CHMY373 BU Physical Chemistry Thermodynamics Partial Pressures Help

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answer the problems in the attachment with full steps and on the side write an explanation with bold

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CHMY 373 2019 – Problem Set 10 Due: Fri Apr. 12 Reading: Sections 26.1‐26.5 1. Mercurial Phase Transition. The equilibrium between a liquid and its vapor can be treated analogously to the equilibrium achieved in a chemical reaction, where the liquid is the sole reactant and the vapor is the sole product. The Gibbs free energy of formation and enthalpy of formation of mercury vapor at ambient temperature 298 K and standard pressure 1 bar are 31.8 kJ mol and 61.4 kJ mol , respectively. a. Determine the Gibbs free energy of formation and enthalpy of formation of liquid mercury at 298 K and standard pressure. ○ ○ b. Determine the ∆ and the ∆ of mercury at 298 K and standard pressure. c. Write a balanced “chemical reaction” for the vaporization of mercury and determine the equilibrium constant at 298 K and standard pressure. d. Determine the equilibrium vapor pressure of liquid mercury at 298 K, assuming the vapor is an ideal gas. In the presence of gases, assume that any liquid species have an activity equal to 1. ○ e. Determine ∆ . f. Estimate the normal boiling point temperature of mercury using the ○ ○ ∆ and ∆ obtained above. 2. Boranes. Diborane, B2H6, is a toxic, pyrophoric, and unstable gas at ambient conditions, decomposing into hydrogen and other higher boranes as, for example: B H 2 → B H 5 6 H 5 When this reaction is carried out starting with 1 mol of B H no initial B H or H , the degree of dissociation, , is found to be 0.020 at 25 °C and a fixed ○ for the decomposition of B H at 25 °C. total pressure of 2 bar. Determine ∆ 3. Cracking Ammonia, Gromit! The cracking of gaseous ammonia is an industrial process used to create so‐called “forming gas”, a mixture of hydrogen and nitrogen: 2NH →N The equilibrium constant for this reaction is standard states at 1 bar standard pressure. 3H 4748 at 673 K, using ideal gas Calculate the final partial pressures of each of the three gaseous species after pumping pure ammonia into a cracking reactor held at fixed temperature and pressure of 673 K and 5 bar . You may need to use a calculator or solver program, and some intuition to deduce the sole physical result. 4. Syngas. “Synthesis gas,” a mixture containing carbon oxides and hydrogen, is a crucial intermediate for several important chemicals e.g., in the production of synthetic liquid fuels . Syngas is directly employed to produce methanol, a reaction that is highly selective when performed over the appropriate catalyst at high pressures. 2H → CH OH CO The equilibrium constant for this reaction is standard states at 1 bar. 0.00627 at 523 K, using ideal gas If carbon monoxide and hydrogen are initially added in a 1:1 ratio, calculate the equilibrium partial pressures of the three gaseous species at 523 K and 50 bar. You may need to use a calculator or solver program, and some intuition to deduce the sole physical result. ...
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Tutor Answer

dynamic_tutor
School: UIUC

Attached.

Surname 1

Your name
Professor’s name
Course
Date
Partial Pressures
Q3
Let's start with one mole of pure NH3, given that the pressure is 5 bars
Let's start with one mole of pure NH3, given that the initial pressure is 1 bar and the final
equalities pressure is 5 bars
2NH3(g) ----> N2 (g) + 3H2 (g)
initial

1 mole

equilibrium 1-2x

0

0

x

3x

1-2x+x+3x=5
X=2
Final Partial pressure Kp
𝐾𝑝 =
𝐾𝑝 =

Q4

𝑃𝐻2 ∗ (𝑃𝑁2 )4
(𝑃𝑁𝐻3 )2

(3 ∗ 2) ∗ (24 )
(1 − 2(2))

2

=

6 ∗ 16
= 10.67
9

Surname 2

Let's start with one mole of pure CH3 OH, given that the initial pressure is 1 bar and the final
equalities pressure is 50 bars
CO(g) + 2H2 (g) ----> CH3 OH(g)
T=0
equilibrium

Po
Po -x

Po -x + Po -2x + x = 50
Po -x = 50/2 = 25
Po -2x = 25 – x

X2

Final Partial pressure Kp

P0
Po -2x

__
x

Surname 3

(𝑃𝐶𝐻3 𝑂𝐻 )2
𝐾𝑝 =
(𝑃𝐶𝑂 )2 ∗ 𝑃𝐻2

Reference

Hi,I am sending you the homework :)

Running head: ENTROPY CHANGE

0

Entropy Change
Name of Author
Institutional Affiliation
Date

ENTROPY CHANGE

1
Question 1
Question a.

When calculating the Standard Gibbs free energy of formation of a certain compound,
there various steps needed to be undertaken. These are standard enthalpy of formation which
is designated as (ΔHƒ°), and absolute standard entropy which is designated as (ΔS°) and
standard temperature which is (298 K).
ΔGƒ° = ΔHƒ° (given compound) - TΔS° (given compound)
In formation of mercury liquid from vapour, the product is liquid while the vapour as
reactants in a reverse equation. Therefore, the given values above of standard Gibbs free
energy of formation will be used in calculating the formation of free energy and enthalpy
formation.
ΔG° (reaction) = ΣGƒ° (products) − ΣGƒ° (reactants)
Therefore energy of formation is 61.4 Kj/mol + 31.8Kj/mol=93.2kJ/mol
While the enthalpy of formation is 29.6kJ/mol.
Question b.
Change in Hvap=H vapor-H liquid
Therefore since we have Hvapor as 61.4 Kj/mol and that of liquid is 93.2kJ/mol. computing the
two values that is;
Is 61.4 Kj/mol-93.2kJ/mol=-31.8 Kj/mol.
Therefore for Gibbs energy of vaporization is
31.8 Kj/mol-93.2kJ/mol= -61.4 Kj/mol

ENTROPY CHANGE

2
Question c.

Vaporization of mercury equation is given below. Mercury is liquid in nature.
2Hg

Hg2.

The equilibrium constant is; 31.8kJ/mol. this is equal to formation of mercury vapour at
constant pressure and temperature

The equilibrium constant of an equation is given by K which symbolizes the value of Q
when the overall reaction of the equation is at equilibrium. K uniquely has a value at a given
reaction equation at constant pressure and temperature.
both Q and K are expressed in terms of concentrations, and partial pressure.
Question d.
Calculating the vapor pressure equilibrium is given by In P1/P2=change H vap /R
(1/T1-1/T2)
Assuming it is an ideal gas, PV=NRT and an activity of 1.
Moles of liquid mercury = 2
P= (2×8.3142×298) ÷1=4955.2632
Therefore, P1= 100000pascals and p2 is 4955pascals
In (100000÷4955.26) ÷ (1/273...

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Anonymous
Excellent job

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