Question

A mixture of 0.09581 mol of C2H4, 0.02766 mol of N2, 0.04516 mol of NH3, and 0.08098 mol of C6H6 is placed in a 1.0-L steel pressure vessel at 731 K.

The following equilibrium is established:

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

At equilibrium 0.006932 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 = .

Answer 1

Since all is in V = 1L, then assume moles = concnetration since M = mol/L and L = 1

then

the reaction: 3 C2H4(g) + 1 N2(g) ---> 2 NH3(g) + 1 C6H6(g)

the K

K = [NH3]^2[C6H6]/[C2H4]^3[N2]

[NH3] = 0.04516

[C6H6] = 0.08098

[C2H4] =0.09581

[N2] = 0.02766

in equilibrium; due to stocihiometric coefficients:

[NH3] = 0.04516 +2x

[C6H6] = 0.08098 +x

[C2H4] =0.09581 -3x

[N2] = 0.02766 -x

and

we know that

[N2] = 0.02766 -x = 0.006932

then

x = -(0.006932-0.02766 ) = 0.020728

substitute in all

[NH3] = 0.04516 +2*0.020728= 0.086616

[C6H6] = 0.08098 +0.020728 = 0.101708

[C2H4] =0.09581 -3*0.020728 = 0.033626

[N2] = 0.02766 -0.020728 = 0.006932

change all to P as follows:

PV = nRT

P = n/V*RT

P = M*RT

R = 0.082 anT = 731

RT = 0.082*731 = 59.942

therefore

P-NH4 = 0.086616*59.942 = 5.191936

P-C6H6 =0.101708*59.942 = 6.0965

P-C2H4 = 0.033626*59.942 = 2.0156

P-N2= 0.006932*59.942 =0.4155

Kp

Kp = PNH3^2*PC6H6/*C2H4^3*PN2

Kp = (5.191936^2)(6.0965)/((2.0156^3)(0.4155)) = 48.30089

Kp = 48.30089