Question

The thermal decomposition of dimethyl ether ( C H 3 ) 2 O ( g ) → C H 4 ( g ) H 2 ( g ) CO ( g ) is to be carried out in an isothermal 2.00-liter laboratory reactor at 600°C. The reactor is charged with pure dimethyl ether at a pressure of 350 torr. After about two hours, the reactor pressure is 875 torr. Has the reaction proceeded to completion at the end of the two-hour period

Answer 1

The reaction did not go to completion.

Step-by-step explanation:

This problem is solved by realizing that we can calculate the number of moles of ( C H 3 ) 2 O ( g ) initially present , doing the stoichiometry of the reaction to find how many moles of products theoretically should be produced if the reaction is 100 percent complete by using the Ideal Gas law and comparing them.

C H 3 ) 2 O ( g ) → C H 4 ( g ) H 2 ( g ) CO ( g )

PV = nRT ∴ n = PV/RT

Calculation for the # moles initially, n₁ :

P = 350 torr = 350 torr x 1 atm/ 760 torr = 0.46 atm

V = 2.00 L

R= 0.08205 Latm/Kmol

T= 600 +273 K = 873 K

n₁ = 0.46 atm x 2.00 L / (0.08205 Latm/Kmol x 873 K) = 0.013 mol

After the two hours =

P = 875 torr x 1 atm/ 760 torr = 1.15 atm

V = 2.00 L

R = 0.08205 Latm/Kmol

T = 600 + 273 K = 873 (We are told is carried out isothermally)

n₂ = 1.15 atm x 2.00 L/ ( 0.08205 Latm/Kmol x 873 K ) = 0.032 mol

If the reaction proceeded to completion given the sotichiometry of the reaction, we would have three times the number of moles of the reactant ( C H 3 ) 2 O since all the products are gases. So,

0.013 mol ( C H 3 ) 2 O x 3 mol products / 1 mol = 0.039 mol

The calculation for the number of moles of gases present at the end of the two hours was 0.032 mol is less than the theoretical yield so the reaction did not go to completion.