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Kp-Kc (RT)An K=(°C+273.15°C)(C) An= moles of gaseous products-moles of gaseous reactants R=0.08206 L'atm mol-K 1. The eq
Posted: Thu Jul 07, 2022 2:17 pm
by answerhappygod
- Kp Kc Rt An K C 273 15 C C An Moles Of Gaseous Products Moles Of Gaseous Reactants R 0 08206 L Atm Mol K 1 The Eq 1 (33.79 KiB) Viewed 54 times
Kp-Kc (RT)An K=(°C+273.15°C)(C) An= moles of gaseous products-moles of gaseous reactants R=0.08206 L'atm mol-K 1. The equilibrium constant K₁ for the reaction: CO₂(g) CO(g) + 1/2O₂(g) The reaction below has an equilibrium constant K₂. 2 CO(g) + O₂(g) 2 CO₂(g) K₂ How are K₁ and K₂ related? Calculate the value of K₂. 2. Using this data, 2 NO(g) + Cl₂(g) 2 NOCI(g) 2 NO₂(g) = 2 NO(g) + O₂(g) calculate a value for K3 for the reaction, NOCI(g) + O2(g) NO₂(g) + Cl₂(g) is 6.66 X 10-¹2 at 1000 K. K₁ = 3.20 x 10-³ K₂ = 15.5