A mixture of 0.02445 mol of C2H6, 0.05019 mol of N2, 0.02122 mol of NH3, and 0.03692 mol of C2H4 is placed in a 1.0-L steel pressure vessel at 1808 K. The following equilibrium is established:3 C2H6(g) + 1 N2(g) 2 NH3(g) + 3 C2H4(g)At equilibrium 0.01391 mol of NH3 is found in the reaction mixture.(a) Calculate the equilibrium partial pressures of C2H6, N2, NH3, and C2H4.Peq(C2H6) = .Peq(N2) = .Peq(NH3) = .Peq(C2H4) = .(b) Calculate KP for this reaction.

The given reaction is : 3C2H6(g) + 1N2(g) ⇌ 2NH3(g) + 3C2H4(g) Inital moles of C2H6=0.02445…

Answered: A mixture of 0.02445 mol of C2H6,…

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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A mixture of 0.02445 mol of C₂H₆, 0.05019 mol of N₂, 0.02122 mol of NH₃, and 0.03692 mol of C₂H₄ is placed in a 1.0-L steel pressure vessel at 1808 K. The following equilibrium is established:

3 C₂H₆(g) + 1 N₂(g) 2 NH₃(g) + 3 C₂H₄(g)

At equilibrium 0.01391 mol of NH₃ is found in the reaction mixture.

(a) Calculate the equilibrium partial pressures of C₂H₆, N₂, NH₃, and C₂H₄.

P_eq(C₂H₆) =  .

P_eq(N₂) =  .

P_eq(NH₃) =  .

P_eq(C₂H₄) =  .

(b) Calculate K_P for this reaction.

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