Thermodynamics, Rate Equations

Question 1

What are the orders for concentration’s influence on rate of reaction?

Answer 1

if no effect —>
Zero order A^0

if rate doubles when concentration doubles (direct proportion) —>
First Order A^1

if rate quadruples when concentration doubles —>
Second Order A^2

Question 2

The rate constant…

Answer 2

• Different for every reaction
• Varies with temperature
• units are different for every reaction

Question 3

What is Enthalpy of Lattice Formation?

Answer 3

Enthalpy change when 1 mole of ionic lattice/solid ionic compound is formed from its gaseous ions.

Question 4

Why may the calculation of enthalpy for lattice formation based on the perfect ionic model give a smaller numerical value than the calculated value? e.g in the case of AgI

Answer 4

The Covalent character of the molecule

They may not be completely ionic.

the forces/bonds holding the matrix together are stronger.

Question 5

Enthalpy change of formation

Answer 5

The enthalpu change when 1 mole of a compound is formed from it’s elements in their standard states under standard conditions.
(2Cs + 2H2aq –> C2H6aq)

Question 6

Lattice enthalpy of formation

Answer 6

The enthalpy change when 1 mole of a solid ionic compound is formed from it’s gaseous ions under standard conditions.
Ca2+g + 2Cl-g –> CaCl2s
- exothermic

Question 7

Lattice enthalpy of dissociation

Answer 7

The enthalpy change when 1 MOLE of solid ionic compound is dissociated into it’s gaseous ions under standard conditions.
CaCl2s –> Ca2+ + 2Cl-g

Question 8

Enthalpy change of dissociation

Answer 8

The enthalpy when 1 mole of bonds of the same type of molecule in the gaseous state is broken
F2g –> 2Fg

Question 9

Enthalpy change of 1st ionisation energy

Answer 9

The enthalpy change of when 1 mole of gaseous 1+ ions are made from 1 mole of gaseous atoms
- endothermic

Question 10

Enthalpy change of atomisation

Answer 10

The enthalpy change when 1 mole of gaseous atoms are made from an element in it’s standard state.
1/2F2(g) –> F(g)
- endothermic

Question 11

Enthalpy change of 2nd ionisation

Answer 11

The enthalpy change when 1 mole of gaseous 2+ ions are made from 1 mole of gaseous 1+ ions.
Ca+(g) –> Ca2+
- endothermic

Question 12

1st electron affinity

Answer 12

The enthalpy change when 1 mole of gaseous atoms form 1 mole of gaseous 1- ions.
e.g. O(g) –> O-(g)
- exothermic

Question 13

Why do we use Born Haber cycles?

Answer 13

Born-Haber cycles are useful to calculate lattice enthalpies. This is because we can’t calculate directly from experiments.

Question 14

2nd electron affinity

Answer 14

• The enthalpy change when 1 mole of gaseous 1- ions form 1 mole of gaseous 2- ions.
  e.g. O-(g) –> O2-(g)
  -endothermic

Question 15

Why may the theoretical Lattice enthalpies and experiment Lattice enthalpies differ?

Answer 15

• The covalent character of the ionic compound ( the amount at which the positive ions polarise the negative ions)
• can result in greater experimental values for lattice enthalpy than theoretical ( when there is high covalent character)
• The larger the ion the greater the distortion

Question 16

Theoretical Lattice Enthalpies and Experimental Lattice Enthalpies

Answer 16

• Theoretical lattice enthalpies can be calculated from data assuming a perfectly ionic model.
• during experimental testing you will find that you will not get the exact value you get from the theoretical lattice enthalpies
• This means that the compound doesn’t follow the perfectly ionic model and has some covalent characteristics.
• Most of the time the positive ion distorts the charge distribution in the negative ion. ( The positive ion polarises the negative ion + the more polarisation the greater the covalent character)

Question 17

Enthalpy change of solution

Answer 17

The enthalpy change when 1 mole of an ionic substance is dissolved in the minimum amount of solvent to ensure NO FURTHER enthalpy change is observed upon further dilution.

Question 18

For a Substance to dissolve… (enthalpy change of solution)

Answer 18

1. substance bonds must break (endothermic)
2. New bonds formed between the solvent and substance (exothermic)

Question 19

Why may an ionic compound not dissolve?

Answer 19

For an ionic compound to dissolve the new bonds formed during the hydration of the freely moving ions must be equivalent or greater in strength than the ionic bonds broken.
- If not then substance is unlikely to dissolve

Question 20

How do you calculate the enthalpy change of solution

Answer 20

1. using lattice dissociation enthalpy
2. using enthalpy of hydration
   (use the hess’ law structure)
   The sum of the two is equal to the enthalpy of solution.

Question 21

Enthalpy of Hydration

Answer 21

• Enthalpy change when 1 mole of aqueous ions is made from 1 mole of gaseous ions. / Enthalpy change when 1 mole of gaseous ions form 1 mole of aqueous ions.
  (exothermic)

Question 22

Define Entropy

Answer 22

Entropy is the measure of disorder in a system.
The more disorder there is the higher the level of entropy

Question 23

things to remember when calculating gibbs free energy

Answer 23

• when calculating D S make sure it’s S = products - reactants and that you include balancing numbers in calculation
• making sure you convert your entropy value from JK-1mol-1 to KJK-1mol-1 ( or you can convert your enthalpy to KJmol-1 but former is easier)
• making sure your temperature is in KELVIN ( it’s not temp change so it matters that it’s kelvin)

Question 24

Why may endothermic reactions still be spontaneous?

Answer 24

• a calculated positive entropy can tell you a reaction is feasible
• increasing entropy is energetically favourable and some reactions that are enthalpically unfavourable (endothermic) can still spontaneously (feasibly) react if changes in entropy overcome changes in enthalpy

Question 25

Why may a reaction be theoretically feasible but not have a spontaneous reaction?

Answer 25

• the reaction may be feasible but not spontaneous
• This could mean it has a high activation energy or it’s an incredibly slow reaction.

Question 26

When is a reaction feasible?

Answer 26

When Delta G </ 0
- Temperature effects feasibility of a reaction
- this is dependent on whether delta H or delta S is positive or negative

Question 27

Why does the rate of reaction change when temperature changes, explain in terms of the rate constant.

Answer 27

• the rate constant is fixed at particular temperatures. If the temperature changes the rate constant also changes.
• As we increase the temperature the particles have more energy and they collide more often. This increases the rate.

Question 28

when calculating the rate constant what is important to remember?

Answer 28

• finding the orders of reaction
• calculating the units using the concentrations.

Question 29

rate equations

Answer 29

• rate equations can only be constructed by conducting an experiment.
• ## the initial rates can be used to work out the rate equation for a reaction.

Question 30

Iodine Clock Experiment

Answer 30

H2O2(aq) + 2H+(aq) + 2I- –> 2H2O(l) + I2(aq)
- in this practical you are looking for colour changes
- add sodium thiosulfate and starch (which acts as an indicator)
- the sodium thiosulfate reacts immediatley with the I2
- when there is no more sodium thiosulfate the I2 reacts with starch, resulting in a deep blue/black colour.

Question 31

Other methods of calculating the rate of an experiment

Question 32

Colorimeter

Answer 32

• measures the absorbance of light by a coloured sample. The more concentrated a sample is, the darker it’s colour and hence the more light absorbed.
• good for any reaction with a colour change.

Question 33

Rate-Concentration Graphs VS Concentration-Time Graphs

Answer 33

Rate-Concentration graphs
- help us identify the order
- the rate on a STRAIGHT HORIZONTAL LINE GRAPH is constant so the reactants are ZERO ORDER. (the change in concentration is not affecting the rate)
- the r ate on a straight DIAGONAL line shows a proportional relationship so FIRST ORDER.
- the rate on a curved line shows shows a SECOND ORDER reactant(s)

CONCENTRATION-TIME
- concentration on a straight diagonal line shows the reactant to be ZERO ORDER
- the concentration on a curved but moderately steep line is FIRST ORDER (more working out will likely be needed)
- the concentration on a STEEPLY curved line would be SECOND ORDER.

Question 34

Rate-determining step

Answer 34

the SLOWEST step in a MULTI-STEP reaction.
- speeding up the slowest step will increase the whole reaction rate.
- substances not in the rate equation won’t be in the rate determining step. THUS all substances in the rate determining step are IN the rate equation.
- catalysts can also be within rate equations
- for rate determining step reactants their rate equation power is their balancing number.
- The reactants that form the reactants in the rate determining step must also be apart of the rate equation.

Question 35

How can we determine the likeliness of a mechanism using the rate determining step?

Answer 35

• when determining which mechanism is more likely to occur for a reaction you can look at the rate equation.
• If certain molecules/reactants that are within the rate equation (including their balancing numbers shown as powers) show up in one of the mechanisms then that mechanism is more likely to occur.

Question 36

Arrhenius Equation

Answer 36

• can be used to find activation when plotting