# CHMY373 BU Physical Chemistry Thermodynamics Partial Pressures Help

*label*Science

*timer*Asked: Apr 9th, 2019

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### Question Description

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

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## Tutor Answer

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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