A mixture of 0.08069 mol of C2H4, 0.04804 mol of N2, 0.06868 mol of NH3, and 0.1210 mol of C6H6 is placed in a 1.0-L steel pressure vessel at 720 K. The following equilibrium is established:3 C2H4(g) + 1 N2(g) 2 NH3(g) + 1 C6H6(g)At equilibrium 0.02900 mol of N2 is found in the reaction mixture.(a) Calculate the equilibrium partial pressures of C2H4, N2, NH3, and C6H6.Peq(C2H4) = .Peq(N2) = .Peq(NH3) = .Peq(C6H6) = .(b) Calculate KP for this reaction.KP = .

The given data is as follows, the mol of C2H4 = 0.08069 mol the mol of N2 = 0.04804 mol the mol of…

Answered: A mixture of 0.08069 mol of C2H4,…

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

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A mixture of 0.08069 mol of C₂H₄, 0.04804 mol of N₂, 0.06868 mol of NH₃, and 0.1210 mol of C₆H₆ is placed in a 1.0-L steel pressure vessel at 720 K. The following equilibrium is established:

3 C₂H₄(g) + 1 N₂(g) 2 NH₃(g) + 1 C₆H₆(g)

At equilibrium 0.02900 mol of N₂ is found in the reaction mixture.

(a) Calculate the equilibrium partial pressures of C₂H₄, N₂, NH₃, and C₆H₆.

P_eq(C₂H₄) =  .

P_eq(N₂) =  .

P_eq(NH₃) =  .

P_eq(C₆H₆) =  .

(b) Calculate K_P for this reaction.

K_P =  .

Expert Solution

Step 1

The given data is as follows,

the mol of C₂H₄= 0.08069 mol

the mol of N₂ = 0.04804 mol

the mol of NH₃ = 0.06868 mol

the mol of C₆H₆ = 0.1210 mol

the volume vessel = 1 L

the temperature is given = 720 K

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