A quantity of liquid methanol, CH 3OH, is introduced into a rigid 3.00-L vessel, the vessel is sealed, and the temperature is raised to 500K. At this temperature, the methanol vaporizes and decomposes according to the reaction CH 3OH(g) CO(g) + 2 H 2(g), K c= 6.90×10 –2. If the concentration of H 2 in the equilibrium mixture is 0.426M, what mass of methanol was initially introduced into the vessel?

74.3g of methanol were introduced into the vessel

Explanation:

In the equilibrium:

CH₃OH(g) ⇄ CO(g) + 2H₂(g)

Kc is defined as the ratio between concentrations in equilibrium of :

Kc = 6.90x10⁻² = [CO] [H₂]² / [CH₃OH]

Some methanol added to the vessel will react producing H₂ and CO. And equilibrium concentrations must be:

[CH₃OH] = ? - X

[CO] = X

[H₂] = 2X

Where ? is the initial concentration of methanol

As [H₂] = 2X = 0.426M; X = 0.213M

[CH₃OH] = ? - 0.213M

[CO] = 0.213M

[H₂] = 0.426M

Replacing in Kc to solve equilibrium concentration of methanol:

6.90x10⁻² = [0.213] [0.426]² / [CH₃OH]

[CH₃OH] = 0.560

As:

[CH₃OH] = ? - 0.213M = 0.560M

? = 0.773M

0.7733M was the initial concentration of methanol. As volume of vessel is 3.00L, moles of methanol are:

3.00L * (0.773 mol / L) = 2.319 moles methanol.

Using molar mass of methanol (32.04g/mol), initial mass of methanol added was:

2.319 moles * (32.04g / mol) =

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