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A LEVEL - Chemistry (Inorganic) > Topic 11 - Equilibrium II > Flashcards

Topic 11 - Equilibrium II Flashcards

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1
Q

What is the equation for Kc?

For aA + bB -> dD + eE

A

Kc = [D]^d x [E]^e / [A]^a x [B]^b

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2
Q

In general, how could the equation for Kc be expressed?

A

The ratio of product concentration to reactant concentration.

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3
Q

When does the Kc equation apply?

A

At equilibrium.

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4
Q

What do you need to know to work out Kc?

A

The molar concentration of each substance at equilibrium.

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5
Q

What are the conditions for a Kc value?

A
  • Quoted at a certain temperature

* Molar concentrations are at equilibrium

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6
Q

Is Kc for a reaction the same at any temperature?

A

No, it is quoted at a certain temperature.

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7
Q

What is the equation for Kc for this reaction:

H2(g) + I2(g) -> 2HI(g)

A

Kc = [HI]² / [H2] x [I2]

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8
Q

What chemicals are included in a Kc equation?

A

It depends on whether the reaction is in homogeneous or heterogeneous equilibrium.

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9
Q

What is a homogeneous equilibrium?

A

One where all the products and reactants are in the same phase.

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10
Q

What is a heterogeneous equilibrium?

A

One where the products and/or reactants are not all in the same phase.

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11
Q

What is included in the Kc equation for a homogeneous equilibrium?

A

Everything.

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12
Q

What is included in the Kc equation for a heterogeneous equilibrium?

A

Everything except solids and pure liquids.

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13
Q

Why are solids and pure liquids not included in the equation for Kc in a heterogeneous equilibrium?

A

Their concentrations stay constant throughout the reaction.

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14
Q

Describe how you work out Kc when given the molar quantities before the reaction and at equilibrium.

A

1) Work out the number of moles of each product and reactant at equilibrium (by looking at the ratios in the equation).
2) Work out the concentration of each (usually by looking at the volume of the container).
3) Choose the appropriate chemicals and put their concentrations into the Kc equation.
4) Find the units of Kc and quote.

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15
Q

How can you work out the units for Kc?

A

Divide the units of everything on top of the Kc equation by everything on the bottom.

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16
Q

0.20 moles of phosphorus(V) chloride decomposes at 600K in a vessel of 5.00dm³. The equilibrium mixture is found to contain 0.080 moles of chlorine. Write the expression for Kc and calculate its value including units.

PCl5(g) -> PCl3(g) + Cl2(g)

A
  • The equation tells you that when 1 mole of PCl5 decomposes, 1 mole of PCl3 and 1 mole of Cl2 are formed. So if 0.080 moles of chlorine are produced at equilibrium, then there will be 0.080 moles of PCl3 as well. 0.080 moles of PCl5 must have decomposed, so there will be (0.20 - 0.080 =) 0.12 moles left
  • [PCl3] = [Cl2] = 0.08 / 5.00 = 0.016 mol/dm³
  • [PCl5] = 0.12 / 5.00 = 0.024 mol/dm³
  • Kc = [PCl3][Cl2] / [PCl5] = 0.016 x 0.016 / 0.024 = 0.011
  • Units = (mol/dm³)² / mol/dm³ = mol/dm³
  • So Kc = 0.011 mol/dm³
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17
Q

Is Kc for a reaction the same at any temperature?

A

No, it is quoted at a given temperature.

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18
Q

What is Kc shorthand for?

A
  • Equilibrium constant
  • It is not the only equilibrium constant though, so it is best thought of as the “equilibrium constant in terms of concentrations”
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19
Q

Describe how you work out the concentration of a reactant or product in an equilibrium when you are given Kc and the concentrations of the other reactants.

A

1) Write the equation for Kc.
2) Plug in the value for Kc and all the concentrations.
3) Rearrange to give the necessary concentration.

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20
Q

When the reaction between ethanoic acid and ethanol was allowed to reach equilibrium at 25C, it was found that the equilibrium mixture contained 2.0 mol/dm³ ethanoic acid and 3.5 mol/dm³ ethanol. Kc of the equilibrium is 4.0 at 25C. What are the concentrations of the other components?

CH3COOH(l) + C2H5OH(l) -> CH3COOC2H(l) + H2O (l)

A
  • Kc = [CH3COOC2H][H2O] / [CH3COOH][C2H5OH]
  • 4.0 = [CH3COOC2H][H2O] / (2.0 x 3.5)
  • 28.0 = [CH3COOC2H][H2O]
  • Since [CH3COOC2H] = [H2O], [CH3COOC2H] and [H2O] = sqrt(28) = 5.3 mol/dm³
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21
Q

Describe how you could work out Kc at 298K for this reaction:

Fe2+(aq) + Ag+(aq) -> Fe3+(aq) + Ag(s)

A

1) Leave a mixture of iron(II) sulfate solution and silver nitrate solution in a stoppered flask at 298K.
2) Take samples from the equilibrium mixture and titrate them.
3) Using this, work out the equilibrium concentration of Fe2+ ions.
4) From this, work out concentrations of the other components, and so Kc.

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22
Q

500cm³ of 0.100 mol/dm³ iron(II) sulfate solution and 500cm³ of 0.100 mol/dm³ silver nitrate solution are placed in a stoppered flask and allowed to reach equilibrium. It’s found that the equilibrium concentration of Fe2+ is 0.0439 mol/dm³ under s.t.p. Calculate Kc for this reaction at s.t.p.

Fe2+(aq) + Ag+(aq) -> Fe3+(aq) + Ag(s)

A
  • The equation tells us that 1 mole of Fe2+ reacts with 1 mole of Ag+ to form 1 mole of Fe3+ and 1 mole of Ag.
  • The silver formed is solid, so you don’t need to include it in the expression for Kc.
  • Starting concentrations of Ag+ and Fe2+ are the same and equal to 0.500 mol/dm³.
  • The equilibrium concentration of Ag+ will be the same as Fe2+, i.e. 0.0439 mol/dm3.
  • The equilibrium concentration of Fe3+ will be 0.0500 - 0.0439 = 0.0061 mol/dm³
  • Kc = [Fe3+] / [Fe2+][Ag+] = 0.0061 / (0.0439 x 0.0439) = 3.17
  • Units = mol/dm³ / (mol/dm³ x mol/dm³) = mol^-1 dm³
  • So at s.t.p., Kc = 3.17 mol^-1 dm³
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23
Q

What is the partial pressure of a gas?

A

The pressure that a gas in a mixture would exert if it were on its own.

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24
Q

How are the total pressure and partial pressures related?

A

The total pressure is the sum of all the partial pressures of the individual gases.

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25
Q

How is the partial pressure of a gas symbolised?

A

p(X) where X is the gas.

There may be no brackets sometimes.

26
Q

When 3.0 moles of the gas PCl5 is heated, it decomposes into PCl3 and Cl2. In a sealed vessel at 500K, the equilibrium mixture contains chlorine with a partial pressure of 2.6atm. If the total pressure of the mixture is 7.0atm, what is the partial pressure of PCl5?

PCl5(g) -> PCl3(g) + Cl2(g)

A
  • From the equation, you know that PCl3 and Cl2 are produced in equal amounts, so the partial pressure of these two gases are the same at equilibrium - they’re both 2.6atm.
  • Total pressure = p(PCl5) + p(PCl3) + p(Cl2)
  • 7.0 = p(PCl5) + 2.6 + 2.6
  • So partial pressure or PCl5 = 7.0 - 2.6 - 2.6 = 1.8atm
27
Q

What is a mole fraction?

A

The proportion of a gas mixture that is a particular gas.

28
Q

What are the two formulas relating to mole fractions that you need to know?

A

1) Mole fraction of a gas = No. of moles of gas / Total no. of moles of gas in the mixture
2) Partial pressure of a gas = Mole fraction of a gas x Total pressure of the mixture

29
Q

What is the equation for the mole fraction of a gas?

A

Mole fraction of a gas = No. of moles of gas / Total no. of moles of gas in the mixture

30
Q

How are partial pressure and mole fraction of a gas related?

A

Partial pressure of a gas = Mole fraction of a gas x Total pressure of the mixture

31
Q

When 3.00 mol of PCl5 is heated in a sealed vessel, the equilibrium mixture contains 1.75 mol of chlorine. If the total pressure of the mixture is 7.0, what is the partial pressure of PCl5?

PCl5(g) -> PCl3(g) + Cl2(g)

A
  • From the equation above, PCl3 and Cl2 are produced in equal amounts, so there’ll be 1.75 moles of PCl3 too.
  • 1.75 moles of PCl5 must have decomposed so (3.00 - 1.75 =) 1.25 moles of PCl5 must be left at equilibrium.
  • This means that the total number of moles of gas at equilibrium = 1.75 + 1.75 + 1.25 = 4.75 mol
  • So the mole fraction is of PCl5 = 1.25 / 4.75 = 0.263…
  • Partial pressure = 0.263… x 7.0 = 1.8atm
32
Q

Is partial pressure the same as mole fraction?

A

No, remember this!!!

33
Q

What is Kp?

A
  • Equilibrium constant
  • It is not the only equilibrium constant though, so it is best thought of as the “equilibrium constant in terms of partial pressures”
34
Q

What is an equilibrium constant?

A

A measure of how far to the right or left an equilibrium for a given reaction will be at a certain temperature.

35
Q

What are the two equilibrium constants and when’s are they used?

A
  • Kc -> Used when concentrations of the products/reactants are known
  • Kp -> Used when the partial pressures of the products/reactants are known
36
Q

How does Kp differ to Kc?

A
  • The equations are the same, but partial pressures are used instead of concentrations.
  • Kp includes round brackets with a p in front, while Kc uses square brackets
37
Q

Give the equation for Kp for this reaction:

aA(g) + bB(g) -> dD(g) + eE(g)

A

Kp = p(D)^d x p(E)^e / p(A)^a x p(B)^b

38
Q

What is it important to remember about the brackets in equilibrium constant equations?

A
  • For Kc, they are square brackets.

* For Kp, they are round brackets with a p in front, since this shows partial pressures

39
Q

How do you work out the units for Kp?

A
  • Just like for Kc

* Divide the units on the top of the equation by those on the bottom

40
Q

Remember to revise the difference between Kc and Kp.

A

Chapter 11 of revision guide.

41
Q

Calculate Kp for the decomposition of PCl5 gas at 500K. The partial pressures of each gas are:
• p(PCl5) = 1.8 atm
• p(PCl3) = 2.6 atm
• p(Cl2) = 2.6 atm

A
  • Kp = p(Cl2)p(PCl3) / p(PCl5)
  • Kp = 2.6 x 2.6 / 1.8 = 3.755 = 3.8 (2 s.f.)
  • Units = atm x atm / atm = atm
  • So Kp = 3.8 atm
42
Q

An equilibrium exists between ethanoic acid monomers, CH3COOH, and dimers, (CH3COOH)2. At 160*C, Kp for the reaction is 1.78 atm. At this temperature, the partial pressure of the dimer (CH3COOH)2 is 0.281 atm. Calculate the partial pressure of the monomer in this equilibrium and state the total pressure exerted by the equilibrium mixture.

(CH3COOH)2(g) -> 2CH3COOH(g)

A
  • Kp = p(CH3COOH)2 / p((CH3COOH)2)
  • This rearranges to p(CH3COOH)² = Kp x p((CH3COOH)2) = 1.78 x 0.281 = 0.500…
  • p(CH3COOH) = sqrt(0.500) = 0.707… atm
  • So the total pressure of the equilibrium mixture = 0.281 + 0.707 = 0.988 atm
43
Q

Does the equation for Kp include all the products/reactants?

A
  • In homogeneous equilibria -> Yes
  • In heterogeneous equilibria -> Only gases

(Note, this is slightly different from Kc)

44
Q

How do the products/reactants included in Kp equations differ to Kc equations?

A
  • Kp only includes gases in heterogenous equilibria, whole Kc includes everything except solids and pure liquids
  • In homogeneous equilibria, they both include everything
45
Q

Write an expression for Kp for the following reaction:

NH4HS(s) -> NH3(g) + H2S(g)

A
  • This is heterogeneous, so solids don’t get included

* Kp = p(NH3)p(H2S)

46
Q

Remember to practise doing the examples calculations for equilibrium constants.

A

Pg 122-125

47
Q

Describe the rate of the forward and backward reactions in a dynamic equilibrium.

A

They are equal.

48
Q

When you alter the concentration, pressure or temperature of a reversible reaction, does the equilibrium position change?

A

Yes, so the amounts of reactants and products change.

49
Q

If a change causes more product to form, which way has the equilibrium shifted?

A

To the right.

50
Q

If a change causes less product to form, which way has the equilibrium shifted?

A

To the left.

51
Q

State Le Chatelier’s principle.

A

If there’s a change in concentration, pressure or temperature in a reversible reaction, the equilibrium will move to counteract that change.

52
Q

What does a small Kc or Kp tell you about the position of the equilibrium?

A

It is more to the left.

53
Q

Does changing the temperature of a reversible reaction change the value of the equilibrium constant?

A

Yes

54
Q

The reaction below is exothermic in the forward direction. If you increase the temperature, what happens to Kp?

2SO2(g) + O2(g) -> 2SO3(g) dH = -197 kJ/mol

A
  • Increasing the temperature causes the equilibrium to shift to the left to absorb the extra heat
  • So less product is formed
  • Looking at the equation for Kp = p(SO3)² / p(SO2)²p(O2) you can see that if there’s less product, then Kp will be smaller
  • So Kp becomes smaller
55
Q

Give the general rule about what happens to Kc/Kp when the temperature is changed.

A
  • If temperature change causes less product to form -> Equilibrium moves to the left -> Kc/Kp decreases
  • If temperature change causes more product to form -> Equilibrium moves to the right -> Kc/Kp increases
56
Q

Do temperature, concentration or pressure affect the equilibrium constant?

A
  • Temperature -> Yes

* Concentration + Pressure -> No

57
Q

Does a change in concentration affect Kc/Kp? Why?

A
  • No
  • Because the value of Kc/Kp is fixed at a given temperature.
  • So if the concentration of one thing in the equilibrium mixture changes, then the concentrations of the others must change to keep the values of Kc the same.
58
Q

In this reaction, explain how increasing the concentration of CH3COOH affects Kc.

CH3COOH(l) + C2H5OH(l) -> CH3COOC2H5(l) + H2O(l)

A
  • If you increase the concentration, then the equilibrium will move to the right to get rid of the extra CH3COOH.
  • So more CH3COOC2H5 and H2O are produced.
  • This keeps Kc/Kp the same.
59
Q

Does a change in pressure (total or partial) affect Kc/Kp? Why?

A
  • No
  • Because the value of Kc/Kp is fixed at a given temperature.
  • So if the pressure changes, then the moles of the others must change to keep mean that overall the value of Kc/Kp stays the same.
60
Q

In this reaction, explain how increasing the pressure affects Kp.

2SO2(g) + O2(g) -> 2SO3(g)

A
  • There are 3 moles on the left, but only 2 on the right .
  • So an increase in pressure would shift the equilibrium to the right.
  • But Kc/Kp does not change.
61
Q

Check idea about ratios.

A

Pg 127

62
Q

How do catalysts affect the position of an equilibrium?

A
  • They have no effect on the position, so they don’t affect Kc or Kp.
  • They can’t increase the yield, but they do mean equilibrium is approached faster.

A LEVEL - Chemistry (Inorganic) (21 decks)

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