A mixture of 0.02233 mol of C2H6, 0.02937 mol of N2, 0.03719 mol of NH3, and 0.03411 mol of C2H4 is placed in a 1.0-L steel pressure vessel at 2646 K. The following equilibrium is established:3 C2H6(g) + 1 N2(g) 2 NH3(g) + 3 C2H4(g)At equilibrium 0.01682 mol of C2H6 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 : Initial moles of C2H6=0.02233 molInitial moles of N2=0.02937 molInitial moles of…

Answered: A mixture of 0.02233 mol of C2H6,…

Principles of Modern Chemistry

8th Edition

ISBN:9781305079113

Author:David W. Oxtoby, H. Pat Gillis, Laurie J. Butler

Publisher:David W. Oxtoby, H. Pat Gillis, Laurie J. Butler

Chapter14: Chemical Equilibrium

Section: Chapter Questions

Problem 76AP: Methanol can be synthesized by means of the equilibriumreaction CO(g)+2H2(g)CH3OH(g) for which the...

See similar textbooks

Related questions

Question

100%

A mixture of 0.02233 mol of C₂H₆, 0.02937 mol of N₂, 0.03719 mol of NH₃, and 0.03411 mol of C₂H₄ is placed in a 1.0-L steel pressure vessel at 2646 K. The following equilibrium is established:

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

At equilibrium 0.01682 mol of C₂H₆ 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 =

Expert Solution

Step by step

Solved in 5 steps

SEE SOLUTION Check out a sample Q&A here

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.

Similar questions