65. A Mercury vapor is dangerous because breathing it brings this toxic element into the lungs. We wish to estimate the vapor pressure of mercury at two different temperatures from the following data: A,H° (kJ/mol) S (J/K mol) A,G° (kJ/mol) Hg(€) 76.02 Hg(g) 61.38 174.97 31.88 Estimate the temperature at which Kp for the process Hg(€) sure of Hg is 1.00 bar). Next, estimate the temperature at whch the vapor pressure is (1/760) bar. (Experi- mental vapor pressures are 1.00 mm Hg at 126.2 °C and 1.00 bar at 356.6 °C.) (Note: The temperature at which P = 1.00 bar can be calculated from thermodynamic data. To find the other temperature, you will need to use the temperature for P 1.00 bar and the Clausius- Clapeyron equation on page 570.) Hg(g) is equal to 1.00 (and the vapor pres
65. A Mercury vapor is dangerous because breathing it brings this toxic element into the lungs. We wish to estimate the vapor pressure of mercury at two different temperatures from the following data: A,H° (kJ/mol) S (J/K mol) A,G° (kJ/mol) Hg(€) 76.02 Hg(g) 61.38 174.97 31.88 Estimate the temperature at which Kp for the process Hg(€) sure of Hg is 1.00 bar). Next, estimate the temperature at whch the vapor pressure is (1/760) bar. (Experi- mental vapor pressures are 1.00 mm Hg at 126.2 °C and 1.00 bar at 356.6 °C.) (Note: The temperature at which P = 1.00 bar can be calculated from thermodynamic data. To find the other temperature, you will need to use the temperature for P 1.00 bar and the Clausius- Clapeyron equation on page 570.) Hg(g) is equal to 1.00 (and the vapor pres
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Please find temperature wherein P=1.00 bar using clausius clapeyron equation. I will put here the partial answers of bartleby qand a.
Transcribed Image Text:65. A Mercury vapor is dangerous because breathing it
brings this toxic element into the lungs. We wish to
estimate the vapor pressure of mercury at two different
temperatures from the following data:
A,H° (kJ/mol)
S° (J/K · mol)
A,G° (kJ/mol)
Hg(€)
0.
76.02
0.
Hg(g)
61.38
174.97
31.88
Estimate the temperature at which Kp for the process
Hg(t)
sure of Hg is 1.00 bar). Next, estimate the temperature
at whch the vapor pressure is (1/760) bar. (Experi-
mental vapor pressures are 1.00 mm Hg at 126.2 °C and
1.00 bar at 356.6 °C.) (Note: The temperature at which
P= 1.00 bar can be calculated from thermodynamic
data. To find the other temperature, you will need to
use the temperature for P= 1.00 bar and the Clausius-
Clapeyron equation on page 570.)
Hg(g) is equal to 1.00 (and the vapor pres-
![4:29
Step1
a)
Civen, The searhion îz –
bar
The value f K, = 1, vabovne fressurce =
vahoune parssure "1
Temberatune (T) 357.6°C
find Out Temperatrze (5) = 9 at whih
Vafour þressuae P2 =
El bar = P,
o barr
Step2
b)
we Know that,
clautius - clap eryen epnation is
%3D
New, for this tisst tind Out the
and temferatuse at whih kp = 1
Ginen, Diti [Hgl] = 61.30, Hgl) = 0 K3/mel
61-38-0 = G1-38 KI|mel
dimillarly , ter As° = Éspodu - Esracat
AC = 174.97 – 76.02 = 9.95 I/K.mol.
so tind the Temporatuse ?
4 ter Kp=1 so De'=0
G1.38 2
98.95 X16%
61-38 - I x 98.95 => T=
%3D
T, = 620.3K](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5c04593f-7d69-4859-a6c6-314ab21b3d0d%2Fca7ab80e-6c35-45f9-817f-8911b7a69e12%2F65vx3pi_processed.jpeg&w=3840&q=75)
Transcribed Image Text:4:29
Step1
a)
Civen, The searhion îz –
bar
The value f K, = 1, vabovne fressurce =
vahoune parssure "1
Temberatune (T) 357.6°C
find Out Temperatrze (5) = 9 at whih
Vafour þressuae P2 =
El bar = P,
o barr
Step2
b)
we Know that,
clautius - clap eryen epnation is
%3D
New, for this tisst tind Out the
and temferatuse at whih kp = 1
Ginen, Diti [Hgl] = 61.30, Hgl) = 0 K3/mel
61-38-0 = G1-38 KI|mel
dimillarly , ter As° = Éspodu - Esracat
AC = 174.97 – 76.02 = 9.95 I/K.mol.
so tind the Temporatuse ?
4 ter Kp=1 so De'=0
G1.38 2
98.95 X16%
61-38 - I x 98.95 => T=
%3D
T, = 620.3K
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