enthalpy and entropy Flashcards

(59 cards)

1
Q

Define lattice enthalpy and give an eg

A

The enthalpy change that accompanies the formation of one mole of an ionic compound from its gaseous ions under standard conditions.
k+(G) +cl(g)- —kcl(s)

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2
Q

What kind of energy change will lattice enthalpy always be?

A

Exothermic

= negative

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3
Q

What is the Born-Haber cycle?

A

A way of indirectly calculating lattice enthalpy from other known energy changes.

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4
Q

whats route 1 of the born haber cycle and route 2

A

formation of gaseous atoms
formation of gaseous ions
lattice formation

converts elemenst in their standard states directly to ionic lattice.uses enthalpy change of formation.

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5
Q

What is the formation of gaseous atoms and what energy change is it?

A

Changing elements in their standard states into gaseous atoms.
Endothermic as it involves bond breaking.

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6
Q

What is the formation of gaseous ions and what energy change is it?

A

Changing gaseous atoms into positive and negative gaseous ions.
Endothermic.

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7
Q

What is the lattice formation and what energy change is it?

A

Changing gaseous ions into the solid ionic lattice.

exothermic

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8
Q

enthalpy changes involved in born haber process

A

stanadard enthalpy change of formation
standard enthalpy change of atomisation
the first ionisation energy
first electron affinity

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9
Q

Define standard enthalpy change of formation triangle fh

A

The enthalpy change that takes place when one mole of a compound is formed from its elements under standard conditions, with all reactants and products in their standard states.
na(g) + 1/2Cl2(g)–nacl(s)

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10
Q

Define standard enthalpy change of atomisation triangle at h

A

The enthalpy change that takes place for the formation of one mole of gaseous atoms from the element in its standard state under standard conditions.
na(s)–na(g)
1/2cl2(g)—–cl(g)
endothermic process as bonds r broken to form gaseous atoms.when elements a gas in its standard state,at its related to the bond enthalpy of bond being broken.

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11
Q

Define first ionisation energy

A

The enthalpy change required to remove one electron from each atom in one mole of gaseous atoms to form one mole of gaseous 1+ ions.

na(G)–na+(g) + e-

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12
Q

Why is ionisation energy always endothermic?

A

Energy is required to overcome the attraction between a negative electron and a positive nucleus.

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13
Q

Define first electron affinity

A

The enthalpy change that takes place when one electron is added to each atom in one mole of gaseous atoms to form one mole of gaseous 1- ions.

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14
Q

Why is first elctron affinity always exothermic?

A

Electron being added is attracted in towards the nucleus.

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15
Q

differences between ionisation enrgy and electron affinity

A

affinity measures the energy need to gain e-
ionisation measures the energy to lose e-

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16
Q

how to calc a full born haber cycle

A
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17
Q

what are sucessive elctron affinities

Why are successive electron affinities endothermic?

A

when an anion has a greater charge than 1-,sucessive elctron aff needed.
first=O(g) + e-=O-(g)
second=O-(g) + e- = O2-(g)

a second electron is being gained by a negative ion,which repels the electron away.so energy must be put in to force the negatively charged electron onto the negtaive ion.

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18
Q

What take places when a solid ionic compound dissolves in water?

A

water molecules cause giant Ionic lattice to break up

and overcome the strong electrsostatic attraction between the oppositley charged ions.

energy change involves is enthslpy chsnge of solution

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19
Q

Define standard enthalpy change of solution
show the equation for of nacl in h20

A

The enthalpy change that takes place when one mole of a solute dissolves in a solvent. if solvents water the ions from the ionic lattice finish up surrounded w water molcules as aqeous ions.
na+cl-(s) + aq–na+aq + cl-(aq)
enthalpy change can be endothermic and exothermic

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20
Q

explains position of ions in lattice when nacl solid n aqeous

A

when solid na+ and cl- ions are attracted in a giant ionic lattioce
when aqeous na+ and cl- ions are seperate but now surrounded by water molecules.
-in the aqeous ions the electrogentaive and electropositive charges in water molecules are attracted toward sthe + n - ions.

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21
Q

calc enthalpy change of solution

A

q=mct
mass/rfm
big/small

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22
Q

why mass used in enthalpy calcs

A

the mass that changes temp. theromemetr is in the solution that changes temp so it shpuld be mass of solution and not mass of water

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23
Q

how does the dissolving process work when an ionic compound dissolves in water

A

1)ionic lattice breaks up
2)water molecules are attracted to and surround the ions.
1)ionic lattice is broken up forming seperate gaseous ions that are free moving .
2)the seperate gaseous ions interact with polar water molecules and ions become hydrated and form bonds to form hydrated aqeous ions.enthalpy change is ecof hydration

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24
Q

whats enthalpy change of hydration

A

the enthalpy change that accompanies the dissolving of gaseous ions in water tof rom one mole of aqeous ions.
na+(g) + aq —-na+
cl-(g) + aq —-cl-

25
energy cycle to show enthalpy changes involved in dissolivng of an ionic compound in water
ecof solution can be endo or exo depending on the relative sizes of lattice enthalpya nd ecof hydration PHONEEEEEEE
26
how to determine unknown enthalpy changes energy cycle;e
phone
27
Properties of ionic compounds again and what’s ionic compounds that are liquid at rtp called
High mp and bp Soluble in polar solvents Conduct electricity when molten or in aqueous solution Ionic liquids
28
Factors that affect lattice enthalpy
Ionic size-ionic radius increases,attraction between ions decreases,lattice energy less negative,melting point decreases. Smaller the ion the stronger the attraction between ions and smaller ions can pack tg more closely and more energy is required to overcome the stronger forces. The MP AND BP ARE HIGHER ionic charge -as ionic charge increases Stronger electrostatic Attraction between ions increases Lattice energy becomes more negative Melting point increasese
29
What happens across period 3
The effects of both ionic size and charge affect attraction of ions Supporting effects:As increasing charge gives more attraction Decreasing size gives more attraction The opposing effects: inc charge gives more attraction Increasing size gives less attraction
30
What ensures jewels are long lasting and used as protective coating for inside or furnaces and refractories
Metal oxides with like very expthermic lattice enthalpues so they have a hugh mp
31
Lattice enthalpy is a very good indicator of..
The size of the melting point and the packing of ions In an ionic lattice
32
Factors affecting hydration
Ionic size-as ionic radius increases the attraction between ions and water molecules decreases and hydration energy is less negative Ionic charge-as ionic charge increases attraction with water molecules increases and hydration energy becomes more negative
33
What allows a ionic compound to fill dissolve in water
For this to happen the attraction between ions in ionic lattice is over come which needs energy equal to or more lattice enthalpy.water molecules are attracted to oppositely charged ions surrounding them releasing energy equal to hydration enthalpy If sum of hydration enthalpy larger than magnitude of lattice enthalpy then the overall energy char w be exothermic and it should dissolved Some compounds can still dissolve w endothermic energy changes but this can be due to entropy n temp
34
What is entropy (S)
It can be used to explain natural processes -A gas spreading trough a room -Heat from a. Side spreading through the room -Ice melting in a hot room Energy’s being dispersed here and becoming more spread out. There’s always a natural tendency for energy to spread out than being concentrated in one place. The greater the entropy the greater the dispersal of energy and the greater the disorder.
35
What’s entropy used for and what’s the units
Dispersal of energy within chemicals that make up the chemical system Units are jk-1 mol. The greater the entropy value the greater energy is spread out per K per mole
36
How does entropies differ between solids liquids and gases
Solids have the smallest entropies Liquids have the greater entropies Gases have the greatest entropies
37
What’s entropy used for and what’s the units
The dispersal of energy within chemicals that make up the chemical system Jk mol-1 The greater the entropy value he greater the energy’s spread out per k per mole.
38
How does entropy vary between solid liquid gases
Solids have the smallest entities Liquids have the greater entropies Gases have the greatest entropies
39
What happens at 0K and above 0K
There would be no energy and all substances have an entropy value of 0. Above 0K energy becomes dispersed among particles and all substances have a positive entropy. Systems that are more chaotic have a higher entropy value.
40
What happens to entropy (ΔS) when a system becomes more random and less random, and why?
AWhen a system becomes more random, entropy increases (ΔS is positive) because energy is more spread out, increasing disorder in the system. When it becomes less random energy becomes more concentrated and entropy change will be negative.
41
How can you predict the sign of entropy change (ΔS) in a chemical equation?
Compare the physical states and the no of gaseous molecules on each side.
42
How does entropy change work when it comes to changes of state
Entropy increases during changes in state that give a more random arrangement of particles : Solid-liquid-gas So when any substance changes state from solid to liquid to gas it’s entropy INCREASES -mp and bp increase the randomness of particles -Energy is spread out more and the 🔼s is positive
43
What do reactions that produce gases result in
An increase in entropy due to increasing of disorder of particles. Energy is spread out more and 🔼S is positive. We can predict the sign of entropy change where R+P have diff no of gas molecules.
44
Explain the entropy sign for N2(g) +3H2(g)—2NH3(g)
4 moles—2moles There’s a decrease in randomness of particles And the energy is spread out less and 🔼s is negative .
45
What’s the standard entropy
The entropy of one mole of a substance under standard conditions (100KPa and 298K) Standard entropies always positive and units is jk mol-1
46
How to calculate standard entropy changes
🔼s° = {S° (products) - {s° (reactants)
47
Calculate entropy change for the reaction below 2NO (g) + 02(g) → 2N02 (g) Standard enthalpies- NO-+211 O2- +205 NO2-+240
(2 x 140)- {(2x211) + 205} = -147 jk mol-1
48
Why do reactions happen
- a reaction can happen if the products have a lower overall energy than the reactants. feasibility used to describe whether a reaction is able to happen and is energetically feasible. Or spontaneous may be used for energetically feasible.
49
What’s free energy change
The overall energy during a chemical reaction and it’s made up two types of energy: 1) enthalpy change 🔼h. Heat transfer between chemical system n surroundings 2) entropy change at temp of reaction T🔼S. this is the dispersal of energy within chemical system itself.
50
Whats the gibbs equation and what does it show
The relationship between the two types of energy 🔼H and T🔼S. 🔼G=🔼G - T🔼S G= free energy change H=energy change w surroundings T= temp in K 🔼S= entropy change of system
51
Conditions for feasibility
there must be a decrease in free energy: • 🔼G < 0 The value for AH is usually much larger than for AS Typically AH values are shown in units of kJ mol- whereas the much smaller AS values are shown in units of J K-' mol-l Value and units of 🔼S must be changed to kjmol by /1000.
52
53
How does temperature affect the relative importance of enthalpy (ΔH) and entropy (TΔS) in determining the Gibbs free energy change (ΔG)?
At rtp, ΔH usually has a much larger magnitude than TΔS, so ΔG depends mainly on ΔH. As temperature increases, the TΔS term becomes more significant, affecting ΔG more.
54
The AH and As values for the decomposition of calcium carbonate are shown below. AH = +178 kJ mol-; CaCOz(s) → CaO(s) + CO, (g) AS = 161J K-' mol-i Determine the feasibility of this reaction at 25°C and at 1000°C.
Step 1: Make units consistent. Convert T to K: Convert AS to kJ K-' mol-1: T = 273 + 25 = 298K 1S = 161 1000 = 0.161 kJK-'mol-! Step 2: Use Gibbs' equation to calculate AG at 25 °C. AG = +178 - 298 × 0.161 = +130 kJ mol-1 As AG > 0, the reaction is not feasible at 25°C Step 3: Use Gibbs' equation to calculate AG at 1000°C. T = 273 + 1000 = 1273K AG = AH - TAS AG = +178 - 1273 × 0.161 = -27.0 kJ mol-1 As AG < O, the reaction is feasible at 1000°C The reaction can proceed at the high temperature of 1000°C. This fits with the process for making calcium oxide (lime) by heating calcium carbonate (limestone) in a furnace or lime kiln. • At 25°C, AG is positive. The reaction is not feasible and does not take place. • Above 833°C (inside the lime kiln), AG is negative. The reaction is feasible and does take place.
55
The balance between 🔼H and T🔼S
Feasibility depends upon the balance between 🔼H and T🔼s.
56
How to calculate minimum temp for feasibility
phone
57
How can endothermic reactions take place at rtp
Some ionic compounds dissolve in water at rtp in an endothermic process, cooling down the water. For eg,the enthalpy change of solution KCL is endothermic (AH = +16 kJmol-4). To find out why endothermic processes can take place at rtp, you need to use entropy and free energy.
58
How an endothermic process can be feasible at rtp: The enthalpy change and standard entropies for KCI(s), K+(aq), and Cl (aq) are shown below. KCI(s) → K+(aq) + Cl (aq) AH = +16 kJ mol-! S/JK-1mol-1 +83 +103 +57 Show that this process is feasible at 25°C.
: Calculate 🔼s at 25 °C. 🔼s = [S (products) - [S (reactants) = (103 + 57) - 83 = +77 JK-1 mol- = +0.077 kJ K-1 mol-1 Step 2: Use Gibbs' equation to calculate AG at 25°C. T = 273 + 25 = 298K AG = +16 - 298 × 0.077 = -7 kJ mol-As AG < O, the process is feasible.
59
Limitations of prediction made for feasibility
Many reactions have a negative 🔼g and do not seem to take place. Eg, the equation for the decomposition of hydrogen peroxide is shown below, together with its AG value at 25°C. H, (1) → H0(1) + 202(8) AG = -117kJ mol-1 But hydrogen peroxide does not decompose spontaneously at 25°C. So why doesn't the reaction take place? The answer lies with the very large activation energy resulting in a very slow rate. If hydrogen peroxide is left for long enough, it would decompose. The reaction does take place by addition of a trace of a MnO, catalyst, which allows the reaction to take place via an alternative route with a lower activation energy. The reduced energy barrier then allows the reaction to take place. So, although the sign of AG indicates the thermodynamic feasibility, it takes no account of the kinetics or rate of a reaction.