Temperature dependence of equilibrium constant K (thermo-42) Calculate ΔGoR for the combustion reaction temperatures of 298K, 900K and 2300K from N2 to NO. The given temperatures of 900 K correspond to the temperature in an exhaust or in the combustion chamber (2300K), Note: Calculate the reaction for 1 particle (respectively 1 mol) NO.

  Exercise: thermo-42    

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Calculate ΔGoR for the combustion reaction temperatures of 298K, 900K and 2300K from N₂ to NO. The given temperatures of 900 K correspond to the temperature in an exhaust or in the combustion chamber (2300K),
Note: Calculate the reaction for 1 particle (respectively 1 mol) NO.

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Solution:

The following reaction equation is considered: ½·O₂(g) + ½·N₂(g) ⇄ NO (g)
Table of values:

	ΔHof(kJ·mol-1)	So(J·K-1·mol-1)
O2 (g)	0	205.03
N2 (g)	0	191.61
NOO (g)	90.37	210.62

1. reaction equation:
       ½·O₂(g) + ½·N₂(g) ⇄ 1·NO (g)
2. ΔHoR = ∑ΔHof(Produkte) – ∑ΔHof(Edukte)
   ΔHoR = (1· ΔHof(NO,g)) – (½· ΔHof(O₂, g) + ½· ΔHof(N₂,g))
   ΔHoR = (90.37) – (½·0 + ½·0) = +90.37 kJ/mol
3. ΔSoR= S^o(Produkte) – S^o(Edukte)
   ΔSoR = (1·S^o(NO,g)) – (½·S^o(O₂, g) + ½·S^o(N₂,g))
   ΔSoR = (210.62)-(½·205.03 + ½·191.61) = +12.3 J·mol^-1·K^-1
4. ΔGoR = ΔHoR – T· ΔSoR
   ΔGoR(T=298K) = 90.37 – (298)·0.0123 = 86.7 kJ/mol
   ΔGoR(T=900K) = 90.37 – (900)·0.0123 = 79.3 kJ/mol
   ΔGoR(T=2300K) = 90.37 – (2300)·0.0123 = 62.08 kJ/mol
5. with dG = -RTln K follows: K = e^(-dG/R·T), R=0.008314 kJ/mol·K
   ΔGoR(T=298K) = 86.7 kJ/mol → K= 6.34·10^-16
   ΔGoR(T=900K) = 79.3 kJ/mol → K= 2.5·10^-5
   ΔGoR(T=2300K) = 62.08 kJ/mol → K= 0.03

Discussion: it acting itself to a endothermal reaction ( ΔHoR > 0). To promote the reaction So an increase in temperature would be necessary. This is through the calculation the equilibrium constant K confirmed: [translation unavailable] higher the temperature is, the more greater is the K-value self. Es shows itself but, that the balance but also at high temperatures always still relative very strong on the page the reactants is, a application of a catalysts offers itself thus at.

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