The thermal decomposition of ethylene occurs in many industrial contexts: during

Question

The thermal decomposition of ethylene occurs in many industrial contexts: during ethylene transit in piplines, formation of polyethylene, drying of the gas and seperating it from impurities. The decomposition reaction is as follows:

CH2 + CH2(g) -> CH4(g) + C (graphite)

If the decomposiiton of 1 mole of ethylene with a gas density of 0.215 g/mL starts at 27C and the temperature increases by 950 K, what is the final pressure of the confined gas? (Ignore the volume of graphite and use the van der Waals equation.)

Explanation / Answer

CH2 + CH2(g) -> CH4(g) + C

Assume only gas final is CH4

for methane

a = 2.283 = 0.2283  m6·Pa/mol2

b = 0.04278 L/mol = 4.278*10^-5 m3/mol

VDW equation

(P+a(n/V)^2*(V - nb) = nRT

we have, a, b, n, R, T, find Volume

MW ethylente = 16 g/gmol

n/V = 0.215 g/ml * (1000 ml/L) / ( 16g/gmol) = 13.437 gmol/L = 13437 gmol/m3

if we use 1 gmol then n/V = 13437 gmol/m3

V = 1/13437 = 7.44*10^-5 m3

Now subsitutein VDW

(P+a(n/V)^2*(V - nb) = nRT

(P + 0.2283 (13437)^2) * (7.44*10^-5 - 1*4.278*10^-5) = (1)*(8.314)(950)

(P + 41220242) * (3.162*10^-5) = 7898

Solve for P

P = 208567866 Pa

P = 208567.8 kPa

P = 208.5 MPa

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