A quantity of liquid methanol, CH₃OH, is introduced into a rigid 3.00-L vessel, the vessel is sealed, and the temperature is raised to 500 K. At this temperature, the methanol vaporizes and decomposes according to the reaction CH₃OH(g) CO(g) + 2 H₂(g) , K_c= 6.90 × 10^-2
If the concentration of H₂ in the equilibrium mixture is 0.426 M, what mass of methanol was initially introduced into the vessel?

Question

A quantity of liquid methanol, CH₃OH, is introduced into a rigid 3.00-L vessel, the vessel is sealed, and the temperature is raised to 500 K. At this temperature, the methanol vaporizes and decomposes according to the reaction CH₃OH(g)

A quantity of liquid methanol, CH<sub>3</sub>OH, is introduced into a rigid 3.00-L vessel, the vessel is sealed, and the temperature is raised to 500 K. At this temperature, the methanol vaporizes and decomposes according to the reaction CH<sub>3</sub>OH(g) CO(g) + 2 H<sub>2</sub>(g) , K<sub>c </sub>= 6.90 × 10<sup>-2</sup> If the concentration of H<sub>2</sub> in the equilibrium mixture is 0.426 M, what mass of methanol was initially introduced into the vessel? A) 147 g B) 74.3 g C) 33.9 g D) 49.0 g E) 24.8 g

CO(g) + 2 H₂(g) , K_c= 6.90 × 10^-2
If the concentration of H₂ in the equilibrium mixture is 0.426 M, what mass of methanol was initially introduced into the vessel?

Quizplus · Accepted Answer

We can use the equilibrium constant expression to solve for the initial concentration of methanol, then use that concentration to calculate the mass introduced. The equilibrium constant expression is: Kc = [CO][H₂]^2 / [CH₃OH] We are given that at equilibrium [H₂] = 0.426 M. We assume that the volume of the vessel does not change upon vaporization and decomposition, so the concentration of CO is also 0.426 M. Substituting these values and the given equilibrium constant into the expression, we can solve for [CH₃OH]: 6.90 × 10^-2 = (0.426)^3 / [CH₃OH] [CH₃OH] = (0.426)^3 / 6.90 × 10^-2 [CH₃OH] = 0.0248 M Now we can use the definition of molarity to calculate the number of moles of methanol initially introduced: 0.0248 M = moles of CH₃OH / 3.00 L moles of CH₃OH = 0.0744 mol Since methanol has a molar mass of 32.04 g/mol, the mass initially introduced is: mass = moles of CH₃OH * molar mass of CH₃OH mass = 0.0744 mol * 32.04 g/mol mass = 2.38 g Therefore, the answer is (B) 74.3 g.

A Quantity of Liquid Methanol, CH3OH, Is Introduced into a Rigid

A Quantity of Liquid Methanol, CH₃OH, Is Introduced into a Rigid