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CHEMISTRY, M5 2024 HSC 37

The relationship between the equilibrium constant, \(K_{eq}\), and \(\Delta G^{\circ}\) for any reaction is shown in the graph, for a limited range of  \(\Delta G^{\circ}\) values.
 

 

The \(\Delta H^{\circ}\) and \(T \Delta S^{\circ}\) values for the reaction between copper\(\text{(I)}\) sulfide and oxygen are provided.

\(\ce{Cu2S(s) +O2(g) \rightarrow 2Cu(s) +SO2(g) \quad \quad}\) \(\Delta H^{\circ}=-217 \text{ kJ mol}^{-1}\)
\(\Delta S^{\circ}=-3 \text{ kJ mol}^{-1}\)

Explain, with reference to the information provided, why this reaction proceeds to completion rather than coming to equilibrium.   (3 marks)

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Calculating \(\Delta G\):

  \(\Delta G = \Delta H-T\Delta S=-217-(-3)=-214\ \text{kJ mol}^{-1}\)

→ The graph shows that large negative values of \(\Delta G\) correspond to large \(K_{eq}\) values. Large \(K_{eq}\) values represent that the concentration of the products is significantly higher than the concentration of the reactants. i.e. the reaction runs to completion.

→ The \(\Delta G\) value for the reaction (see above) is significantly negative and would correspond to a significantly high \(K_{eq}\) value. 

→ This reaction proceeds to completion primarily because of its significantly large, negative \(\Delta H\) value, while the \(T\Delta S\) term is relatively small and has minimal influence.

Show Worked Solution

Calculating \(\Delta G\):

  \(\Delta G = \Delta H-T\Delta S=-217-(-3)=-214\ \text{kJ mol}^{-1}\)

→ The graph shows that large negative values of \(\Delta G\) correspond to large \(K_{eq}\) values. Large \(K_{eq}\) values represent that the concentration of the products is significantly higher than the concentration of the reactants. i.e. the reaction runs to completion.

→ The \(\Delta G\) value for the reaction (see above) is significantly negative and would correspond to a significantly high \(K_{eq}\) value. 

→ This reaction proceeds to completion primarily because of its significantly large, negative \(\Delta H\) value, while the \(T\Delta S\) term is relatively small and has minimal influence.

CHEMISTRY, M5 EQ-Bank 21

Potassium chloride readily dissolves in water. With the use of a labelled diagram, describe the changes in bonding and entropy that occurs during this process.  (4 marks)

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→ Potassium chloride has a high tendency to dissociate into \(\ce{K^+}\) and \(\ce{Cl^{-}}\) ions when mixed with water (i.e. it is highly soluble).

→ Water is a dipolar molecule because each atom has a partial charge, as shown in the diagram. 

→ The oxygen dipole in water has a partial negative charge and is attracted to the potassium ion. The hydrogen dipoles have a partial negative charge and are attracted to the chloride ion. 

→ This attraction breaks the ionic bonds and forms ion-dipole bonds.

→ The entropy of the system is increased as the ionic bonds of the \(\ce{KCl}\) are broken and the \(\ce{K^{+}}\) and \(\ce{Cl^{-}}\) ions disperse throughout the solution.

Show Worked Solution

→ Potassium chloride has a high tendency to dissociate into \(\ce{K^+}\) and \(\ce{Cl^{-}}\) ions when mixed with water (i.e. it is highly soluble).

→ Water is a dipolar molecule because each atom has a partial charge, as shown in the diagram. 

→ The oxygen dipole in water has a partial negative charge and is attracted to the potassium ion. The hydrogen dipoles have a partial negative charge and are attracted to the chloride ion. 

→ This attraction breaks the ionic bonds and forms ion-dipole bonds.

→ The entropy of the system is increased as the ionic bonds of the \(\ce{KCl}\) are broken and the \(\ce{K^{+}}\) and \(\ce{Cl^{-}}\) ions disperse throughout the solution.

CHEMISTRY, M5 2018 HSC 30

Over the last 50 years, scientists have recorded increases in the following:

  • the amount of fossil fuels burnt
  • atmospheric carbon dioxide levels
  • average global air temperature and ocean temperature
  • the volume of carbon dioxide dissolved in the oceans.

Analyse the factors that affect the equilibrium between carbon dioxide in the air and carbon dioxide in the oceans. In your answer, make reference to the scientists' observations and include relevant equations.  (7 marks)

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Fossil fuel combustion:

→ Combustion of fossil fuels releases \(\ce{CO2}\) and heat energy, both of which are released into the atmosphere.

→ Increased burning of fossil fuels will contribute to further rises in atmospheric \(\ce{CO2}\), as described in the equation for the combustion of octane

\(\ce{2C8H18 + 25O2 -> 16CO2 + 18H2O\qquad \triangle H –ve}\)
 

Carbon dioxide and other climate interactions:

→ \(\ce{CO2}\) combines with water according to the following equilibrium in an exothermic reaction.

\(\ce{CO2(g) + H2O(l) \rightleftharpoons H2CO3(aq) \rightleftharpoons H+(aq) + HCO^3-(aq)\qquad   \triangle H –ve}\)

→ This is an equilibrium and by Le Chatelier’s principle when a system is changed, the system will adjust to oppose the change.

→ Factors that affect equilibrium in this system are temperature, pressure and concentration of reactants and products.

→ The increase of \(\ce{CO2}\) in the air due to the combustion of fossil fuels described above, increases the pressure due to \(\ce{CO2}\) in the system.  By Le Chatelier’s principle, the system will oppose this by absorbing more \(\ce{CO2}\) into the oceans.

→ Scientists have been measuring the level of \(\ce{CO2}\) in oceans due to this effect and confirmed the increase in \(\ce{CO2}\).

→ However, this equilibrium is exothermic and as it causes temperature rises, by Le Chatelier’s principle, the reverse reaction may be subsequently favoured. This would have the effect of decreasing the amount of \(\ce{CO2}\) dissolving in the oceans.

→ In summary, if global temperatures continue to rise and \(\ce{CO2}\) in the atmosphere becomes stable or reduces, the system may adjust so that oceans may release \(\ce{CO2}\) rather than absorbing it.

Show Worked Solution

Fossil fuel combustion:

→ Combustion of fossil fuels releases \(\ce{CO2}\) and heat energy, both of which are released into the atmosphere.

→ Increased burning of fossil fuels will contribute to further rises in atmospheric \(\ce{CO2}\), as described in the equation for the combustion of octane

\(\ce{2C8H18 + 25O2 -> 16CO2 + 18H2O\qquad \triangle H –ve}\)
 

Carbon dioxide and other climate interactions:

→ \(\ce{CO2}\) combines with water according to the following equilibrium in an exothermic reaction.

\(\ce{CO2(g) + H2O(l) \rightleftharpoons H2CO3(aq) \rightleftharpoons H+(aq) + HCO^3-(aq)\qquad   \triangle H –ve}\)

→ This is an equilibrium and by Le Chatelier’s principle when a system is changed, the system will adjust to oppose the change.

→ Factors that affect equilibrium in this system are temperature, pressure and concentration of reactants and products.

→ The increase of \(\ce{CO2}\) in the air due to the combustion of fossil fuels described above, increases the pressure due to \(\ce{CO2}\) in the system.  By Le Chatelier’s principle, the system will oppose this by absorbing more \(\ce{CO2}\) into the oceans.

→ Scientists have been measuring the level of \(\ce{CO2}\) in oceans due to this effect and confirmed the increase in \(\ce{CO2}\).

→ However, this equilibrium is exothermic and as it causes temperature rises, by Le Chatelier’s principle, the reverse reaction may be subsequently favoured. This would have the effect of decreasing the amount of \(\ce{CO2}\) dissolving in the oceans.

→ In summary, if global temperatures continue to rise and \(\ce{CO2}\) in the atmosphere becomes stable or reduces, the system may adjust so that oceans may release \(\ce{CO2}\) rather than absorbing it.


♦♦ Mean mark 41%.

CHEMISTRY, M5 2019 HSC 30

The following data apply to magnesium fluoride and magnesium chloride dissolving in water at 298 K.
 


 

Compare the effects of enthalpy and entropy on the solubility of these salts.   (3 marks)

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→ Magnesium chloride dissolves in water spontaneously as it has a negative \(\triangle_{\text {sol }} G^{\ominus}\) ( – 125 kJ mol ¯1).

→ Magnesium flouride however does not dissolve in water spontaneously which is shown by its corresponding \(\triangle_{\text {sol }} G^{\ominus}\) of +58.6 kJ mol ¯1.

→ Both salts have a negative \(\triangle_{\text {sol }} S^{\ominus}\), resulting in a net positive \(-T \triangle_{\text {sol }} S^{\ominus}\) contribution to \(\triangle_{\text {sol }} G^{\ominus}\).

→ Both salts have a negative \(\triangle_{\text {sol }} H^{\ominus}\). It should be noted however that magnesium chloride’s negative value is significantly more negative at \(\triangle_{\text {sol }} H^{\ominus}\) ( – 160 kJ mol ¯1). This is greater than the \(-T \triangle_{\text {sol }} S^{\ominus}\) contribution (+34.2 kJ mol ¯1), resulting in a negative \(\triangle_{\mathrm{sol}} G^{\ominus}\).

→ This can be compared to magnesium fluoride that has a relatively small negative \(\triangle_{\text {sol }} H^{\ominus}\) ( – 7.81 kJ mol ¯1) which is smaller than the \(-T \triangle_{\text {sol }} S^{\ominus}\) contribution (+66.4 kJ mol ¯1), resulting in a positive \(\triangle_{\text {sol }} G^{\ominus}\).

Show Worked Solution

→ Magnesium chloride dissolves in water spontaneously as it has a negative \(\triangle_{\text {sol }} G^{\ominus}\) ( – 125 kJ mol ¯1).

→ Magnesium flouride however does not dissolve in water spontaneously which is shown by its corresponding \(\triangle_{\text {sol }} G^{\ominus}\) of +58.6 kJ mol ¯1.

→ Both salts have a negative \(\triangle_{\text {sol }} S^{\ominus}\), resulting in a net positive \(-T \triangle_{\text {sol }} S^{\ominus}\) contribution to \(\triangle_{\text {sol }} G^{\ominus}\).

→ Both salts have a negative \(\triangle_{\text {sol }} H^{\ominus}\). It should be noted however that magnesium chloride’s negative value is significantly more negative at \(\triangle_{\text {sol }} H^{\ominus}\) ( – 160 kJ mol ¯1). This is greater than the \(-T \triangle_{\text {sol }} S^{\ominus}\) contribution (+34.2 kJ mol ¯1), resulting in a negative \(\triangle_{\mathrm{sol}} G^{\ominus}\).

→ This can be compared to magnesium fluoride that has a relatively small negative \(\triangle_{\text {sol }} H^{\ominus}\) ( – 7.81 kJ mol ¯1) which is smaller than the \(-T \triangle_{\text {sol }} S^{\ominus}\) contribution (+66.4 kJ mol ¯1), resulting in a positive \(\triangle_{\text {sol }} G^{\ominus}\).


♦♦ Mean mark 34%.

CHEMISTRY, M5 2020 HSC 26

Nitric oxide gas (`text{NO}`) can be produced from the direct combination of nitrogen gas and oxygen gas in a reversible reaction.

  1. Write the balanced chemical equation for this reaction.   (1 mark)


  2. The energy profile diagram for this reaction is shown. 
     

   

  1. Explain, using collision theory, how an increase in temperature would affect the value of `K_{eq}` for this system. Refer to the diagram in your answer.  (4 marks)

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a.   \(\ce{N2(g) + O2(g) \rightleftharpoons 2NO(g) }\)
 

b.    → From the graph, the forward reaction is endothermic.

→ The activation energy of the forward endothermic reaction is greater than the activation energy of the reverse exothermic reaction.

→ An increase in temperature would cause the rates of both the forward and reverse reaction due to the higher average kinetic energy, resulting in a larger likelihood of a successful collisions.

→ However, the rate of the forward reaction would increase to a higher extent than the reverse reaction, since it is an endothermic reaction.

→ Using  `K_(eq) = ([text{NO}]^2)/([text{N}_2][text{O}_2])`, as the equilibrium shifts right, the equilibrium constant would increase.

Show Worked Solution

a.   \(\ce{N2(g) + O2(g) \rightleftharpoons 2NO(g) }\)
 

b.    → From the graph, the forward reaction is endothermic.

→ The activation energy of the forward endothermic reaction is greater than the activation energy of the reverse exothermic reaction.

→ An increase in temperature would cause the rates of both the forward and reverse reaction due to the higher average kinetic energy, resulting in a larger likelihood of a successful collisions.

→ However, the rate of the forward reaction would increase to a higher extent than the reverse reaction, since it is an endothermic reaction.

→ Using  `K_(eq) = ([text{NO}]^2)/([text{N}_2][text{O}_2])`, as the equilibrium shifts right, the equilibrium constant would increase.

CHEMISTRY, M5 2022 HSC 36

Consider the equilibrium system shown.

 \( \ce{H2O(l) \rightleftharpoons H2O(g)} \)

In a laboratory at 23°C, a 100 mL sample of water is held in a beaker and another 100 mL sample is held in a sealed bottle.

Explain the differences in evaporation for these TWO samples. In your answer, consider changes in enthalpy and entropy for this process.   (4 marks)

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→ The evaporation of water absorbs energy, hence is an endothermic reaction and results in a positive change in enthalpy (`ΔH > 0`).

→ Additionally, the process converts a liquid into a gaseous state, and thus increases the disorder of the system, as a result, entropy increases (`ΔS > 0`).

→ Since enthalpy and entropy are both positive, according to `ΔG = ΔH − T ΔS`, the evaporation of water is spontaneous at high temperatures, ie when `ΔG < 0`.
 

Beaker sample:

→ The evaporation of water in a beaker represents an open system, where vapour molecules are able to escape the system.

→ As a result, there would be a continuous disturbance to the equilibrium, and according to Le Chatelier’s Principle, the equilibrium will shift to counteract the change, and thus produce more gaseous water until there is no liquid water left.

→ Thus, dynamic equilibrium will not be established in a beaker.
 

Sealed bottle sample:

→ On the other hand, the evaporation of water in a sealed bottle represents a closed system where the water vapour cannot escape from the system.

→ In this reaction liquid water would evaporate, shifting the equilibrium to the right until the rate of the forward reaction and the rate of the reverse reaction is equal.

→ At this point, there would be virtually no change in the concentration of liquid water and gaseous water, and thus dynamic equilibrium will be established.

Show Worked Solution

→ The evaporation of water absorbs energy, hence is an endothermic reaction and results in a positive change in enthalpy (`ΔH > 0`).

→ Additionally, the process converts a liquid into a gaseous state, and thus increases the disorder of the system, as a result, entropy increases (`ΔS > 0`).

→ Since enthalpy and entropy are both positive, according to `ΔG = ΔH − T ΔS`, the evaporation of water is spontaneous at high temperatures, ie when `ΔG < 0`.
 

Beaker sample:

→ The evaporation of water in a beaker represents an open system, where vapour molecules are able to escape the system.

→ As a result, there would be a continuous disturbance to the equilibrium, and according to Le Chatelier’s Principle, the equilibrium will shift to counteract the change, and thus produce more gaseous water until there is no liquid water left.

→ Thus, dynamic equilibrium will not be established in a beaker.
 

Sealed bottle sample:

→ On the other hand, the evaporation of water in a sealed bottle represents a closed system where the water vapour cannot escape from the system.

→ In this reaction liquid water would evaporate, shifting the equilibrium to the right until the rate of the forward reaction and the rate of the reverse reaction is equal.

→ At this point, there would be virtually no change in the concentration of liquid water and gaseous water, and thus dynamic equilibrium will be established.


♦ Mean mark 49%.

CHEMISTRY, M5 2021 HSC 33

The relationships between `Delta H` and `TDelta S` with temperature for a chemical system are displayed in the graph.
 
   

  1. Calculate `Delta G` for this system at 300 K.   (2 marks)

  2. What can be deduced about the system when the temperature is `T_1`, `T_2` and `T_3`? Support your answer with reference to the graph.  (4 marks)

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a.  `-15\ text{kJ/mol}`

b.   → For all three reactions `ΔH < 0` (all are exothermic).

→ The entropy of the reaction, `ΔS` is negative as `TΔS` is negative (`T>0`).

→ From the relationship `ΔG = ΔH − TΔS`, we can deduce whether the reaction will be spontaneous (`ΔG<0`) or non-spontaneous (`ΔG>0`). 

→ At `text{T}_1`:  `ΔH` is more negative than `– TΔS`, and so `ΔG<0`. Thus the reaction is spontaneous.

→ At `text{T}_2`: `ΔH` is equal to `– TΔS`, and so `ΔG=0`. Thus the system is in equilibrium.

→ At `text{T}_3`: `ΔH` is less negative than `– TΔS`, and so `ΔG<0`. Thus the reaction is non-spontaneous.

Show Worked Solution

a.   At Temperature = 300K:

`TΔS = – 78\ text{kJ/mol},  ΔH = –93\ text{kJ/mol}`

`ΔG` `=ΔH-TΔS `  
  `=-93-(-78)`  
  `=-15\ text{kJ/mol}`  

 
b.   → For all three reactions `ΔH < 0` (all are exothermic).

→ The entropy of the reaction, `ΔS` is negative as `TΔS` is negative (`T>0`).

→ From the relationship `ΔG = ΔH − TΔS`, we can deduce whether the reaction will be spontaneous (`ΔG<0`) or non-spontaneous (`ΔG>0`). 

→ At `text{T}_1`:  `ΔH` is more negative than `– TΔS`, and so `ΔG<0`. Thus the reaction is spontaneous.

→ At `text{T}_2`: `ΔH` is equal to `– TΔS`, and so `ΔG=0`. Thus the system is in equilibrium.

→ At `text{T}_3`: `ΔH` is less negative than `– TΔS`, and so `ΔG<0`. Thus the reaction is non-spontaneous.


Mean mark (b) 55%.