A mixture of 0.01076 mol of Cl₂, 0.04718 mol of H₂0, 0.02718 mol of HCl, and 0.01676 mol of O₂ is placed in a 1.0-L steel pressure vessel at 1846 K. The following equilibrium is established: 2 Cl₂(g) + 2 H₂O(g) 4 HCl(g) + 1 O₂(g) At equilibrium 0.001782 mol of Cl₂ is found in the reaction mixture. (a) Calculate the equilibrium partial pressures of Cl₂, H₂O, HCl, and 0₂. Peq(Cl₂)= Peq(H₂O) = Peq (HCI) = Peq(O₂) = (b) Calculate Kp for this reaction. Kp =

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Chapter14: Chemical Equilibrium
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A mixture of 0.01076 mol of Cl₂, 0.04718 mol of H₂0, 0.02718 mol of HCI, and 0.01676 mol of O₂ is placed in a 1.0-L steel
pressure vessel at 1846 K. The following equilibrium is established:
2 Cl₂(g) + 2 H₂O(g) = 4 HCl(g) + 1 O₂(g)
At equilibrium 0.001782 mol of Cl₂ is found in the reaction mixture.
(a) Calculate the equilibrium partial pressures of Cl₂, H₂O, HCl, and 0₂.
Peq(Cl₂)=
Peq(H₂O) =
Peq (HCI)
Peq(O₂) =
=
(b) Calculate Kp for this reaction.
Kp
=
Transcribed Image Text:A mixture of 0.01076 mol of Cl₂, 0.04718 mol of H₂0, 0.02718 mol of HCI, and 0.01676 mol of O₂ is placed in a 1.0-L steel pressure vessel at 1846 K. The following equilibrium is established: 2 Cl₂(g) + 2 H₂O(g) = 4 HCl(g) + 1 O₂(g) At equilibrium 0.001782 mol of Cl₂ is found in the reaction mixture. (a) Calculate the equilibrium partial pressures of Cl₂, H₂O, HCl, and 0₂. Peq(Cl₂)= Peq(H₂O) = Peq (HCI) Peq(O₂) = = (b) Calculate Kp for this reaction. Kp =
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