to Chemical Kinetics I. Calculate concentrations as a function of time for the second-order reaction A +B - C for which -d[A]/dt = -d[B]/dt d[C]/dt = k[A][B]). Use [A]o = 0.02000, [B], = 0.02000, k = 0.050 s'. Calculate concentrations over the time range from 0 to 500 seconds. %3! %3D
to Chemical Kinetics I. Calculate concentrations as a function of time for the second-order reaction A +B - C for which -d[A]/dt = -d[B]/dt d[C]/dt = k[A][B]). Use [A]o = 0.02000, [B], = 0.02000, k = 0.050 s'. Calculate concentrations over the time range from 0 to 500 seconds. %3! %3D
Introduction to Chemical Engineering Thermodynamics
8th Edition
ISBN:9781259696527
Author:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Publisher:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Chapter1: Introduction
Section: Chapter Questions
Problem 1.1P
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![to, Chemical Kinetics I. Calculate concentrations as a function of time for the
sccond-order reaction
A+B C
for which -d[A]/dt = -d[B]/dt d[C]/dt = k[A][B]. Use [A]o = 0.02000,
[B], = 0.02000, k = 0.050 s'. Calculate concentrations over the time range
from 0 to 500 seconds.
%3D](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe011370a-bb5f-4181-a6f9-8ef21d2088ba%2Fe6358dd7-128d-4a80-b9c0-e8d97a1eaebc%2Fscuqayh_processed.jpeg&w=3840&q=75)
Transcribed Image Text:to, Chemical Kinetics I. Calculate concentrations as a function of time for the
sccond-order reaction
A+B C
for which -d[A]/dt = -d[B]/dt d[C]/dt = k[A][B]. Use [A]o = 0.02000,
[B], = 0.02000, k = 0.050 s'. Calculate concentrations over the time range
from 0 to 500 seconds.
%3D
![i-2.5 sint"e 2 5 2.5 sin(2.51 -
4
4
Kind the first time after t0 when the current reaches zero.)
aD'+ bD +en )
is encountered. Solve for D, if a 700, b=-2.9, e300.
When the sparingly soluble salt BACO, is dissolved in water, the following
simultaneous equilibria apply:
BaCO, Ba" + Co,
co, + H;0 HCo, + OH
Employing mass- and charge-balance equations, the following relationship
can be obtained for a saturated solution of BaCO, in water:
Kyp = [Ba"][CO,']= 5.1 x 10*
K, = [OH][ HCO, JM Co,')=2.1 x 10+
[Ba"j - K,K, (Ba²]" - K, = 0
Find the concentration of free Ba" in the saturated solution.
A solution of 0.10 M nitric acid (HNO,) is saturated with silver acetate
(AgAc), a sparingly soluble salt. The system of mass- and charge-balance
equations describing the system is
[NO,] = 0.10
[Ag') = S
[Ac] + [HAc) = S
JAg']+ [H') = [Ac]+ INO,]
[Ag'][Ac] = 4.0 x 10
[H*][Ac/[HAc] =1.8 x 10
(mass balance)
(mass balance)
(mass balance)
(charge balance)
Kp
K.
2 In pipe flow problems the relationship
escillates as a of time t to the following:
A circuit of a a and a has a i that](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe011370a-bb5f-4181-a6f9-8ef21d2088ba%2Fe6358dd7-128d-4a80-b9c0-e8d97a1eaebc%2Fc68483_processed.jpeg&w=3840&q=75)
Transcribed Image Text:i-2.5 sint"e 2 5 2.5 sin(2.51 -
4
4
Kind the first time after t0 when the current reaches zero.)
aD'+ bD +en )
is encountered. Solve for D, if a 700, b=-2.9, e300.
When the sparingly soluble salt BACO, is dissolved in water, the following
simultaneous equilibria apply:
BaCO, Ba" + Co,
co, + H;0 HCo, + OH
Employing mass- and charge-balance equations, the following relationship
can be obtained for a saturated solution of BaCO, in water:
Kyp = [Ba"][CO,']= 5.1 x 10*
K, = [OH][ HCO, JM Co,')=2.1 x 10+
[Ba"j - K,K, (Ba²]" - K, = 0
Find the concentration of free Ba" in the saturated solution.
A solution of 0.10 M nitric acid (HNO,) is saturated with silver acetate
(AgAc), a sparingly soluble salt. The system of mass- and charge-balance
equations describing the system is
[NO,] = 0.10
[Ag') = S
[Ac] + [HAc) = S
JAg']+ [H') = [Ac]+ INO,]
[Ag'][Ac] = 4.0 x 10
[H*][Ac/[HAc] =1.8 x 10
(mass balance)
(mass balance)
(mass balance)
(charge balance)
Kp
K.
2 In pipe flow problems the relationship
escillates as a of time t to the following:
A circuit of a a and a has a i that
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