A mixture of 0.02431 mol of C2H6, 0.02411 mol of N2, 0.03217 mol of NH3, and 0.02028 mol of C2H4 is placed in a 1.0-L steel pressure vessel at 1844 K. The following equilibrium is established:3 C2H6(g) + 1 N2(g) 2 NH3(g) + 3 C2H4(g)At equilibrium 0.01316 mol of C2H4 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.KP =.

Given that : Moles of C2H6=0.02431 molMoles of N2=0.02411 molMoles of NH3=0.03217 molMoles of…

Answered: A mixture of 0.02431 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...

See similar textbooks

Related questions

Question

A mixture of 0.02431 mol of C₂H₆, 0.02411 mol of N₂, 0.03217 mol of NH₃, and 0.02028 mol of C₂H₄ is placed in a 1.0-L steel pressure vessel at 1844 K. The following equilibrium is established:

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

At equilibrium 0.01316 mol of C₂H₄ 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.

K_P =.