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Chemsitry Year 2 > Energetics II > Flashcards

Energetics II Flashcards

1
Q

Define lattice enthalpy of formation

A

Enthalpy change when 1 mole of solid conic compound is formed from its gaseous ions under standard conditions Ca 2+ (g) + 2Cl- —> CaCl2. It’s exothermic

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2
Q

Define enthalpy change of atomisation

A

Enthalpy change when 1 mole of gaseous atoms is made from an element in its standard states.
Its endothermic

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3
Q

Define 1st electron affinity

A

Enthalpy change when 1 mole of gaseous 1- ions are made from 1mole of gaseous atoms. It’s exothermic

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4
Q

Define 2nd electron affinity

A

Enthalpy change when 1 mole of gaseous 2- ions are made from 1 mole of gaseous 1-ions

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5
Q

Define ionic bonding. What are the 2 factors that influences the strength of ionic bonding. What physical properties does it give and why?

A

Ionic bonding is the strong electrostatic F. A between oppositely charged ions. The size of the charges and ionic radii influences the strength of the bond. Bigger the charge - stronger ionic bonding. Bigger the ionic radii weaker the ionic bond - inverse relationship. So bigger charge and small radii (High charge density)→ strong ionic bond (strong electrostatic attraction)→ More energy required to break it → high m.p

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6
Q

Why is born haber cycles useful

A

To calculate lattice enthalpies which otherwise cannot be calculated directly from experiments

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7
Q

Define enthalpy of formation

A

Enthalpy change when 1 mole of solid compound is formed from its elements in standard states under standard conditions. Its exothermic

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8
Q

Define 1st ionisation energy

A

Energy required to remove the most loosely held electron from one mole of gaseous atoms to produce one mole of gaseous ions with 1+ charge

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9
Q

Born haber cycle and Hess cycle use the same principle. What is that

A

Total enthalpy change of a reaction is the same no matter what route is taken

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10
Q

What are the 2 conditions of a perfectly ionic model

A

1- ions that are prefectly spherical
2- The charge is evenly distributed in the sphere

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11
Q

Why is the experimental lattice enthalpy always different from theoretical

A

Its because compounds being experimented on doesn’t follow the perfectly ionic model and has some covalent characteristics

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12
Q

How to find if something has more covalent characteristics

A

Most of the time the positive ion distorts the charge distribution of negative ion. We say positive ion polarises negative ion. The more polarisation we get, more covalent character there is.

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13
Q

What does lattice enthalpy tell us

A

How much of a substance is purely ionic

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14
Q

What does the difference in lattice enthalpy show

A

Higher the difference → higher the polarisation and greater covalent character.
The compound with the smallest difference in theoretical and experimental lattice enthalpy in the entire data→ most perfectly ionic.

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15
Q

Smaller cations are more polarising than larger ones. Why

A

Smaller cations have higher charge density as the charge is concentrated in small area. The cations pull electrons towards itself

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16
Q

Are large or small anions polarised easily. And why

A

Large anions with large charge are polarised more easily. As electrons are further away from the nucleus and there is more repulsion b/w the electrons in the ion. The electrons can be pulled away towards the cation

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17
Q

Define electronegativity. How is the electronegativity of elements in the periodic table?

A

Electronegativity is the ability to attract electrons towards itself in a covalent bond.in the periodic table as you go furthe up and right electronegativity increases

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18
Q

What scale measures electronegativity

A

The Pauling scale. Fluorine→ most electronegative with a score of 4. After F its o→ 3.4

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19
Q

What does electronegativity tell about ionic and covalent characters

A

Bigger the difference in electronegativity more ionic is the compound. A difference of zero means its purely covalent

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20
Q

Define enthalpy change of solution

A

Enthalpy change when 1 mole of ionic compound is dissolved in minimum amount of solvent to ensure no further enthalpy change is observed upon further dilution

21
Q

What are the 2 conditions which needs to occur for a substance to dissolve

A

I- substance bonds must break (endothermic)
2-New bonds formed b/w solvent and substance(exothermic)

22
Q

How does ionic compound break when dissolved. What is hydration

A

Most ionic compounds dissolve in water. The delta +ve H is attracted to negative ions and delta -ve is attracted to positive ions. So the structure starts to break down when water molecules surround the ions in this process is called hydration

23
Q

What is required for hydration process to occur. Why do soluble substance have exothermic enthalpies?

A

The new bonds formed b/w water and substance ions must be the same strength or greater than those broken. If not then the substance is unlikely to dissolve. Soluble substance tends to have exothermic enthalpies of solution for this reason.

24
Q

Define enthalpy of hydration

A

Enthalpy change when 1 mole of aqueous ions is made from 1 mole of gaseous ions.

25
Q

What are the 2 things that can affect the enthalpy change of hydration

A

1- charge→higher the charge, water molecules attracted more strongly because of stronger electrostatic attraction→ more exothermic enthalpy of hydration - Larger the charge greater the enthalpy of hydration.
2-size→ smaller→higher change density and therefore attracts water molecules more strongly→ more exothermic enthalpy of hydration. Smaller the ion → Greater the enthalpy of hydration

26
Q

How to calculate H(solution)

A

It’s H(lattice dissociation) + sum of H(Hydration enthalpies)

27
Q

What is entropy(S)?

A

Measure of disorder in a system. it is the number of ways energy can be shared out between particles. More disorder→ higher the level of entropy

28
Q

In solids, liquids and gas which has the highest and lowest entropy and why?

A

Solids have the lowest level of disorder as particles are neatly arranged in rows→ lowest entropy. Liquids and then gases as there is lots of space so more random motion so therefore more disordered, so gas→ highest entropy

29
Q

What’s the other factor which increases entropy

A

Number of particles. If a reaction is the same state but more moles are produced then entropy increases. There are more ways energy can be distributed→More disorder. Eg→N2O4→2NO2 → 1 mole of reactants produces 2 moles of products → More no. Of particles → entropy increases

30
Q

Why are endothermic reactions enthalpically unfavourable

A

Because products have higher energy than reactants → not very common → therefore not enthalpically favourable

31
Q

Will reactions usually tend towards more entropy or less? And why

A

A reaction will tend towards more disorder and hence increases entropy. Increasing entropy is energetically favourable and some reactions that are enthalpically unfavourable (endothermic) can react spontaneously if changes in entropy overcome changes in enthalpy

32
Q

How does reaction between hydrated barium hydroxide and ammonium chloride show how entropy changes overcome enthalpy changes. Write the equation

A

Ba(OH)2.8H2O(s) + 2NH4Cl(s) → 2NH3(g) + 10H2O(l) + BaCl2(s)
2 solids are mixed and a gas, liquid and solid is produced.
1-This reaction is very endothermic and enthalpically not favourable.
2-Then 3 moles on LHS and 13 moles on RHS → entropically favourable
3-starting with 2 solids but making a gas and liquid. Increased disorder so entropically favourable

33
Q

How is entropy of system calculated and what’s it’s units

A

Change in S=Entropy(products) — Entropy(reactants)
Units - JK-1mol-1

34
Q

What are standard entropy values

A

1mole of substance
100kPa
298K

35
Q

How to know if reaction is entropically feasible(calculation)

A

If after the calculation of total change in entropy if the answer is positive then the reaction is entropically feasible

36
Q

How to calculate total entropy change

A

Total change in entropy → change in entropy (system) + change in entropy (surroundings)

37
Q

How to calculate change of entropy of the surroundings

A

—enthalpy change/ Temperature
Enthalpy change in Jmol-1 not KJ
Temperature in K
REMEMBER TO CONVERT TO CORRECT UNITS

38
Q

What does gibs free energy(G) tell us? How to calculate? What’s its units

A

If a reaction is feasible or not.
Triangle G = change in enthalpy(H) - Temperature(kelvin)* change in entropy(system)
Units is Jmol-1 so is the units of (H)

39
Q

What’s the basic rule of Gibbs free energy

A

A reaction is feasible in theory it (triangle)G is negative or zero

40
Q

Even after reaction is said to be feasible in theory why is no reaction observed

A

Activation energy being too high or the rate of reaction being very slow→ eg rusting of iron

41
Q

Hows the reaction feasibility affected when there’s a change in temperature and entropy

A

Endothermic and decrease in entropy - never feasible at any temp
Endothermic and increase in entropy - feasible only at high temp
Exothermic and decrease in entropy - feasible only at low temp
Exothermic and increase in entropy - always feasible at any temperature

42
Q

When is a reaction just feasible? How to calculate the temperature at which the reaction is just feasible?

A

A reaction is just feasible when (delta)G<0
(Delta)H- T(delta)S<0 a d solve to find t
Or use delta G =0

43
Q

Why is it better to use less energy

A

Less fossil fuels burnt therefore good for environment
Less cost

44
Q

What happens to the equilibrium constants when (delta)G value changes

A

Equilibrium constants get larger when reversible reactions are feasible, so when (delta)G is -ve equilibrium constants > 1
When (delta)G is +ve, equilibrium constants < 1

45
Q

How’s the relationship between (delta)G and equilibrium constants show? What’s the formula

A

(Delta) G = -RT * lnK
K- equilibrium constant

46
Q

What’s an equilibrium constant

A

It’s the concentration ratio between reactants and products at equilibrium at a certain temperature

47
Q

How to calculate equilibrium constant from (delta)G

A

LnK = (delta)G/-RT
Then take inverse of ln to find k

48
Q

Which part of Gibbs’s free energy corresponds to equation of straight line

A

Y= delta G
m= -delta S
X= temperature
C= delta H

49
Q

What’s the condition of Gibbs’s free energy for reversible physical changes at const temperature eg melting and boiling

A

Delta G=0
Which means Delta S= Delta (H)/Temperature

Chemsitry Year 2 (2 decks)

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