thermodynamics

Will the reaction take place under certain conditions?

How much useful energy a reaction can generate

At what ratio of reactants to products is the reaction at equilibrium. Ie what products will form and what will the equilibrium ratios of reactants to products be

Open: Exchanges mass (matter) And energy

Closed: Energy not matter

Isolated: neither

Energy can't be created or destroyed only converted between forms

Through mass and or heat and or work

The transfer of energy that causes disorderly molecular motion (Thermal motion). Heat Transfer from the system to the surroundings causes random motion of molecules in the surroundings

The transfer of energy that causes organise molecular motion. Work done by the system on surroundings causes motion Of molecules in the surroundings in the same direction

Extensive variables depend on the amount of material for example grams

Intensive variables don't e.g. molar energy

State functions describe the systems condition. They are independent of path taken to reach that condition.

Temperature T, Pressure P, Volume V, Internal energy U, enthalpy H, Entropy S, gibbs free energy G

Heat and Work aren't state functions (They are path functions) this is because it matters how you get from initial to final states to work out their change

How energy and temperature change when gases expand

We look at a infinitesimally small change in volume instead an integrate between the final and initial values

Pext<Pint for expansion to occur

The most work has done when the external pressure is as large as possible i.e. the max amount of work done is when the external pressure is as close to the internal pressure as possible

Bomb contains sample and high Pressure O2

Measure the temperature change before and after ignition (Use extrapolation to account for heat loss To surroundings)

Then use Δq=cΔT

If Δq=Δu, heat capacity can also be written as du/dt. This means that the heat capacity at a constant volume tells us the change in U Needed to change temperature by a certain amount.

We can express heat capacity at a constant volume as

We've used the partial derivative of U, as U can also change with pressure

It is easier to measure in standard reactions than U (To measure U We need to make sure that the system doesn't lose heat or change In At constant pressure sovolume)

KNOW THAT ΔH=Δq at constant pressure so

Δq = Δu + pΔv

So we can just measure temperature changes

Remember Δq=cΔT

1 bar pressure 101325pa

25°c = 298.15k

Hθ or Hθ298 mean standard enthalpy of a sample at T=298k

At Non-standard conditions:

H= Hθ+ correction (for conditions)

The enthalpy change associated with forming that molecule from its constituent elements In the standard state. This is why the enthalpy of Formation for elements is zero

ΔvH: eg: H2O(l)-> H2O(g)

ΔcH[x]: oxidation of X with O2 to give carbon dioxide and water

Gas-> aqueous (ions)

Enthalpy is the state function so the Standard enthalpy Of a given reaction can be obtained as the sum of the standard enthalpies For a sequence of reactions that lead from the same reactions to the same product.

Ie It doesn't matter what path you take If starting reactants and final products are the same, enthalpy Is the same

Adjust a reaction in enthalpy for temperature changes using the heat capacity of the substances involved. Ie a “heat correction” for Different temperatures

Step 1: Heat the Sample and Reference

• Sample: The substance you’re testing (e.g., plastic, chocolate, drug).

• Reference: An inert material (like an empty pan) that doesn’t change when heated.

• Both are heated at the same rate. The reference acts as a "baseline" to compare against the sample.

Step 2: Track Heat Flow

• As temperature rises, the instrument measures tiny differences in heat flow (energy needed to heat the sample vs. the reference).

• If the sample absorbs extra heat (e.g., melting) or releases heat (e.g., crystallization), the DSC detects this.

Step 3: Heat Capacity and Enthalpy

• Heat capacity (CpCp​) is how much heat energy the sample needs to warm up by 1°C.

• The DSC graph plots heat flow (related to CpCp​) vs. temperature.

• Enthalpy change (total heat absorbed/released during a process like melting) is the area under the peaks in the graph. Example: A melting peak’s area tells you how much heat energy was used to melt the sample.

Atomisation: eg 1/2 cl2(g)-> cl(g)

Produces one mole of gaseous ions from the element in standard state

Lattice enthalpy: enthalpy taken to from one mole of a lattice from gaseous ions

A measure of how disordered the system is

We can increase it by heating as heat increases random motion. Ie Heating leads to random motion of molecules/atoms that will increase disorder

We can also increase disorder by changing physical state

For surroundings:

Q/T

For system:

Q(reversible)/T

Where ΔQ(reversible) = -Δw= nRTln(Vf/Vi)

When calculating the change in entropy for a system undergoing an irreversible process, we can't track temperature or pressure throughout because the system is not in equilibrium — these variables are not well-defined during the process. However, calculating entropy involves integrating heat over temperature, so we need to know the temperature at each step. To solve this, we imagine a hypothetical reversible path where the system stays in equilibrium and temperature and pressure are well-defined throughout. This allows us to use the reversible heat qrevq_{\text{rev}}qrev​ to calculate entropy change accurately, since entropy is a state function and depends only on the initial and final states — not the path taken

The universe is at constant temperature and pressure as It is so large.

The temperature and pressure does not always remain constant for a system so we use q reversible because it allows us to calculate entropy change Using a reversible path, what temperature and pressure are defined at every point, even if the actual process is irreversible.

Entropy of an isolated system increases in the course of a spontaneous change

This means that the entropy of an isolated system ie The universe can never decrease

ΔSuniverse=ΔSsystem+ΔSsurroundings> or equal to 0

The system entropy can decrease, but only if entropy of the surroundings increases by a greater amount

Heat flows equally opposite (When a hot object loses heat, A cold object gains the same amount of heat)

However, change in entropy isn't the same

ΔS=q/T. A cold object has a smaller T, Causing a larger entropy increase. A hot object has a high T, Causing a small entropy decrease. Ie the effects of temperature change on the change of entropy are different

If a cold object loses heat, the entropy is a large negative and if a hot object gains heat, the entropy is A small positive. This leads to a overall decrease in entropy which is not possible therefore heat must flow from hot to cold

Reversible: The process is idealised, infinitesimally Slow and the system is always in Equilibrium. Change can be undone without leaving any effect on the system or the surroundings.

Irreversible: Happens quickly or spontaneously. They can't be reversed without leaving a net effect for example, heat loss in the system or the surroundings

At very low T, solids don't absorb heat linearly

S(T=0k)=0

This is only true for perfect crystals

At zero Kelvin, it's atoms can't change arrangements so it only has one way of ordering atoms and it can't decrease in disorder any further

The sum of entropy changes to get to that temperature

Ie entropy at 0k+ entropy(0k->T)

Measure used to predict whether a chemical reaction will happen spontaneously

ΔG=ΔH-TΔS

ΔG<0 spontaneous

=0 system is at equilibrium

>0 reaction won’t happen unless you add energy

The values of H and S Increase, and it isn't obvious what the overall contribution to gibbs is

V & s are always positive

We have to assume enthalpy change doesn’t change with temperature

You can also use this to work out at what temperature is a reaction no longer spontaneous

Gases have the greatest entropy change at higher temperatures because their particles have the most freedom to explore new configurations as thermal energy increases.

Gas has a higher entropy than solids or liquids so as you increase temperature, Gibbs gets more negative more quickly

Clapeyron equation

Using the Clausius-clapeyron equation

Some substances require energy to overcome molecular interactions which may decrease surrounding entropy. Meaning that the universal entropy may decrease

The change in Gibbs energy with respect to the number of moles.

If there is 1 substance

Ie The chemical potential is equal to the molar Gibbs energy

They are equal

For example, ice and water coexist when their chemical potentials are equal

For ideal gases (As they don't interact with each)

for non-ideal gases (ie There are inter-component interactions): we use gibbs free energy of mixing (equation in booklet)

If one chemical potential changes, it effects all other values of chemical potential