A mixture of 0.04289 mol of C2H4, 0.08501 mol of N2, 0.02401 mol of NH3, and 0.05848 mol of C6H6 is placed in a 1.0-L steel pressure vessel at 1258 K. The following equilibrium is established: 3 C2H4(g) + 1 N2(g) 2 NH3(g) + 1 C6H6(g) At equilibrium 0.005819 mol of NH3 is found in the reaction mixture. (a) Calculate the equilibrium partial pressures of C2H4, N2, NH3, and C6H6.
A mixture of 0.04289 mol of C2H4, 0.08501 mol of N2, 0.02401 mol of NH3, and 0.05848 mol of C6H6 is placed in a 1.0-L steel pressure vessel at 1258 K. The following equilibrium is established:
3 C2H4(g) + 1 N2(g) 2 NH3(g) + 1 C6H6(g)
At equilibrium 0.005819 mol of NH3 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 chemical reaction is as follows:
3 C2H4(g) + 1 N2(g) -----------> 2 NH3(g) + 1 C6H6(g).............................(1).
The initial concentrations of the equilibrium species are,
[C2H4(g)]=
[N2(g)]=
[NH3(g)]=
[C6H6(g)]=
We can also say that moles and the concentrations are equal for simplified calculation.
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