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Chemistry Tutorial 5 Chemical energetics and Thermodynamics

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: N 2(g) + O 2(g)  2NO (g) ΔH = +180kJmol -1

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Chemistry Tutorial 5 Chemical energetics and Thermodynamics

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  1. Chemistry Tutorial 5Chemical energetics and Thermodynamics Question 3: Leslie Tan Zheng Yu 12S7F

  2. The question • Dinitrogenpentoxide 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. 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.

  4. 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. • Meaning reaction is spontaneous • This means that ΔH must be positive.

  5. 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).

  6. 3 b) ii) Write an equation which corresponds to the enthalpy change of formation of dinitrogenpentoxide. • 2N2(g) + 5O2(g) 2N2O5(g) • Since • Standard enthalpy change of formation ΔHfO: The enthalpy change when ONE MOLEof a substance is formed from its constituent elements in their standard state at 1 atm and 298k (standard conditions). • Hence, • N2(g) + O2(g)  N2O5(g)

  7. 3 b) iii) Use the data given to calculate the enthalpy change of formation of dinitrogenpentoxide. • 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

  8. Thankyou! Questions?

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