Standard entropies can be used to calculate the entropy change of a reaction, ΔS. For example, for the formation of nitrogen monoxide from nitrogen and oxygen.
NO (g) + O3 (g) → NO2 (g) + O2 (g)
Substance
Entropy value (J K-1 mol-1)
NO (g)
210.8
O2 (g)
205.2
NO2 (g)
240.0
O3 (g)
238.9
Use the data given to calculate the entropy change of the reaction between nitric oxide and ozone at 298K.
The contact process is a method used industrially to form sulfur trioxide, by reacting sulfur dioxide and oxygen together over a vanadium(V) oxide catalyst.
The equation for this reaction is shown below:
2SO2 (g) + O2 (g) ⇌ 2SO3 (g)
Substance
Formation enthalpy values (kJ mol-1)
SO2 (g)
-297
SO3 (g)
-395
i)
Calculate the standard enthalpy change of the contact process reaction using the data provided.
ii)
The standard entropy change of this reaction is –189 J K-1 mol-1. Use this value and your enthalpy value from part (i) calculate a value for the free energy change for this reaction at 298K.
iii)
Use your answer to part(ii)to explain whether the reaction is feasible at 298 K.
Question 1d
Marks: 2
d)
The value for the free energy change is an indication whether the forward or backwardreaction is favoured.
The curve that we would expect to see for the reaction between sulfur dioxide and oxygen is shown below.
Explain why the curve for this reaction is shifted to the right hand side.
The enthalpy of solution of sodium chloride is +4 kJ mol-1. Explain why the free energy change for dissolving sodium chloride in water is negative, despite the enthalpy change being a positive value.
Some ionic compounds such as potassium chloride, KCl, will dissolve in water at room temperature in an endothermic process.
KCl (s) → K+ (aq) + Cl- (aq) ΔH = +16 kJ mol-1
Substance
Entropy value J K-1 mol-1
KCl (s)
+83
K+ (aq)
+103
Cl- (aq)
+57
i)
Using the data provided, prove that this process is feasible at 298 K.
ii)
Use your knowledge of structure and bonding to explain why ΔHΘ is positive for this process.
Question 2d
Marks: 3
d)
Diamond and graphite are both allotropes of carbon.
The conversion of graphite into diamond is represented as follows
Carbon (graphite) → Carbon (diamond)
Use this data below to calculate values forΔH and ΔS for the reaction. Use these values to explain why this reaction is not feasible under standard pressure at any temperature.
C (graphite)
C (diamond)
ΔH (kJ mol-1)
0
+1.9
ΔS (J K-1 mol-1)
+5.7
+2.4
Question 3a
Marks: 3
a)
Ethanol is used in large quantities in the production of alcoholic beverages and as a fuel.
The combustion of ethanol is represented by the equation
CH3CH2OH (l) + 3O2 (g) → 2CO2 (g) + 3H2O (g)
The standard entropy, SΘ, of O2 (g) is 205.2 J K-1 mol-1
Using the data given and Section 12 in the Data Booklet, determine the entropy change, ΔSΘ, for the combustion of ethanol at 298K.
Using the enthalpy of combustion for ethanol from Section 13 in the Data Booklet and the ΔSϴ determined in part (a), calculate the standard free energy for the combustion of ethanol.
The boiling point of a liquid is the temperature at which its solid and liquid phases are in equilibrium as shown in the equation for the vaporisation of water.
H2O (l) → H2O (g)
Use Section 12 of the Data Booklet to determine values for the enthalpy change, ΔHϴ, and entropy change, ΔSϴ, for the reaction at 298 K.
