The thermal decomposition of dimethyl ether (CH3)2O(g) → CH4(g) + H2(g) + CO(g) is to be...

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The thermal decomposition of dimethyl ether (CH3)2O(g) → CH4(g) + H2(g) + CO(g) is to be...

The thermal decomposition of dimethyl ether

(CH3)2O(g) → CH4(g) + H2(g) + CO(g)

is to be carried out in an isothermal, 3.00-liter laboratory reactor at 700.0°C.

The reactor is charged with pure dimethyl ether at a pressure of 350.0 torr. After about two hours, the reactor pressure is 925.0 torr.

What percentage of the dimethyl ether has decomposed? (round to three decimal places)

science chemistry

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Answer 1

Answer #1

Value of gas constant R is 62.364 L Torr mol^-¹K^-¹.

Given

T=700C= 973K

V=3L

initial Pressure =350 Torr

final Pressure =925 Torr

Using ideal gas equation

PV=nRT

n=PV/RT

So, initial moles= 350*3/(62.364*973)

=.0173

and final moles= 925*3/(62.364*973)

= .04573

If x moles of pure dimethyl ether has dissociated, then number of final moles in the system=

.173-x + x(CH4 formed) +x(H2 formed) +x(CO formed)

=.0173+2x which is equal to .04573

On solving x=.014215

Therefore % decomposition = (no of moles dissociated/ no of initial moles)*100

=82.167%

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Answer 2

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