Chemistry Tutorial 5
Chemical energetics
and Thermodynamics
Question 3:
Leslie Tan Zheng Yu 12S7F
The question
• Dinitrogen pentoxide N2O5 can be produced by
the following reaction sequence:
• I: N2(g) + O2(g)  2NO(g) ΔH = +180kJmol-1
• II: NO(g) + ½ O2(g)  NO2(g) ΔH = -57kJmol-1
• III: 2NO2(g) + ½ O2(g)  N2O5(g) ΔH = -55kJmol-1
3 a) i) Explain why reaction I
occurs in car engines.
• Approximately 70% of air consists of nitrogen gas.
• Nitrogen gas can be oxidized at high temperature
and pressure in internal combustion engine of
vehicles.
3 a) ii) Suggest why reaction I is
endothermic.
• I: N2(g) + O2(g)  2NO(g) ΔH = +180kJmol-1
• ∆Go = ∆Ho - T∆So
• Reaction can only take place at high temperature.
• ΔG < 0 when temperature is sufficiently high.
o Meaning reaction is spontaneous
• This means that ΔH must be positive.
3 b) i) Explain what is meant by the standard
enthalpy change of formation of a compound?
• Standard enthalpy change of formation ΔHfO : The
enthalpy change when one mole of a substance is
formed from its constituent elements in their
standard state at 1 atm and 298k (standard
conditions).
3 b) ii) Write an equation which corresponds to the
enthalpy change of formation of dinitrogen pentoxide.
• 2N2(g) + 5O2(g)  2N2O5(g)
• Since
• Standard enthalpy change of formation ΔHfO : The
enthalpy change when ONE MOLE of a substance is
formed from its constituent elements in their
standard state at 1 atm and 298k (standard
conditions).
• Hence,
𝟓
𝟐
• N2(g) + O2(g)  N2O5(g)
3 b) iii) Use the data given to calculate the enthalpy
change of formation of dinitrogen pentoxide.
• I: N2(g) + O2(g)  2NO(g) ΔH = +180kJmol-1
• II: NO(g) + ½ O2(g)  NO2(g) ΔH = -57kJmol-1
• III: 2NO2(g) + ½ O2(g)  N2O5(g) ΔH = -55kJmol-1
• ΔHfO = +180kJmol-1 + (2 x -57kJmol-1) + (-55kJmol-1)
• ΔHfO = +11kJmol-1
Thankyou! Questions?