Question

The thermal decomposition of dimethyl ether

(CH₃)₂O(g) -> CH₄(g) + H₂(g) + CO(g)

Is to be carried out in an isothermal 2.0 L lab reactor at 600 ^oC. The reactor is charged with pure dimethyl ether at a pressure of 349 torr. After about two hours, the reactor pressure is 840 torr.

Calculate the percentage of the dimethyl ether has decomposed.

Answer 1

PV = nRT

n = PV/(RT)

n = (349)(2)/(62.3*(600+273)) = 0.0128337 mol of (CH3)2O

after t = 2 h

then

P = 840 torr..

this must be then

Ptotal = P(CH3)2O + PCH4 + ¨PH2 + PCO2

moles:

n = (840)(2)/(62.3*(600+273)) = 0.03088

assume

mol of ethyl ether = 0.0128337 - x

mol of CH4 = 0 + x

mol of H2 = 0 + x

mol of CO = 0 + x

solfe for x

Ptotal = P(CH3)2O + PCH4 + ¨PH2 + PCO

total mol = mol of (CH3)2O + mol of CH4 + mol of H2 + mol of CO

note that mol of CH4 = mol of H2 = mol of CO

so

0.03088 = ( 0.0128337 - x) + 3x

2x = 0.03088 -0.0128337

x = ( 0.03088 -0.0128337 )/2

x = 0.00902315

so..

mol of ethyl ether = 0.0128337 - x = 0.0128337 -0.00902315 = 0.00381055 mol left..

% decomposition = (0.0128337 -0.00381055)/0.0128337 *100 = 70.3082% is already reacted