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Equilibrium Constant Flashcards

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1
Q

Which change would alter the value of the equilibrium constant ( Kp ) for this reaction?

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

A Increasing the total pressure of the system.

B Increasing the concentration of sulfur trioxide.

C Increasing the concentration of sulfur dioxide.

D Increasing the temperature.

A
  • D Increasing the temperature
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2
Q

This question is about the reaction given below.

CO ( g ) + H2O ( g ) < - > CO2 ( g ) + H2 ( g )

Enthalpy data for the reacting species are given in the table below.

Substance

1 ) CO ( g )

2 ) H2O ( g )

3 ) CO2 ( g )

4 ) H2 ( g )

Delta H / kj mol^-

1 ) - 110

2 ) - 242

3 ) - 394

4 ) 0

Which one of the following statements is not correct?

A The value of Kp changes when the temperature changes.

B The activation energy decreases when the temperature is increased.

C The entropy change is more positive when the water is liquid rather than gaseous.

D The enthalpy change is more positive when the water is liquid rather than gaseous.

A
  • B The activation energy decreases when the temperature is increased.
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3
Q

The equation for the combustion of butane in oxygen is

C4H10 + 13 / 2O2 - > 4CO2 + 5H2O

The mole fraction of butane in a mixture of butane and oxygen with the minimum amount of oxygen required for complete combustion is

A 0.133

B 0.153

C 0.167

D 0.200

A
  • A 0.133
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4
Q

This question relates to the equilibrium gas-phase synthesis of sulphur trioxide:

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

Thermodynamic data for the components of this equilibrium are:

Substance

1 ) SO3 ( g )

2 ) SO2 ( g )

3 ) O2 ( g )

Delta H / kj mol^-

1 ) - 396

2 ) - 297

3 ) 0

S / J K^- mol^-

1 ) 257

2 ) 248

3 ) 204

This equilibrium, at a temperature of 585 K and a total pressure of 540 kPa, occurs in a vessel of volume 1.80 dm^3.
At equilibrium, the vessel contains 0.0500 mol of SO2 ( g ), 0.0800 mol of O2 ( g ) and 0.0700 mol of SO3 ( g ).

The mole fraction of SO3 in the equilibrium mixture is

A 0.250

B 0.350

C 0.440

D 0.700

A
  • B 0.350

- 0.07 / 0.07 + 0.05 + 0.08

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5
Q

This question relates to the equilibrium gas-phase synthesis of sulphur trioxide:

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

Thermodynamic data for the components of this equilibrium are:

Substance

1 ) SO3 ( g )

2 ) SO2 ( g )

3 ) O2 ( g )

Delta H / kj mol^-

1 ) - 396

2 ) - 297

3 ) 0

S / J K^- mol^-

1 ) 257

2 ) 248

3 ) 204

This equilibrium, at a temperature of 585 K and a total pressure of 540 kPa, occurs in a vessel of
volume 1.80 dm^3.
At equilibrium, the vessel contains 0.0500 mol of SO2 ( g ), 0.0800 mol of O2 ( g ) and 0.0700 mol of SO3 ( g ).

With pressures expressed in MPa units, the value of the equilibrium constant, Kp, is

A 4.90

B 6.48

C 9.07

D 16.8

A

C 9.07

mol fractions:

  • SO2 = 0.05 / 0.2 = 0.25
  • O2 = 0.08 / 0.2 = 0.4
  • SO3 = 0.07 / 0.2 = 0.35
  • 540 kpa = 0.54 mpa
  • Kp = ( 0.35 x 0.54 )^2 / ( 0.4 x 0.54 ) x ( 0.25 x 0.54 )^2
  • Kp = 9.07
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6
Q

This question relates to the equilibrium gas-phase synthesis of sulphur trioxide:

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

Thermodynamic data for the components of this equilibrium are:

Substance

1 ) SO3 ( g )

2 ) SO2 ( g )

3 ) O2 ( g )

Delta H / kj mol^-

1 ) - 396

2 ) - 297

3 ) 0

S / J K^- mol^-

1 ) 257

2 ) 248

3 ) 204

This equilibrium, at a temperature of 585 K and a total pressure of 540 kPa, occurs in a vessel of
volume 1.80 dm^3.
At equilibrium, the vessel contains 0.0500 mol of SO2 ( g ), 0.0800 mol of O2 ( g ) and 0.0700 mol of SO3 ( g ).

With pressures expressed in MPa units, the value of the equilibrium constant, Kp, is

A 4.90

B 6.48

C 9.07

D 16.8

A

C 9.07

mol fractions:

  • SO2 = 0.05 / 0.2 = 0.25
  • O2 = 0.08 / 0.2 = 0.4
  • SO3 = 0.07 / 0.2 = 0.35
  • 540 kpa = 0.54 mpa
  • Kp = ( 0.35 x 0.54 )^2 / ( 0.4 x 0.54 ) x ( 0.25 x 0.54 )^2
  • Kp = 9.07`
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7
Q

This question relates to the equilibrium gas-phase synthesis of sulphur trioxide:

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

Thermodynamic data for the components of this equilibrium are:

Substance

1 ) SO3 ( g )

2 ) SO2 ( g )

3 ) O2 ( g )

Delta H / kj mol^-

1 ) - 396

2 ) - 297

3 ) 0

S / J K^- mol^-

1 ) 257

2 ) 248

3 ) 204

This equilibrium, at a temperature of 585 K and a total pressure of 540 kPa, occurs in a vessel of volume 1.80 dm^3.
At equilibrium, the vessel contains 0.0500 mol of SO2 ( g ), 0.0800 mol of O2 ( g ) and 0.0700 mol of SO3 ( g ).

Possible units for the equilibrium constant Kp include

A no units

B kPa

C Mpa^-1

D kPa^-2

( Previous answer was in Mpa )

( 2 moles on the top and 3 moles at the bottom )

A
  • C Mpa^-1
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8
Q

This question relates to the equilibrium gas-phase synthesis of sulphur trioxide:

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

Thermodynamic data for the components of this equilibrium are:

Substance

1 ) SO3 ( g )

2 ) SO2 ( g )

3 ) O2 ( g )

Delta H / kj mol^-

1 ) - 396

2 ) - 297

3 ) 0

S / J K^- mol^-

1 ) 257

2 ) 248

3 ) 204

This equilibrium, at a temperature of 585 K and a total pressure of 540 kPa, occurs in a vessel of
volume 1.80 dm^3.
At equilibrium, the vessel contains 0.0500 mol of SO2 ( g ), 0.0800 mol of O2 ( g ) and 0.0700 mol of SO3 ( g ).

At equilibrium in the same vessel of volume 1.80 dm^3 under altered conditions, the reaction
mixture contains 0.0700 mol of SO3 ( g ), 0.0500 mol of SO2 ( g ) and 0.0900 mol of O2 ( g ) at a total pressure of 623 kPa.
The temperature in the equilibrium vessel is

A 307 °C

B 596 K

C 337 °C

D 642 K

A
  • D 642 K ( Look at the ratio of temperature to pressure )
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9
Q

The following information concerns the equilibrium gas-phase synthesis of methanol.

CO ( g ) + 2H2 ( g ) < - > CH3OH ( g )

At equilibrium, when the temperature is 68 °C, the total pressure is 1.70 MPa.
The number of moles of CO, H2 and CH3OH present are 0.160, 0.320 and 0.180, respectively.

Thermodynamic data are given below.

Substance

1 ) CO ( g )

2 ) H2 ( g )

3 ) CH3OH ( g )

Delta H f / kj mol^-

1 ) - 110

2 ) 0

3 ) - 201

S / J K^- mol^-

1 ) 198

2 ) 131

3 ) 240

Possible units for the equilibrium constant, Kp, for this reaction are

A no units

B kPa

C MPa^-1

D kPa^-2

A
  • D kPa^-2 ( 1 mol on top and 3 at the bottom )
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10
Q

The following information concerns the equilibrium gas-phase synthesis of methanol.

CO ( g ) + 2H2 ( g ) < - > CH3OH ( g )

At equilibrium, when the temperature is 68 °C, the total pressure is 1.70 MPa.
The number of moles of CO, H2 and CH3OH present are 0.160, 0.320 and 0.180, respectively.

Thermodynamic data are given below.

Substance

1 ) CO ( g )

2 ) H2 ( g )

3 ) CH3OH ( g )

Delta H f / kj mol^-

1 ) - 110

2 ) 0

3 ) - 201

S / J K^- mol^-

1 ) 198

2 ) 131

3 ) 240

The mole fraction of hydrogen in the equilibrium mixture is

A 0.242

B 0.485

C 0.653

D 0.970

A
  • B 0.485

- 0.320 / 0.160 + 0.320 + 0.180

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11
Q

The following information concerns the equilibrium gas-phase synthesis of methanol.

CO ( g ) + 2H2 ( g ) < - > CH3OH ( g )

At equilibrium, when the temperature is 68 °C, the total pressure is 1.70 MPa.
The number of moles of CO, H2 and CH3OH present are 0.160, 0.320 and 0.180, respectively.

Thermodynamic data are given below.

Substance

1 ) CO ( g )

2 ) H2 ( g )

3 ) CH3OH ( g )

Delta H f / kj mol^-

1 ) - 110

2 ) 0

3 ) - 201

S / J K^- mol^-

1 ) 198

2 ) 131

3 ) 240

With pressures expressed in MPa units, the value of the equilibrium constant, Kp, under these conditions is

A 1.37

B 1.66

C 2.82

D 4.80

A

B 1.66

Mol fractions:

  • CO = 0.160 / 0.660 = 0.242
  • H2 = 0.320 / 0.660 = 0.485
  • CH3OH = 0.180 / 0.660 = 0.273
  • Kp = ( 0.273 x 1.70 ) / ( 0.242 x 1.70 ) x ( 0.485 x 1.70 )^2
  • Kp = 1.66
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12
Q

The following information concerns the equilibrium gas-phase synthesis of methanol.

CO ( g ) + 2H2 ( g ) < - > CH3OH ( g )

At equilibrium, when the temperature is 68 °C, the total pressure is 1.70 MPa.
The number of moles of CO, H2 and CH3OH present are 0.160, 0.320 and 0.180, respectively.

Thermodynamic data are given below.

Substance

1 ) CO ( g )

2 ) H2 ( g )

3 ) CH3OH ( g )

Delta H f / kj mol^-

1 ) - 110

2 ) 0

3 ) - 201

S / J K^- mol^-

1 ) 198

2 ) 131

3 ) 240

Which one of the following statements applies to this equilibrium?

A The value of Kp increases if the temperature is raised.

B The value of Kp increases if the pressure is raised.

C The yield of methanol decreases if the temperature is lowered.

D The yield of methanol decreases if the pressure is lowered.

A
  • D The yield of methanol decreases if the pressure is lowered.
  • ( Decreasing pressure favours the side with more moles )
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13
Q

An experiment was carried out to determine the equilibrium constant, Kc, for the following
reaction.

CH3CH2COOH + CH3CH2CH2OH < - > CH3CH2COOCH2CH2CH3 + H2O

A student added measured volumes of propan-1-ol and propanoic acid to a conical flask.
A measured volume of concentrated hydrochloric acid was added to the flask, which was then sealed.

After 1 week, the contents of the flask were poured into water and the solution was made up to a known volume.
This solution was titrated with standard sodium hydroxide solution.

Explain how the student could determine the amount, in moles, of propan-1-ol added to the flask.

A
  • Multiply volume of propan-1-ol by density
  • ( Density x volume = mass )
  • Divide the mass by the Mr of propan-1-ol
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14
Q

The titration described above gives the total amount of acid in the equilibrium mixture.
Explain how, by carrying out a further experiment, the student could determine the
amount of propanoic acid in the equilibrium mixture.

( CH3CH2COOH + CH3CH2CH2OH < - > CH3CH2COOCH2CH2CH3 + H2O )

( A student added measured volumes of propan-1-ol and propanoic acid to a conical flask.
A measured volume of concentrated hydrochloric acid was added to the flask, which was then sealed. )

A
  • Titrate a measured volume of the concentrated HCl to determine the moles of HCl used in the experiment
  • Subtract this number of moles of HCl from the total moles of acid at equilibrium
  • ( Total amount of acid in the mixture, since there are only 2 acids, you can subtract one from the other )
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15
Q

In a repeat experiment, the student failed to seal the flask that contained the equilibrium mixture.

Explain why this error would lead to the student obtaining an incorrect value for the equilibrium constant Kc

( CH3CH2COOH + CH3CH2CH2OH < - > CH3CH2COOCH2CH2CH3 + H2O )

( Propanoic acid + propan-1-ol < - > Ethyl propanoate + water )

A
  • Ester will evaporate

- So you will get incorrect values to determine the Kc value

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16
Q

Many chemical processes release waste products into the atmosphere.
Scientists are developing new solid catalysts to convert more efficiently these emissions into useful products, such as fuels.
One example is a catalyst to convert these emissions into methanol.
The catalyst is thought to work by breaking a H - H bond.

An equation for this formation of methanol is given below.

CO2 ( g ) + 3H2 ( g ) < - > CH3OH ( g ) + H2O ( g ) ∆H = −49 kJ mol^-1

Some mean bond enthalpies are shown in the following table.

Bond

1 ) C = O

2 ) C - H

3 ) C - O

4 ) O - H

Mean bond enthalpy / kj mol^-1

1 ) 743

2 ) 412

3 ) 360

4 ) 463

Use the enthalpy change for the reaction and data from the table to calculate a value for the H - H bond enthalpy.

A
  • ( ( 2 x 743 ) + 3x ) - ( ( 3 x 412 ) + 360 + ( 3 x 463 ) ) = - 49
  • ( Mean bond enthalpies, so arrows for the equation go downwards, making the left side positive and the right side negative )
  • 1486 + 3x - 2985 = - 49
  • 3x = 1450
  • x = 483 kj mol^-1
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17
Q

A data book value for the H–H bond enthalpy is 436 kJ mol^-1.

Suggest one reason why this value is different from your answer to part ( a ).

( CO2 ( g ) + 3H2 ( g ) < - > CH3OH ( g ) + H2O ( g ) )

A
  • Mean bond enthalpies are not the same as the actual bond enthalpies in CO2 ( or any other compounds in the equation )
18
Q

Suggest one environmental advantage of manufacturing methanol fuel by this reaction.

( CO2 ( g ) + 3H2 ( g ) < - > CH3OH ( g ) + H2O ( g ) )

A
  • The carbon dioxide is used up in this reaction
19
Q

Use Le Chatelier’s principle to justify why the reaction is carried out at a high pressure rather than at atmospheric pressure.

( CO2 ( g ) + 3H2 ( g ) < - > CH3OH ( g ) + H2O ( g ) )

A
  • 4 mol of gas form 2 mol of gas
  • At high pressure the position of equilibrium shifts to the right in order to oppose increasing pressure
  • This increases the yield of methanol
20
Q

Suggest why the catalyst used in this process may become less efficient if the carbon dioxide and hydrogen contain impurities.

( CO2 ( g ) + 3H2 ( g ) < - > CH3OH ( g ) + H2O ( g ) )

A
  • Impurities block the active sites
21
Q

In a laboratory experiment to investigate the reaction shown in the equation below, 1.0 mol of carbon dioxide and 3.0 mol of hydrogen were sealed into a container.
After the mixture had reached equilibrium, at a pressure of 500 kPa, the yield of methanol was 0.86 mol.

CO2 ( g ) + 3H2 ( g ) < - > CH3OH ( g ) + H2O ( g )

Calculate a value for Kp
Give your answer to the appropriate number of significant figures.

Give units with your answer.

A
  • CO2 ( g ) + 3H2 ( g ) < - > CH3OH ( g ) + H2O ( g )

I ) 1 3 0 0

C ) 0.86 0.86 x 3 0.86 0.86

E ) 0.14 0.42 0.86 0.86

  • Total moles = 2.28

mol fraction:

  • CO2 = 0.14 / 2.28 = 0.06140350877
  • H2 = 0.42 / 2.28 = 0.1842105263
  • CH3OH = 0.86 / 2.28 = 0.3771929825
  • H2O = 0.86 / 2.28 = 0.3771929825

pP:

  • CO2 = 0.06140350877 x 500 = 30.70175439
  • H2 = 0.1842105263 x 500 = 92.10526315
  • CH3OH = 0.3771929825 x 500 = 188.5964913
  • H2O = 0.3771929825 x 500 = 188.5964913
  • Kp = ( pP CH3OH ) x ( pP H2O ) / ( pP CO2 ) x ( pP H2 )^3
  • Kp = ( 188.5964913 ) x ( 188.5964913 ) / ( 30.70175439 ) x ( 92.10526315 )^3
  • Kp = 1.482691102 x 10^-3 kPa^-2
  • Kp = 1.5 x 10^-3 kPa^-2
  • ( Appropriate s.f means 2 s.f )
22
Q

The manufacture of methanol can be achieved in two stages.

In the first stage, methane and steam react according to the following equation.

CH4 ( g ) + H2O ( g ) < - > CO ( g ) + 3H2 ( g )
∆Hο = +210 kJ mol^-1

Discuss, with reasons, the effects of increasing separately the temperature and the pressure on the yield of the products and on the rate of this reaction.

A
  • An higher temperature shifts equilibrium to the right because forward reaction is endothermic
  • Rate is increased
  • An higher pressure shifts the equilibrium to the left because there are 2 moles on the left and 3 moles on the right side of the equation
  • Rate is increased
23
Q

In the second stage, carbon monoxide and hydrogen react according to the following equation.

CO ( g ) + 2H2 ( g ) < - > CH3OH ( g )

A 62.8 mol sample of carbon monoxide was added to 146 mol of hydrogen.
When equilibrium was reached at a given temperature, the mixture contained 26.2 mol of methanol at a total pressure of 9.50 MPa.

Write an expression for the equilibrium constant, Kp, for this reaction.
Calculate a value for Kp at this temperature and give its units.

A
  • CO ( g ) + 2H2 ( g ) < - > CH3OH ( g )

I ) 62.8 146 0

C ) 26.2 26.2 x 2 26.2

E ) 36.6 93.6 26.2

  • Total moles = 156.4

Mol fraction

  • CO = 36.6 / 156.4 = 0.2340153453
  • H2 = 93.6 / 156.4 = 0.5984654731
  • CH3OH = 26.2 / 156.4 = 0.1675191816

pP

  • CO = 0.2340153453 x 9.50 = 2.22314578
  • H2 = 0.5984654731 x 9.50 = 5.685421994
  • CH3OH = 0.1675191816 x 9.50 = 1.591432225
  • Kp = ( pP CH3OH ) / ( pP CO ) x ( pP H2 )^2
  • Kp = ( 1.591432225 ) / ( 2.22314578 ) x ( 5.685421994 )^2
  • Kp = 0.0221459745 mPa^-2
  • Kp = 0.022 mPa^-2
24
Q

A sealed flask containing gases X and Y in the mole ratio 1:3 was maintained at 600K until the following equilibrium was established.

X ( g ) + 3Y ( g ) < - > 2Z ( g )

The partial pressure of Z in the equilibrium mixture was 6.0 MPa when the total pressure
was 22.0 MPa.

Write an expression for the equilibrium constant, Kp, for this reaction.

A
  • Kp = ( pP Z )^2 / ( pP X ) x ( pP Y )^3
25
Q

Calculate the partial pressure of X and the partial pressure of Y in the equilibrium mixture.

( X ( g ) + 3Y ( g ) < - > 2Z ( g ) )

( The partial pressure of Z in the equilibrium mixture was 6.0 MPa when the total pressure
was 22.0 MPa. )

A
  • 22 - 6 = 16
  • 1 : 3 ratio
  • X = 16 / 4 = 4 mPa
  • Y = 4 x 3 = 12 mPa
26
Q

Calculate the value of Kp for this reaction under these conditions and state its units.

( X ( g ) + 3Y ( g ) < - > 2Z ( g ) )

Partial pressures:

  • X = 4 mPa
  • Y = 12 mPa
  • Z = 6 mPa
A
  • Kp = ( 6.0 )^2 / ( 4.0 ) x ( 12.0 )^3
  • Kp = 5.208333333x 10^-3 mPa^-2
  • Kp = 5.21 x 10^-3 mPa^-2
27
Q

When this reaction is carried out at 300 K and a high pressure of 100 MPa, rather than at 600 K and 22.0 MPa, a higher equilibrium yield of gas Z is obtained.

Give two reasons why an industrialist is unlikely to choose these reaction conditions.

A
  • High pressure is expensive

- Low temperature means rate is slow

28
Q

Sulphur dioxide and oxygen were mixed in a 2:1 mol ratio and sealed in a flask with a
catalyst.
The following equilibrium was established at temperature T1

2SO2 ( g ) + O2 ( g ) < - > 2SO3 ( g ) ΔH = -196 kJ mol^-1

The partial pressure of sulphur dioxide in the equilibrium mixture was 24 kPa and the total pressure in the flask was 104 kPa.

Deduce the partial pressure of oxygen and hence calculate the mole fraction of oxygen in the equilibrium mixture.

A
  • 2 : 1 ratio
  • 24 / 2 = 12
  • pP O2 = 12
  • pP = mole fraction x total pressure
  • Mole fraction = pP / total pressure
  • 12 / 104 = 0.115
  • Mole fraction = 0.115
29
Q

Calculate the partial pressure of sulphur trioxide in the equilibrium mixture.

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

Partial pressures

SO2 = 24

O2 = 12

Total pressure = 104

A
  • 104 - ( 24 + 12 )
  • = 68
  • pP SO3 = 68
30
Q

Write an expression for the equilibrium constant, Kp, for this reaction.
Use this expression to calculate the value of Kp at temperature T1 and state its units.

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

Partial pressures

  • SO2 = 24
  • O2 = 12
  • SO3 = 68
A
  • Kp = ( pP SO3 )^2 / ( pP SO2 )^2 x ( pP O2 )
  • Kp = ( 68.0 )^2 / ( 24.0 )^2 x ( 12.0 )
  • Kp = 0.6689814815 kPa^-1
  • Kp = 0.669 kPa^-1
31
Q

When equilibrium was established at a different temperature, T2, the value of Kp was
found to have increased.
State which of T1 and T2 is the lower temperature and explain your answer.

A
  • T2
  • Exothermic reaction
  • Reduce temperature to shift equilibrium to the right
32
Q

In a further experiment, the amounts of sulphur dioxide and oxygen used, the catalyst and the temperature, T1, were all unchanged, but a flask of smaller volume was used.

Deduce the effect of this change on the yield of sulphur trioxide and on the value of Kp.

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

A
  • Yield of SO3 increases

- No effect on Kp

33
Q

The gaseous reactants W and X were sealed in a flask and the mixture left until the following equilibrium had been established.

2W ( g ) + X ( g ) < - > 3Y ( g ) + 2Z ( g ) ΔH = -200 kJ mol^-1

Write an expression for the equilibrium constant, Kp, for this reaction.
State one change in the conditions which would both increase the rate of reaction and decrease the value of Kp.
Explain your answers.

A
  • Kp = ( pP Z )^2 x ( pP Y )^3 / ( pP W )^2 x ( pP X )
  • Increasing temperature increases rate and decreases Kp
  • Particles have more energy
  • More collisions with particles that have an energy greater than the activation energy
  • Reaction is exothermic
  • So equilibrium shifts to the left
34
Q

Ethyl ethanoate can be prepared by the reactions shown below.

Reaction 1

CH3COOH ( l ) + C2H5OH ( l ) < - > CH3COOC2H5 ( l ) + H2O ( l ) ∆H = -2.0 kJ mol^-1

Reaction 2

CH3COCl ( l ) + C2H5OH ( l ) - > CH3COOC2H5 ( l ) + HCl ( g ) ∆H = -21.6 kJ mol^-1

Give one advantage and one disadvantage of preparing ethyl ethanoate by Reaction 1 rather than by Reaction 2.

A
  • Reaction is not reversible

- However reaction is vigorous

35
Q

Use the information given above and the data below to calculate values for the standard entropy change, ∆S , and the standard free - energy change, ∆G , for Reaction 2 at 298 K.

1 ) CH3COCl ( l )

2 ) C2H5OH ( l )

3 ) CH3COOC2H5 ( l )

4 ) HCl ( g )

S /JK^1 mol^1

1 ) 201

2 ) 161

3 ) 259

4 ) 187

( CH3COCl ( l ) + C2H5OH ( l ) - > CH3COOC2H5 ( l ) + HCl ( g ) ∆H = -21.6 kJ mol^-1 )

A
  • ∆S = ΣS products – ΣS reactants
  • ∆S = ( 259 + 187 ) - ( 161 + 201 )
  • ∆S = 84 JK^-1 mol^-1
  • ΔG = ΔH – TΔS
  • ΔG = -21.6 - 298 x ( 84 x 10^-3 )
  • ΔG = - 46.6 kj mol^-1
36
Q

When a mixture of 0.345 mol of PCl3 and 0.268 mol of Cl2 was heated in a vessel of fixed volume to a constant temperature, the following reaction reached equilibrium.

PCl3 ( g ) + Cl2 ( g ) < - > PCl5 ( g ) ∆H = -93 kJ mol^-1

At equilibrium, 0.166 mol of PCl5 had been formed and the total pressure was 225 kPa.

Calculate the number of moles of PCl3 and of Cl2 in the equilibrium mixture.

A
  • PCl3 ( g ) + Cl2 ( g ) < - > PCl5 ( g )

I ) 0.345 0.268 0

C ) 0.166 0.166 0.166

E ) 0.179 0.102 0.166

  • Moles of PCl3 = 0.179
  • Moles of Cl2 = 0.102
37
Q

State the effect on the mole fraction of PCl3 in the equilibrium mixture if

( PCl3 ( g ) + Cl2 ( g ) < - > PCl5 ( g ) ∆H = -93 kJ mol^-1 )

i ) the volume of the vessel were to be increased at a constant temperature,

ii ) the temperature were to be increased at constant volume.

A

i ) Increased
- ( Increasing volume means pressure is decreased, favouring the side with more moles )

ii ) Increased
- ( Increased temperature favours the endothermic reaction side, equilibrium shifts left )

38
Q

At high temperatures, SO2Cl2 dissociates according to the following equation.

SO2Cl2 ( g ) < - > SO2 ( g ) + Cl2 ( g ) ΔH = +93 kJ mol^-1

When 1.00 mol of SO2Cl2 dissociates, the equilibrium mixture contains 0.75 mol of Cl2 at 673 K and a total pressure of 125 kPa.

Write a general expression for the partial pressure of a gas in a mixture of gases in terms of the total pressure.

( Mole fractions

  • SO2Cl2 = 0.1428571429
  • Cl2 = 0.4285714286 )

( Total pressure = 125 )

A
  • pP = Total pressure x mole fraction
  • pP SO2Cl2 = 125 x 0.1428571429 = 17.85714286
  • pP SO2Cl2 = 17.9 kPa
  • pP Cl2 = 125 x 0.4285714286 = 53.57142857
  • pP Cl2 = 53.6 kPa
39
Q

State the effect, if any, of an increase in temperature on the value of Kp for this reaction.
Explain your answer.

( SO2Cl2 ( g ) < - > SO2 ( g ) + Cl2 ( g ) ΔH = +93 kJ mol^-1 )

A
  • Increased

- Increasing temperature shifts the equilibrium to the right as the forward reaction is endothermic

40
Q

State the effect, if any, of an increase in the total pressure on the value of Kp for this reaction.

A
  • No effect

Chemistry (13 decks)

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