Solutions and Mixtures Flashcards

1
Q

What is a phase?

A

A region with uniform chemical composition and physical properties

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
2
Q

What is a phase diagram?

A

A diagram that shows the most stable phase under different thermodynamic conditions

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
3
Q

What is the critical point in a phase diagram?

A

The point on the phase diagram where the liquid and vapour phases of a substance merge into a single phase

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
4
Q

What is the supercritical region of a phase diagram?

A

The region of a phase diagram where the temperature and pressure is higher than the liquid-vapour phase diagram

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
5
Q

What degree of freedom does the triple point have?

A
  • F (Degrees of freedom) = 0
  • This means that only one set of temperature and pressure values are allowed, if you change temperature and/or pressure, you’re no longer at the triple point
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
6
Q

What degree of freedom does the boundary between two phases have?

A
  • F = 1

- If you change T then P must change accordingly to stay on the same line

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
7
Q

What degree of freedom does one phase region have, either liquid solid or vapour?

A
  • F = 2

- You can change T and P independently and still be in the same area

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
8
Q

What is the Gibbs Phase rule?

A
  • F = C-P+2

- Where F is the degrees of freedom, C is the number of chemical components and P is the number of phases

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
9
Q

How is the Gibbs phase rule derived?

A
  • 2 = the maximum degrees of freedom
  • CP = in each phase there are C components
  • -P = Total mole fraction for each phase = 1
  • -C(P-1) = Equilibrium condition for each component, subtract P-1 degrees of freedom x C components
  • Add together, 2 + CP - P -C(P-1)
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
10
Q

How do you determine the boiling point for a mixture from a phase diagram?

A

At the given composition of the mixture draw a line up until you hit the bottom of the l+v region

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
11
Q

How do you estimate the composition of the vapour that boils off from a mixture of a certain composition?

A

Find the boiling point and draw a line from the liquid line of the l+v region to the vapour line, then draw a line down from the vapour line to get the composition of the vapour

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
12
Q

What is an azeotropic mixture?

A

Mixtures of liquids where the liquid and vapour line merge into one and the mixture now has a constant boiling point

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
13
Q

What degrees of freedom does the dew line of a phase diagram of a mixture have?

A

2, it has 2 phases (liquid and vapour) and 2 components assuming its a mixture of 2 components, 2-2+2 = 2

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
14
Q

Why is the degree of freedom on a dew line 2, when the rule is that on the line degrees of freedom is 1?

A
  • In the phase diagram of a mixture, pressure isn’t included in the graph, we can see from the graph that you can’t change temperature and mole fraction is independent so we would assume from the graph that degrees of freedom is 1
  • Because the calculated value is 2, the extra degree of freedom must be accounting for something else and in this case it is accounting for the phase diagram having constant pressure
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
15
Q

What is a liquid-liquid miscibility diagram?

A
  • A diagram that has a curve where one side is a solvent rich phase and one side is a solute rich phase
  • Between these two sides, the composition separates into two phases
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
16
Q

What is the lever rule?

A
  • From your starting composition, trace up to the given temperature, you will have a distance to (A) rich phase and (B) rich phase
  • The (A) rich distance is the distance from your starting composition to the (B) rich phase
  • The (B) rich distance is the distance from your starting composition to the (A) rich phase
  • If the question asks for the ratio of (A) rich phase to (B) rich phase then the equation is: (A) rich phase / (B) rich phase
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
17
Q

What is the upper critical solution temperature?

A

The temperature above which when the mixture passes, the phases combine, no matter what the starting composition of the mixture is

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
18
Q

Why is a eutectic mixture good for solubility?

A

The eutectic point is the composition of the mixture where melting point is at its

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
19
Q

What is the Clapeyron equation?

A
  • dP/dT = Δs/Δv = ∆h / T∆v, ∆s can be replaced with ∆h/T as on the slope ∆G = 0 as it is the equilibrium between 2 phases, therefore 0 = ∆h - T∆s, T∆s = ∆h, ∆s = ∆h/T
  • Where dP/dT is the gradient of the slope separating two phases
  • ∆s is the molar entropy change
  • ∆v is the molar volume change
  • ∆h is the molar entropy change
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
20
Q

When a dP/dT slope is negative, what does this tell us about melting point as pressure increases?

A
  • As dP/dT is negative, we know that dT/dP is also negative

- This means that as you increase pressure, melting point also decreases

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
21
Q

What is the Clausius-Clapeyron equation?

A
  • dP/dT = P∆s / RT = P∆h
  • The phase above the slope is liquid/solid, the phase below the slope is gas
  • ∆s is the molar entropy change
  • ∆v is the molar volume change
  • ∆h is the molar entropy change, ∆h(gas) - ∆h(liquid)
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
22
Q

In the Clausius-Clapeyron equation, ∆v is the molar volume change, gas volume - (liquid/solid volume). Why can we take ∆v as simply the gas volume?

A

The liquid/solid volume is much smaller than the gas volume, hence can be omitted and we can use the gas volume as ∆v

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
23
Q

Why does decane have a higher ∆h than hexane?

A
  • Decane is a larger molecule, this leads to stronger Van der Waals interactions h(liquid) of decane is therefore lower than that of hexane
  • The h(gas) of both molecules is similar meaning that ∆h is larger for decane as there is larger difference in enthalpy between the liquid and vapour forms of decane
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
24
Q

Why is it impossible for a liquid-vapour boundary phase to be negative?

A
  • Negative slope means negative ∆h
  • ∆h = h(gas) - h(liquid) < 0
  • hence h(gas) < h(liquid) which is impossible because gases have more energy than liquids
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
25
Q

What is the linearised Clausius-Clapeyron equation?

A

dlnP/d(1/T) = - ∆h/R

26
Q

When using the linearised Clausius-Clapeyron equation, how do you find dlnP and d(1/T) from a graph?

A
  • To find these values, pick two points on the slope (x1,y1) and (x2,y2)
  • dlnP is y1 - y2 and d(1/T) is x1 - x2
27
Q

What is the calculation for activity (a1 & a2)?

A
  • Activity (a1 & a2) is a measure of affinities of solute and solvent to solution compared to their pure systems
  • a1 = p1/pº1, a2 = p2/pº2 where p1 and p2 are the partial vapour pressures of the solution and pº1 and pº2 are the vapour pressures of the pure system
28
Q

What does a high activity mean?

A

The solute/solvent vaporises more when combined, compared with when in its pure phase meaning it has a weaker affinity for being in solution when combined compared to being in solution in its pure phase

29
Q

What does a low activity mean?

A

The solute/solvent vaporises less when combined, compared with when in its pure phase meaning it has a stronger affinity for being in solution when combined compared to being in solution in its pure phase

30
Q

What is the activity coefficient?

A
  • γi = ai/xi
  • Where γi is the activity coefficient for one of the components of the solution, ai is the actual activity for one of the components and xi is the mole fraction
  • Activity coefficient tells us how much more affinity a species has for being in solution compared with what we expect, coefficient of < 1 means it has a higher affinity for solution as it has a lower activity than expected
  • If xi (mole fraction) is larger than ai (activity) then the solution has a lower activity than expected
31
Q

What does an activity coefficient of < 1 tell us?

A
  • The solvent/solvent has a stronger affinity for being in solution when combined than when in its pure phase as the actual activity must be lower than mole fraction to give an activity coefficient of < 1
  • P1 ≤ (x1)(Pº1), this equation is telling us that the actual partial pressure of the solute/solvent in the mixture is less than the expected partial pressure
  • a1 ≤ x1 meaning that the actual activity is lower than the mole fraction meaning the actual activity is lower than the expected activity
32
Q

What does the activity multiplied by the partial pressure of solvent/solute in its pure system tell you?

A

The partial pressure of the solvent/solute in the mixture

33
Q

If the partial pressure of the solvent/solute in the mixture is greater than expected, what does this tell us about the activity?

A

The activity will be greater than the activity we expect from that mole fraction in an ideal solution, this means the activity coefficient will be greater than 1

34
Q

What is the value of the activity coefficient in an ideal mixture?

A

1

35
Q

What is the relationship between the activity and the expected activity, when the solution is ideal?

A

ai = xi, as γi =1

36
Q

What is the relationship between a1 and a2 in a perfect solution?

A
  • a1+a2 = 1

- x2 = 1-x1

37
Q

What is the expected activity of a solvent/solute equal to?

A

The mole fraction of that solute/solvent, as the expected activity is the activity calculated from the mixture if it was an ideal solution and we know that in an ideal solution, the activity = mole fraction

38
Q

What is the expected activity of a solvent/solute?

A

The activity calculated assuming the mixture is an ideal solution

39
Q

What is a perfect solution?

A

A solution that is ideal at all concentrations, a solution is likely to be perfect if the solvent is very similar to the solute

40
Q

What is the reason for a solvent/solute having a higher activity than expected?

A

A solute/solvent has a higher activity than expected because it has weaker intermolecular interactions in the solution than in the ideal mixture

41
Q

If partial pressure increases linearly with mole fraction, what does this tell us about the solution?

A

It’s ideal as Pi = Pºi(xi)

42
Q

What is a positive deviation from ideality?

A

When γ > 1 (weaker affinity to solution)

43
Q

What is a negative deviation from ideality?

A

When γ < 1 (stronger affinity to solution)

44
Q

When a solvent/solution has a negative deviation from ideality, are the intermolecular forces in the solution stronger or weaker than the ideal mixture?

A

Stronger as a negative deviation from ideality means γ < 1 which means the solvent/solute has a stronger affinity to the solution than the ideal mixture as it evaporates less in the actual solution

45
Q

What is Raoult’s Law?

A
  • When the solute (x2
46
Q

When calculating the mole fraction of O2 dissolved in water at a pressure of 1 atm, what do you need to multiply the pressure by?

A

0.2 as this is the partial pressure of oxygen in air and we have the pressure of air

47
Q

What is Henry’s Law?

A
  • When the solute is dilute, the mole fraction (x2) is proportional to the partial pressure of the solute P2 in the gas phase
  • P2= kH(x2) where kH is the Henry’s Law constant
48
Q

How do you calculate the pressure at depth X in water?

A

Pressure = (Density of liquid x Gravitational acceleration x Depth of the water) + Pressure of the air above the water (1atm)

49
Q

When calculating the number of moles of a solute released from a solvent when undergoing a pressure change what equation should you use?

A
  • ∆x2 = ∆P2/ kH
  • Change in the mole fraction = change in pressure / Henry’s constant
  • You then need to calculate the moles of the solvent that the solute is being released from, you then multiply your calculated mole fraction by the moles of the solvent, this gives us the moles of solute released
50
Q

What is the boiling point elevation equation?

A
  • ∆Tb = Kb(m2)
  • Where ∆Tb is Tb(solution) - Tb(pure solvent), Kb is the ebullioscopic constant and m2 is the solute molality (molkg^-1)
51
Q

When calculating a boiling point elevation, what units is the elevation in?

A

Because it is a change in boiling point, it doesn’t matter if you use Kelvin or Celcius as the difference between 2 values in Kelvin is equal to the change in the values if they’re converted into Celcius

52
Q

When calculating the molality of NaCl dissolved in water, what must you do to the value for the molality to convert it to the amount of NaCl salt?

A

Divide your molality by 2 as for every 2 moles of NaCl in water there is 1 mol of NaCl salt

53
Q

What is the freezing point depression equation?

A
  • ∆Tf = -Kf(m2)

- Where ∆Tf is the freezing point depression, m2 is solute molality and Kf is the cryoscopic constant

54
Q

What is molality?

A

The moles of solute in a kg of solvent

55
Q

What is molarity?

A

Moles of solute in a litre of solution

56
Q

What is van Hoff’s law

A
  • Osmotic pressure = RT(c2)
  • Where Osmotic pressure is the minimum pressure to prevent influx of solvent molecules, R is the gas constant, T is temperature and c2 is molarity
57
Q

How do you calculate the grams of a dilute solute from the wt% ?

A

Multiply the wt% as a a decimal by the mass of the solvent

58
Q

What assumption can we make about molality when the solution is dilute and water based?

A

The molarity is approximately equal to molality

59
Q

How do you convert molarity to molality for non-dilute solutions?

A
  • Molarity is the moles of solute in a litre of solution, you can multiply molarity by molar mass of the solute to get the grams of solute in a litre of solution
  • You can use the density of the solution to calculate the mass of 1 litre of solution by multiplying by 1000
  • You now have the mass of 1 litre of solution and the mass of the solute in 1 litre of solution, if you do (mass of 1 litre of solution) -(mass of solute in 1 litre of solution) you get the mass of the solvent in 1 litre of solution
  • If you divide the original molarity (the moles of solute in 1 litre of solution) by the mass (kg) of the solvent in 1 litre of solution, you get the molality (moles of solute per kg of solvent)
60
Q

How do you convert molality to molarity?

A
  • You have the molality, this is the moles of solute in a kg of solvent, you can work out the grams of solute in a kg of solvent by multiplying by the molar mass of the solute
  • Now you calculate the total grams of the solution by adding the mass of the solvent to the mass of solute that is contained in that particular mass of solvent, eg. if you have 400 grams of solute per kg of solvent and the mass of solvent is 2kg, you would add 800 grams to the mass of the solvent
  • You then divide the total grams of solution by the density of the solution to obtain the volume of the solution
  • You then divide the original molality by the volume of the solution, this is moles of solute (in 1kg of solvent) / volume of solution (containing 1kg of solvent)
  • This gives you the molarity (moles of solute per litre of solution)