Question

A mixture of 0.05778 mol of C2H4, 0.1509 mol of N2, 0.1433 mol of NH3, and...

A mixture of 0.05778 mol of C2H4, 0.1509 mol of N2, 0.1433 mol of NH3, and 0.1387 mol of C6H6 is placed in a 1.0-L steel pressure vessel at 742 K. The following equilibrium is established: 3 C2H4(g) + 1 N2(g) 2 NH3(g) + 1 C6H6(g) At equilibrium 0.02469 mol of C2H4 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 = .

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Answer #1

a)

3 C2H4(g) + N2(g) <-----> 2 NH3(g) + C6H6(g)

0.05778___0.1509______0.1433___0.1387

0.05778-3x__0.1509-x___0.1433+2x__0.1387+x

0.05778-3x= 0.02469 ---> x= 0.01103mol

mol C2H4= 0.02469mol ---> P=0.02469mol x 0.082L.atm/mol.K x 742K/1L= 1.5 atm

mol N2=0.1509 - 0.01103= 0.13987mol ---> P=0.13987mol x 0.082L.atm/mol.K x 742K/1L= 8.51atm

mol NH3= 0.1433 + 2 x 0.01103= 0.16536mol ----> P= 0.16536mol x 0.082L.atm/mol.K x 742K/1L= 10.06atm

mol C6H6= 0.1387 + 0.01103= 0.14973mol -----> P= 0.14973mol x 0.082L.atm/mol.K x 742K/1L= 9.11atm

b) Kp = PC6H6xPNH32/PC2H43 x PN2= 32.1

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