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

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Presentation transcript:

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 II: NO (g) + ½ O 2(g)  NO 2(g) ΔH = -57kJmol -1 III: 2NO 2(g) + ½ O 2(g)  N 2 O 5(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: N 2(g) + O 2(g)  2NO (g) ΔH = +180kJmol -1 ∆G o = ∆H o - T∆S o 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 Δ H f O : 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.

3 b) iii) Use the data given to calculate the enthalpy change of formation of dinitrogen pentoxide. I: N 2(g) + O 2(g)  2NO (g) ΔH = +180kJmol -1 II: NO (g) + ½ O 2(g)  NO 2(g) ΔH = -57kJmol -1 III: 2NO 2(g) + ½ O 2(g)  N 2 O 5(g) ΔH = -55kJmol -1 Δ H f O = +180kJmol -1 + (2 x -57kJmol -1 ) + (-55kJmol -1 ) Δ H f O = +11kJmol -1

Thankyou! Questions?