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Chemistry A- Level > Thermodynamics. > Flashcards

Thermodynamics. Flashcards

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1
Q

Define Enthalpy change of formation and give an exampe equation that would measure this enthalpy change using sodium and chlorine.

A

The standard enthalpy change of formation of a compound is the energy transfered when 1 mole of the compound is formed from its elements under standard conditions with all reactats and products in their standard states.
Na(s)+0.5Cl2(g) into NaCl (s)

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2
Q

Define enthalpy of atomization.

A

The enthalpy of atmomization of an element is the enthalpy change when 1 mole of gaseous atoms is formed from the element in its standard state.

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3
Q

Give an equation where the enthalpy of atomization of sodium would be measured.

A

For metal: Na(s) into Na(g)

All relative to forming 1 mole of atmoms.

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4
Q

Give an equation where the enthalpy of atomisation of oxygen would be measured.

A

For diatomic: 0.5O2(g) goes into O(g)

All relative to forming 1 mole of atmoms.

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5
Q

What will the enthalpy change where a solid metal turns to gaseous atoms be called and what is it numerically the same as.

A

It is called the enthalpy of sublimation and will be numerically the same as the enthalpy of atomization for that metal. Same thing is happening going from solid metal atom to gaseous metal atom as enthalpy of atomozation requites 1 mole of gaseous atom by definition.

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6
Q

Give an equation for the enthalpy of sublimation of sodium.

A

Na(s) into Na(s).

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7
Q

What is the relationship between the bond dissociation enthalpy of the diatiomic molocule and the enthalpy of atomization of that same element.

A

The bond dissocation enthalpy is twice that of the enthalpy of atomization. Since breaking one mole of covalent bonds will produce two moles not one of gaseous atoms.

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8
Q

Give equations for the enthalpy of atomization and bond dissociation enthlapy of chlorine using X to represent the enthalpy change of each.

A

Cl2(g) goes into 2Cl(g) will be 2X Kjmol for Diss

0.5Cl2 goes into Cl(g) will be X Kjmol for At

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9
Q

Define bond dissociation enthalpy.

A

The bond dissociation enthalpy is the standard molar enthalpy change when one mole of a covalent bond is broken into two gaseous atoms or FREE RADICALS.

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10
Q

Give an equation where bond dissociation enthalpy would be measured. For Chlorine and methane.

A

Cl2(g) goes into 2Cl(g).

CH4(g) goes into CH3(g) + H(g).

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11
Q

Define first ionsiation enthalpy.

A

The enthalpy change required to remove one mole of electrons from one mole of gseous atoms to form one mole of gaseous ions with a +1 charge.

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12
Q

Give an equation where the first ionaisation enthalpy would be measured.

A

Mg(g) goes into Mg+(g) + e- or

Cl(g) into Cl+(g) + e-

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13
Q

Define second ionaisaton enthalpy and give an equation where it would be measured.

A

The

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14
Q

Define first electron affinity.

A

The first electron affinity is the enthalpy change that occurs when 1 mole of gaseous atoms gain 1 mole of electrons to form one mole of gaseous ions with a 1- charge.

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15
Q

State why the first electron affinity is exothermic for atoms that usually form negative ions; a negative enthalpy change.

A

This is because the ion is more stable than the atom and there is an atraction between the nucleus and new electron and this is a bond forming process which gives off energy.

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16
Q

Define second electron affinity.

A

The second electron affinity is the enthalpy change when one mole of gaseous 1- ions gain one electron each to to produce 1 mole of gaseous 2- ions.

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17
Q

Explain why the second electron affinity for oxygen is exothermic , has a possitive energy change.

A

This is as for oxygen it takes energy to overcome the repulsive forces between the negative ion and the second electron that you are trying to add. Remember the oxygen ion is now negitavly chatged so will require energy to fuse the electron with it since they repel. This takes in energy as so the internal chemical energy of the system will increase and it is endothermic.

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18
Q

Give an equation for the first electron affinity of oxygen.

A

O(g) + e- goes into O-(g)

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19
Q

Give an equation for the second electron affinity of oxygen.

A

O-(g) +e- goes into O2- (g)

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20
Q

Define Enthalpy of lattice formation.

A

The standard enthalpy change when when mole of of an ionic crystal lattice is formed from its constituent IONS in gaseous from.

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21
Q

Define enthalpy of lattice dissociation.

A

The standard enthalpy change when one mole of solid ionic crystal lattice is seperated into its constituent ions in gaseous form.

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22
Q

Give an equation where the enthalpy of lattice dissociation would be measured.

A

NaCl(s) into Na+(g) + Cl-(g)

23
Q

Give an equation where the enthalpy of lattice formation would be measured.

A

Na+(g) + Cl-(g) into NaCl(s)

24
Q

Define the enthalpy of hydration and give two possible euations where this enthalpy change could be measured.

A

Enthalpy of hydration is the enthalpy change when one mole of gaseous ions become aqeous ions.
X+(g) + aq into X+(aq)
or the same equation with an anion is possible.

25
Q

Explain what type of reaction and why the enthalpy of hydration will be.

A

Will always be exothermic because ions are made between the ions and the water molocules. Bond forming will give out energy. Refer to “attractive forces between polar water molecules and the ion “.

26
Q

What is the numerical relationship between enthalpy of lattice disociation and enthalpy of lattice formation.
What do you need to do in exams.

A

They have the opposite sign but same magnitude.
DISS will always be endothermic.
FOR will always be exothermic.
Consider if you need to flip a sign if bonds are being broken or made.

27
Q

By describing the atractive forces involved explain why the enthalpy of hydration for the chloride ion is greter than that of the bromide ion.

A

They both have the same ionic charge. Yet chlorine has a lesser ionic radius than bromine does and so chlorine has a greater charge density. This higher charge density means that the polar water molocule will be more strongly atracted to the chloride ion than the bromide ion and so a stronger bond will form and more energy will be given off.

28
Q

Give the definition of enthalpy of solution.

A

The enthalpy of solution is the standrd enthalpy change when One mole of an ionic solid disolves in a large enough amount of water to ensure that the disolved ions are well seperated and do not interact with each other.

29
Q

Give an equation where the enthalpy of soution would be measured.

A

NaCl(s) + aq goes into Na+(aq) and Cl-(aq)

30
Q

How do chemists know what the lattice enthalpy is.

A

The lattice enthalpy cannot be determined directly. We calculate it indirectly by making use of changes forwhich data are available and link them together in an enthalpy cycle the Born-Habercycle

31
Q

Draw and label a Born-Haber Cycle for sodium Chloride.

And state how you would use the quoted data values.

A

Can just put the data in as it is. SEE PIC

32
Q

How could data be given given in a wierd way for your Born Haber cycle and how would you deal with this.

A

The data for theatH(Cl) could also be given as the bond energy for E(Cl-Cl ) bond. Remember :E(Cl-Cl ) = 2 xatH(Cl)

33
Q

Construct a labeled Born Haber cycle for Magnesium Chloride.

A

SEE PIC

34
Q

What adaptations would you need to make to your Born Haber Cycle for MgCl2 or any compound that doesnt have equal charges in this form of XY2.

A

The metal will need to have 2 IE and so 1 and then 2 electrons will be required to make sure the charges balance as the cation gains charges going up.
The enthalpy of atomization
and the electron affinity
will be double as you will be forming two product moles not 1 so its double the definition.

35
Q

Construct a Born-Haber cycle for Calcium oxide.

A

SEE PIC

36
Q

What is a give away that a Haber cycle will need multiple electrons , a doubled enthalpy of atomization and a doubled electron affinity of quoted data or
that the data can go streight in.

A

If the ions have matching charges the data can go in if it has non matching charges the former is required.
This will mean that from the formation step the diatomic anion will not be balanced bye a half to meet the definitions later up the halbert cycle.

37
Q

Why is the second electron afinity for oxygen endothermic and so what does this mean for the direction of its arrow in a Born-Haber cycle.

A

Notice the second electron affinity for oxygen is endothermic because it takes energy to overcome the repulsive force between the negative ion and the electron.
This means that the arrow will point Upwards inlike the first electron affinity which goes down.

38
Q

Why is the first electron affinity an exothermic process.

A

The first electron affinity is exothermic for atoms that normally form negativeions. This isbecause the ion is more stable than the atom,and there is an attraction between the nucleus and the electron.

39
Q

What two factors does the strength of lattice formation depend on. Explain the effect that thease factors have.

A

The strength of a enthalpy of lattice formation depends on the followingfactors:1.
The sizes of the ions . The larger the ions, the less negative the enthalpies of lattice formation(i.e. a weaker lattice). As the ions are larger the charges become further apart and so have a weaker attractive force between them.
2. The charges on the ionThe bigger the charge of the ion, the greater the attraction between the ions so the stronger the lattice enthalpy (more negative values).

40
Q

When thinking about comparing the lattice enthalpies of different compounds and then comparing theretical and experimental values what must you not confuse.

A

Covalent charicter and polarisation is only relevent when considering comparisons between the theoretical and experimental values of the same compound.
All you need to worry about for comparing the same compounds is greater charge density means more negative lattice enthalpies.

41
Q

Define the perfect ionic model.

A

Theoretical lattice enthalpies assume a perfect ionic model where the ions are 100 percent ionic and spherical and the atractions are purely electrostatic.

42
Q

Give four features that will creat a tendency for covalent charicter in Ionic substances. Explain with the aid of a diagram what is polarising what.

A

•the positive ion is small•
the positive ion has multiple charges•
the negative ion is large•
the negative ion has multiple negative charges

The small nippy positive ion will distort the big fat charge cloud of the negative ion.

43
Q

Explain the reason that a Bohr-Haber value for lattice enthalpy would be larger than a theoretical value. Refer to covalency.

A

BORN HABER LARGER (rhymes). The theoretical value assumes a perfect ionic model. When a compound has some covalent character-it tends towards giant covalent so the lattice is stronger than if it where 100% ionic and had purely electrostatic interactions. Therefore the Born-Haber value would be larger than the theoretical value.

44
Q

Define Polarisation.

A

When the metal ion distorts the charge cloud of the anion by atracting it towords itself. making it more covalent. The metal Ion is said to be polarising if it polarised the negative anion.

45
Q

When will the Born Haber experimental and theoretical lattice enthalpies be the same.

A

When the ions are 100 percent ionic and spherical.

The greater the polarisation and covalent character the greater the difference in the two values.

46
Q

What determins the magnitude of the diffrence in experimental and theoretical lattice enthalpies.

A

The more the covalent character the biggerthe difference between the values.

47
Q

Define enthalpy change.

A

Definition: Enthalpy change is the amount of heat energy taken in or given out during any change in a system provided the pressure is constant.

48
Q

Define mean bond enthalpy.

A

Definition: The mean bond energy is the enthalpy needed to break one mole of covalent bond into two gaseous atoms, averaged over different molecules.

49
Q

Why are enthalpy change of reactions calculated from mean bond enthalpies less accurate than those calculated from combustion and formation data. Draw a Hess cycle to illustrate the process above for mean enthalpies.

A

Delta H values calculated using this method will be less accurate than using formation or combustion data because the mean bond energies are not exact.

50
Q

Explain in terms of the strength of bonds why a reaction would be an exothermic process.

A

In an exothermic reaction the sum of the bonds in the reactant molecules will be less than the sum of the bonds in the product molecules.

51
Q

What is used for the bond enthalpy of the C-H bond for a hydrocarbon.

A

We use an average value for the C-H bond for all hydrocarbons.

52
Q

Define specific heat capacity.

A

Amount of energy required to raise the temperature of 1 gram of substance by 1K.

53
Q

What does a discrepancy in experimental and theoretical attic enthalpies imply , explain why there would be a large discrepancy between ZnSe and less between CsF , and state why in principle there would be no difference between two atoms of the same atom bonded together.

A

Discrepancies occur due to covalent character.
The negative anion in a lattice will be polarised by the positive cation. Negative is what gets distorted since not is the electron density that is distorted.
More negatively charged anions are easy to distort because there is more negative charge to distort. Small nippy cations with higher charge will distort the anion to a greater extent. Degree of electron sharing causes discrepancy.
The Zn and Se ion are more highly charged and the Zn is small and the Se ion is big. More than expected covalent character.
The Cs ion is large and the F ion is small. They are both singly charged , so the F is less easily polarises than the Se by the relative anions.

High Charges on both ions
Small nippy cation
Big fat anion means more polarisation, greater discrepancy.

54
Q

The reaction with graphite and oxygen to produce carbon dioxide is spontaneous why don’t you see loads of carbon dioxide coming off your pencil in lessons. State what this says about thermodynamic stability and kinetic stability.

A

The activation energy is too high, so the reaction happens so slowly that in practice it doesn’t take place.

Chemistry A- Level (18 decks)

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