Physical chemistry 1 project

Physical chemistry 1 projectPhysical chemistry 1 projectPhysical chemistry 1 projectPhysical chemistry 1 projectPhysical chemistry 1 Physical chemistry 1 projectPhysical chemistry 1 project

PHYSICAL CHEMISTRY 1 PROJECT

NOVEMBER 25,2021- HO CHI MINH CITY

Problem 1

1A) Try to formulate 6 statements concerning the concept of thermodynamically pure substances.

Solution:

Pure substances: A pure substance has a homogenous and fixed chemical composition throughout, and may exist in more than one phase

6 statements concerning the concept of thermodynamically:

Statement 1:

Superheated vapor: In the region to the right of the saturated line and at the temperature above the saturation temperature substances exist at superheated vapor

Statement 2:

Change process of pure substances: at this point, it is important to consider the liquid to solid phase change process Not so much solid to the liquid because thermodynamics deals only with liquid to gases to generate power

Statement 3:

Saturation temperature: At a given pressure the temperature at which a pure substance

changes phase

Statement 4:

Saturation pressure: likewise at a given temperature the pressure at which a pure substance changes phase

Statement 5:

Triple point: we have defined the eq between two phases However under certain conditions water can exist at the same time as ice, liquid, and vapor These conditions define the so-called triple point

Statement 6:

Compressed liquid water: A substance is said to be a compressed liquid when the pressure is greater than the saturation pressure

1B) Compare a thermodynamically pure substance with a chemically pure

substance What for is the concept of thermodynamically pure substances?

Comparision of thermodynamics pure substance and a chemically pure substance

A chemically pure substance

A chemically pure substance is the

substance that are made up of only one kind

of particles and has a fixed or constant

structure (it is one made of only one kind of

molecule for covalent substances, one kind

of crystal for ionic compounds, one kind of

atoms for rare gases and metals)

Examples: Hydrogen, oxygen, nitrogen,

chlorine, iodine are pure substances

(elements) Water, sodium chloride, are pure

substances (compounds)

A thermodynamically pure substance

A thermodynamically pure substance does not have to be of a single chemical element just as long as it is homogeneous throughout

Examples: A mixture of water liquid and

water vapor is a pure substance because both phases have the same chemical composition A mixture of liquid and gaseous water is also a pure substance, nitrogen, helium,…

The concept of thermodynamically pure substance

A substance that has a fixed chemical composition throughout is called pure substance For example, water, carbon dioxide,oxygen, hydrochloric acid are all pure substances

A thermodynamics pure substance does not have to be of a single chemical element or compound, however A mixture of various chemical elements or compounds also qualifies as

a pure substance as long as the mixture is homogeneous A mixture of phases of two or more substance is can still a pure substance if it is homogeneous, like ice and water (solid and liquid) or water and steam (liquid and gas) However, a mixture of oil and water is not a pure substance because oil is not soluble in water then it will be on top of the water, forming two chemically dissimilar regions Thus, a mixture of two phases of a pure substance is still a pure substance as long as the chemical composition of all phases is the same

In thermodynamics, the term pure substance carries a special meaning A pure substance is a system that is (1) homogeneous in composition, (2) homogeneous in chemical aggregation, and (3) invariable in chemical aggregation This definition clarifies that two systems comprising pure steam and pure steam + water are both pure substances having chemical formula H2O that will not change if we mix them together Similarly, ice + water will be a pure substance But when we consider air - consists of N2 and O2 in the ratio 79/21 by volume

- when in gaseous state at normal temperatures and pressures At very low temperatures, one component will condense out whereas the other in a gaseous state, so that the system will not satisfy condition (2) Similarly, at very high temperatures, O2 and N2 may dissociate to form

O, N, NO leading to condition (3) is not satisfied Thus, air is not a pure substance in general Nevertheless, since in most practical applications extreme temperatures are not encountered and air essentially preserves its chemical composition, it is still treated as a pure substance

Problem 2

2A) Try to formulate 5 statements concerning the equilibrium of a chemical reaction system.

Solution:

5 statements concerning the equilibrium of a chemical reaction system:

Statement 1: The concentration of each of the reactants and the products is constant at the

equilibrium stage

For example: Let us consider a reversible reaction

A + B = C +D

In the starting when t = 0, the concentration of A and B is maximum and the concentration of

C and D is are minimum As the reaction proceeds, the concentration of C and D starts increasing and the concentration of A and B starts decreasing Therefore, the rate of

forwarding reaction keeps on decreasing and the rate of backward reaction keeps on

increasing Finally, a stage comes where the rate of forwarding reaction becomes equal to the rate of backward reaction This stage is known as equilibrium and at this stage, the

concentration of reactants (A and B) and concentration of products (C and D) are constant as both the rate of reactions are equal

- Statement 2: The rate of the forward reaction is equal to the rate of backward reaction at

equilibrium That's why equilibrium is known to have dynamic nature

For example: Let us consider a very simple reaction

H20(liquid)=H20(gas)

When water is heated, the water molecules start shifting into the gaseous phase and gaseous phase particles also start coming back into the liquid phase due to the increased vapor

pressure of the system Eventually, a stage comes when the gaseous phase particles moving back into the liquid phase (Condensation) becomes equal to the liquid phase particles shifting into the gaseous phase (evaporation) This shows the dynamic nature of the equilibrium

- Statement 3: A chemical equilibrium can only be established if none of the products escape

from the system that is the formed product should not separate out as a solid or gas

This is a very obvious thing If the product formed gets escaped from the system then the whole equilibrium gets disturbed as the rate of the forward reaction is no longer equal to the rate of backward reaction

For example: See this irreversible reaction

AgN O3+KCl−→AgCl + KN O3

In the above reaction, AgCl formed get participated Thus, this reaction does not attain an equilibrium

- Statement 4: A chemical equilibrium can be achieved from either side that is from the side

of the reactants or from the side of the products

H2+I2=2HI

Here I2 is purple in color and HI is colorless If 1 mole of H2 and 1 mole of I2 are taken into container A and 2 moles of HI are taken into container B, the intensity of color in container A decreases while that of container B increases, and finally both containers have the same intensity of color This shows that equilibrium can be achieved in either direction

- Statement 5: A chemical equilibrium does not get affected by a catalyst.

This is so because a catalyst increases the rate of forward and backward reaction by same rate Thus, the overall equilibrium is not disturbed A catalyst only helps in attaining

equilibrium quickly

2B) Compare the concepts of ΔG 0 and ΔG of a chemical reaction How could you access their values?

● Compare

Is the Gibbs free energy of a reaction at

equilibrium

Is the Gibbs free energy of a reaction at any given state

Is used for the standard conditions

(Temperature: 25 degree C

Is used for the given conditions (other conditions)

Pressure: 1 bar/ 1 atm

Concentration: 1M)

Is always the same because it is referring to

when the reactants/products are at standard

temperature/pressure

So as the chemical reaction approaches

equilibrium, it remains the same because it

is still referring to when the reaction is at

standard conditions

Is changes because the reaction is proceeding

So as the chemical reaction approaches equilibrium, it approaches zero

● Access their values

1/ According to equilibrium constant

Equation 1:

∆G° = – RT ln K

Since K is the equilibrium constant, we are at equilibrium, the amounts of products and reactants in the mixture are fixed, ∆G° can be thought of as a guide to the ratio of the amount of products to the amount of reactants at equilibrium and therefore the thermodynamic favorability of the reaction If it so happens that products and

reactants are equally favored at equilibrium, then ∆G° is zero

Equation 2:

∆G = ∆G° + RT ln Q

Since Q is not the K, we are not necessarily at the equilibrium position, ∆G can be thought of as a predictor about which way the reaction (has reactants and products defined by Q) will go

If ∆G° is negative at equilibrium, then we will have lots of products at equilibrium, meaning Q needs to be bigger (greater than 1) to approach K As Q gets larger (i.e., as

we get more products), the term ‘RT ln Q’ gets increasingly positive, and eventually adding that term to a negative ∆G°, will make ∆G = 0, equilibrium will be established and no further change occurs

It is possible that Q could already be too large and therefore ∆G is positive IF so, then the reaction will need to from more reactants, reduce the value of Q, and allow

∆G to reach zero, i.e., allow equilibrium to be established

If ∆G° is positive at equilibrium, then we will have lots of reactants at equilibrium, meaning Q needs to be smaller (less than 1) to approach K As Q gets smaller (i.e., as

we get more reactants), the term ‘RT ln Q’ gets increasingly negative, and eventually adding that term to a positive ∆G°, will make ∆G = 0, equilibrium will be established and no further change occurs

It is possible that Q could already be too small and therefore ∆G is negative, IF so, then the reaction will need more products, increase the value of Q, and allow ∆G to reach zero, i.e., allow equilibrium to be established

In short, it is ∆G that will be zero at equilibrium and the sign of it (generated by the combination of ∆G° and RT ln Q in Equation 2 will define which way the reaction proceeds

2/ According to Gibbs - Helmholtz equation

∆G = ∆H - T∆S

With:

- ∆H: enthalpy change

- ∆S: entropy change

- T: temperature

And the Gibbs- Helmholtz equation under standard conditions:

products reactants

∆G0 = ∆H0 - T∆S0

With:

- ∆H: standard enthalpy change

- ∆S: standard entropy change

- T: standard temperature (298K)

3/ According to electrical work done in a cell

∆G0 = nFEo

With:

- Eo is the emf of the cell

- n is the number of moles of electrons transferred in the balanced equation for the reaction occurring in the cell

- F is Faraday’s constant, F=96485C/mol

2C) Try to access the effects of temperature on the chemical reaction equilibrium bymeans of Gibbs energy and its property as a state function

Gibbs free energy and Temperature: The Gibbs-Helmholtz Equation

We also know that there are two components to ΔrG o That is, ΔrG o =ΔrH o − TΔrS where

the ΔrH o term is only weakly dependent on temperature, but the TΔrS o term is strongly

dependent on temperature due to the presence of the T in the term.

The dependence of ΔrG o on temperature is the derivative of G with respect to T.

(1)

If we apply Equation 1 to ΔrG o = G o − G o we get,

(2a, b, c) Equation 2c shows that if ΔrS o is positive ΔrG o decreases with temperature, but if

ΔrS o is negative ΔrG o increases with temperature This will tell us whether

ΔrG o increases or decreases with increasing temperature

However, if we want to know whether or not a reaction is favored by an increase or decrease in temperature we really need to be looking at the equilibrium constant If the equilibrium constant increases with temperature the reaction becomes more favored, but if the equilibrium constant decreases with temperature the reaction becomes less favored

We will show that the equilibrium constant for a chemical reaction depends

on ΔrG o /T and not on ΔrG o The Gibbs-Helmholtz equation explains the

question as to how ΔrG o /T changes with temperature The Gibbs-Helmholtz

equation can be presented in two different, but equivalent forms

(3a) and

(3b)

Or, applying the procedure we used in Equations 2a, b, and c, we can write,

(3c) and

(3d)

Start with the left-hand side of Equation 3a and show that it is equivalent to the definition of Gibbs free energy:

We could easily substitute ΔG for G and end up with ΔH in the above sequence of

equations

So we will see that it is useful after we know the relationship between the equilibrium constant and ΔrG o /T, but for now let's use the Gibbs-Helmholtz equation to calculate ΔrG o

at a temperature, T2 other than 25oC (which we will call T1)

Set up Equation 3c for integration

(4)

(5) Equation 5 integrates to give,

(6)

To carry out the integration on the right-hand side of Equation 6 we would need to know how the heat of reaction changes with temperature This information can be obtained if

we know the heat capacities of the reactants and products as functions of temperature However, usually the heat of reaction varies slowly with temperature so that it is a approximation constant to regard the heat of reaction

Notice that Equation 5 is trivial to integrate if we make the approximation that ΔrH o

is constant With this approximation, Equation 5 integrates to,

(7) or,

We will apply Equation 8 to calculate ΔG for the freezing of super-cooled water at

− 20oC The process is, H2O(l,− 20oC) → H2O(s, − 20oC)

Do we know ΔG for this process at 0oC? Note that freezing water at its normal melting point

is a reversible process so that the heat of fusion is a reversible heat at constant temperature Therefore,

(9) Then

(10a, b, c) Equation 8, for our problem, becomes,

Then ΔG is − 440 J which is what we would expect because freezing super-cooled water

is a spontaneous (and irreversible) process at − 20oC

Why ΔG/T ?

The equilibrium constant is the key factor of how strongly the reaction wants to go, depends on ΔrG o /T Consider any process at constant temperature and pressure, T and p

are constant then,

(11)

Let's divide Equation 11 by T to see what ΔG/T looks like,

(12)

Note that ΔH and ΔS in Equation 12 refer to things happening in the system Rewrite

Equation 12 to indicate that this is true,

(13)

Now, ΔHsys is heat absorbed by the system and heat had to come from the surroundings so

(14a, b)

Most likely our process is irreversibly, but that doesn't matter because H is a state function

so that ΔH is independent of path We can always find a reversibly path to change the enthalpy of the surroundings by an amount ΔHsurr If we add heat ΔHsurr to the surroundings isothermally and reversibly then the entropy change in the surroundings is

(15a, b) or