Reply To:

Temperature dependence The equilibrium constant K (thermo-42)

Calculate ΔGoR for The Temperatures 298K, 900K and 2300K The Combustion reaction of N₂ to NO. The (c) Temperatures of 900 K corresponding to The Temperature in a Exhausted Responsibility. In the Combustion chamber (2300K),
Notes: Calculate The reaction for 1 Particles (Responsibility. 1 mol) NO.

---

Es will be the following: reaction equation Considered: ½·O₂(g) + ½·N₂(g) ⇄ NO (g)
Table of values:

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 following:: 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: Es Acting The Commission to One Endothermal reaction ( ΔHoR > 0). In order to The reaction to to promote would be So what do you mean? One Temperature increase necessary. This will be by: The calculation The equilibrium constant K Confirmed: The Commission higher The Temperature is, the more Greater will be The K-Value self. Es shows The Commission But, that This Balance But also In the case of high Temperatures Always Not yet Relative very strong on The Page The Educated is, a Application One of them Catalysts offers The Commission So what do you mean? to.