A mixture of 0.03185 mol of C 2 H 6 , 0.03696 mol of N 2 , 0.03924 mol of NH 3 ,
ID: 816376 • Letter: A
Question
A mixture of 0.03185 mol of C2H6, 0.03696 mol of N2, 0.03924 mol of NH3, and 0.03880 mol of C2H4 is placed in a 1.0-L steel pressure vessel at 2455 K. The following equilibrium is established:
At equilibrium 0.02588 mol of C2H6 is found in the reaction mixture.
A) Calculate the equilibrium partial pressures of C2H6, N2, NH3, and C2H4.
B) Calculate KP for this reaction.
A mixture of 0.03185 mol of C2H6, 0.03696 mol of N2, 0.03924 mol of NH3, and 0.03880 mol of C2H4 is placed in a 1.0-L steel pressure vessel at 2455 K. The following equilibrium is established: At equilibrium 0.02588 mol of C2H6 is found in the reaction mixture. A) Calculate the equilibrium partial pressures of C2H6, N2, NH3, and C2H4. B) Calculate KP for this reaction.Explanation / Answer
3 C2H6(g) + 1 N2(g) <--------------------> 2 NH3(g) + 3 C2H4(g)
I 0.03185 0.03696 0.03924 0.03880
C -3x -x +2x +3x
E 0.03185-3x 0.03696-x 0.03924+2x 0.03880+3x
Given, equilibrium concentration of C2H6 = 0.02588 mol
Thus, 0.03185-3x = 0.02588 mol
3x = 5.97 x 10^-3 mol
x = 1.99 x 10^-3 mol
Thus,
A)
[C2H6] = 0.03185-3x = 0.02588 mol
PV = nRT
P = (0.02588)(0.0821)(2455) / (1.0) = 5.22 atm
[P C2H6] = 5.22 atm
[N2] = 0.03696-x = 0.03696 - (1.99 x 10^-3) = 0.03497 mol
P = (0.03497)(0.0821)(2455) / (1.0) = 7.05 atm
[P N2] = 7.05 atm
[NH3] = 0.03924+2x = 0.03924 + 2(1.99 x 10^-3) = 0.04322 mol
P = (0.04322)(0.0821)(2455) / (1.0) = 8.71 atm
[P NH3] = 8.71 atm
[C2H4] = 0.03880+3x = 0.03880 + 3(1.99 x 10^-3) = 0.04477 mol
P = (0.04477)(0.0821)(2455) / (1.0) = 9.02 atm
[P C2H4] = 9.02 atm
------------------------
B)
Kp = [P NH3]^2 [P C2H4]^3 / [[P C2H6]^3 [N2]
Kp = (8.71)^2 (9.02)^3 / (5.22)^3 (7.05)
Kp = 55.52
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