&J

    H !e/*(%'$ %&$)! eN%OH %& HC' %&J

    @ @ .4 >?.@@@ 0.4432

    Ta 0a ?5 a01!e& e6'e!.-en a0e& de 0a& en a0'.a& de d.&108c./n ne8 !a0.9ac./n

    N" ee53e($#e!*"

    H $0"'/&$)!e N%OH 0

    J

    H !e/*(%'$ %&$)! eN%OH 0 HC' %&

    J

    H !e/*(%'$ %&$)! eN%OH %& HC' %&

    J@ 4. @. @ @.

    AN@LISIS DE RESULTADOSSe #$ $) '% &%!*$ % e &%'"( /e 0e %20"(2e " 0e e03(e! e e! (e%&&$"!e0 /e *$e!e!'/+%( e! $0"'/&$)! %&/"0%8 /*$'$ %! " /! &%'"(,#e*(".

    L% &%!*$ % e &%'"( %0"&$% % % '%0 (e%&&$"!e0 /,#$&%0 7 % '"0 &%#2$"0 -,0$&"0 0ee0*%2'e&$) #$ $e! " '"0 &%#2$"0 e *e#3e(%*/(% e '% $0"'/&$)!.

    C/%! " 0e ''e9% % &%2" /!% (e%&&$)! e !e/*(%'$ %&$)! e! /! &%'"(,#e*(" #e &'%! " /!%$0"'/&$)! e /! :&$ " &"! "*(% e /!% 2%0e8 e' &%'"( '$2e(% " e! '% (e%&&$)! e!e/*(%'$ %&$)! e0 %20"(2$ " 3"( e' %+/% 7 3"( '%0 3%(e e0 e' &%'"(,#e*(".E! '% 3(:&*$&% 0"'" 0e /*$'$ ) /! 0"'" &%'"(,#e*(" 7% /e '% &%3%&$ % &%'"(,-$&% e /!" ee''"0 0%'$) !e+%*$9" '" &/%' 0e 3/ " e2e( % /e e0*e 7% *e!,% /!% -$0/(% %/! /e !" e(%%3(e&$%2'e % 0$#3'e 9$0*%.

    P%(% e*e(#$!%( '"0 #"'e0 /e (e%&&$"!%! e! &% % (e%&&$)! &"((e03"! $e!*e 0ee*e(#$!% &/%' e0 e' (e%&*$9" '$#$*%!*e 7% /e %' *6(#$!" e e0*e 7% !" e5$0*e (e%&&$)!. E!'% (e%&&$)! e &%'"( e $0"'/&$)! e' N%OH '% e03e&$e/ *$e!e /!% e!*%'3$% e-"(#%&$)! e 7% /e %&* % &"#" e'e#e!*".

    CONCLUSIONESE' &%'"( e (e%&&$)! e0 '% &%!*$ % e e!e(+,% &%'"(,-$&% /e e' 0$0*e#% 1% e &e e( "%20"(2e( 3%(% /e '% *e#3e(%*/(% e' 0$0*e#% (e%&*$9" 3e(#%!e &% $!%'*e(% % e! e'&%#2$" (e%&*$9"> 3(" /&*". C/%! " H e0 /!% (e%&&$)! e! "*6(#$&% 7 &/%! " H (e%&&$)! e5"*6(#$&%.

    T" %0 '%0 (e%&&$"!e0 $!9"'/&(% %0 e! '% 3(:&*$&% 0"! e5"*6(#$&%0.

    L"0 (e0/'*% "0 e53e($#e!*%'e0 */9$e("! /! #%(+e! e e(("( e2$ " %' e /$3" 7 % 0/#%!$3/'%&$)!..

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    BIBLIOGRAFIA

    EIQ8 BUI8 TQ F%2$" Ge(#:! B"(+"+!" C"#3e! $" e P("3$e % e0 T%2'% eE!*%'3,% e F"(#%&$)!8 E!e(+,% L$2(e e G$220 7 E!*("3,% e F"(#%&$)! eC"#3/e0*"0 I!"(+:!$&"0

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    +O A DE CALULOS

    Ta 0a N15 # N15 $

    P%(% &%'&/'%( '% &%3%&$ % &%'"(,-$&% e '"0 &%'"(,#e*("0 0e /*$'$ " '% 0$+/$e!*e -"(#/'%

    m2 Cp H 2 O (T 2 T final )= C (T final T 1 )+m1 Cp H 2 O (T final T 1 )

    De03e=%! " % C *e!e#"0

    C =m2 Cp H 2 O (T 2 T final) m1 Cp H 2 O (T final T 1 )

    (T final T 1 )

    S/0*$*/7e! " %*"0 "2*e!$ "0 e53e($#e!*%'#e!*e 7 *"#%! " % @#L e %+/% @+ e %+/%8 *e!e#"03%(% e' &%'"(,#e*(" e $0"'/&$)!8

    C =50 g( 4.180 J gK )(317.15 304.15 k ) 50 g(4.180 J gK )(304.15 295.15 k )

    (304.15 295.15 k )

    C = 92.89 J / K

    P%(% e' &%'"( e !e/*(%'$ %&$)! *e!e#"0

    C =50 g( 4.180 J gK )(317.15 306.15 k ) 50 g(4.180 J gK )(306.15 294.15 k )

    (306.15 294.15 k )

    C = 17.42 J / K

    Ta 0a 7: Pa!a e0 c 0c801 de -10e& NaO+

    E6'e!.-en 1#

    n= w PM

    n=1.0437 g

    40 gmol

    . ? #"'

    E6'e!.-en 1$ n= w PM

    n=1.0632 g

    40 gmol

    . ? #"'

    Ta 0a 4: Pa!a e0 c 0c801 de -10e& +C0 NaO+

    M10e& NaO+ M10e& de +C0

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    E6'e!.-en 1#

    n= w PM

    n=1.0527 g

    40 gmol

    . ? ?

    #"'

    n= w PM M =

    w PM V MWPMWV

    .?mol L .

    1 Lg

    mol .@? + e! @L

    n= 0.9125 g

    36.5 gmol

    0.025molE6'e!.-en 1

    $

    n= w PM

    n=1.0317 g

    40 gmol

    . ?

    #"'

    Ta 0a ;: Pa!a e0 c 0c801 de -10e& +C0 NaO+ 5$;M

    NaO+ 5$;M M =

    w PM V MWPMWV

    .?mol L 4

    1 Lg

    mol @ + e! @L

    n= 1 g

    40 gmol

    . ? #"'

    +C0 5$;M M = w

    PM V MWPMWV

    .?mol L .

    1 Lg

    mol .@? + e! @L

    n=0.9125 g

    36.5 gmol

    . ? #"'

    C%'&/'%! '"0 9%'"(e0 *e)($&"0 e &%'"( e

    N%OH0 >4? . KJ mol

    HC'%& >@ .4

    KJ mol

    N%C'%& > 411.119 H 2 o ' >? .

    .@? >>>@LX >>>> .@L5 . @? +

    @ + >>>>>@LX >>>> .@L 5 @+

    .@? >>>@LX >>>> .@L 5 . @? +

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    KJ mol

    KJ mol

    OH%&>?? . 4 KJ mol N%OH%&> @@.4

    KJ mol N%%&

    0 KJ mol N% >?4 .4 ?

    KJ mol

    H reaccin = H productos H reacti os

    DISOLUCION(A) !a (ac )

    +

    !aO H (s)+ H 2 O (l )" OH (ac )+ H 2 O (l )

    EXPERIMENTO 1

    H reaccin =[0.02609 mol OH ( 229.94 KJ mol )+0.02609 mol H 2 O( 285.5 KJ mol )]>[0.02609 mol !aOH ( 426.7 KJ mol )+5.6 mol H 2 O( 285.5 KJ mol )]

    H reaccin = 13 #4478 KJ /#%0'.'&2% KJ #5'%.)' KJ

    EXPERIMENTO 2

    H reaccin =[0.02658 molOH ( 229.94 KJ mol)+0.02658 mol H 2 O( 285.5 KJ mol )]>

    [0.02658 mol !aOH

    ( 426.7 KJ

    mol )+5.6 mol H 2 O

    ( 285.5 KJ

    mol )]

    H reaccin = 13.7004 KJ /#%#0.#)#$ KJ #5'%.)) KJ

    NEUTRALI,ACION(B) !aO H (s)+ HCl (l)" H 2 O (l)+ !aC l (ac )EXPERIMENTO 1

    H reaccin =[0.025 mol !aCl ( 411.119 KJ mol)+0.025 mol H 2 O( 285.5 KJ mol)]>

    [0.02632 mol !aOH ( 426.7 KJ mol )+0.025 mol HCl( 167.456 KJ mol)] H reaccin = 17.4155 KJ +15.4171 KJ = 1.9984 KJ

    EXPERIMENTO 2

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    H reaccin =[0.025 mol !aCl ( 411.119 KJ mol)+0.025 mol H 2 O( 285.5 KJ mol)]>[0.02579 mol !aOH ( 426.7 KJ mol )+0.025 mol HCl( 167.456 KJ mol)] H

    reaccin= 17.4155 KJ +15.1909 KJ = 2.2245 KJ

    NEUTRALI,ACION(C) !aO H (ac )+ HC l (l)"H 2 O (l )+ !aC l (ac )

    EXPERIMENTO 1 y 2

    H reaccin =[0.025 mol !aCl ( 411.119 KJ mol)+0.025 mol H 2 O( 285.5 KJ mol)]>

    [0.025 mol !aOH ( 511.435 KJ mol )+0.025 mol HCl( 167.456 KJ mol)] H reaccin = 17.4155 KJ +16.9723 KJ = 0.4432 KJ

    De e0*% #%!e(% 0e "2*$e!e! 3%(% '"0 *(e0 e53e($#e!*"0.

    P%(% e*e(#$!%( '"0 9%'"(e0 e53e($#e!*%'e0 0e *$e!e '% e&/%&$)!

    C +C p H 2 O mtotal (T final T inicial )= J Hneutrali$acin =

    C +C p H 2 O mtotal (T final T inicial )= J Hdisolucin =

    S/0*$*/7e! " %*"0 3%(% '% #$0#% (e%&&$)! e' e=e#3'" %!*e($"( *e!e#"0

    A)

    Hneutrali$aci n = 92.88

    J K

    +(4.180 J

    gK 101.04373 g)(304.15 K 295.15 K ) 4 .@ @J

    K4. ? J

    Hneutrali$aci n = 92.88

    J K

    +( 4.180 J gK

    101.0317 g)(302.15 K 296.15 K ) 4@?@. 4 @JK

    4.@?? J

    B)

    Hneutrali$aci n = 92.88

    J K

    +(4.180 J

    gK 4.6987 g)(301.15 K 296.15 K ) @ 4. JK@.

    4 J

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    Hneutrali$aci n = 92.88

    J K

    +(4.180 J

    gK 4.6777 g)(300.15 K 294.15 K ) 4. JK .

    4 J

    C)

    Hneutrali$aci n = 92.88

    J K

    +(4.180 J

    gK 103.734 g)(297.15 K 295.15 K ) @ ?. ?JK@.

    J