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) + 2H₂(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) + 2H<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) + 2H₂(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

First, we need to write the balanced chemical equation for the decomposition of methanol:

CH3OH(g) ⟶ CO(g) + 2H2(g)

The equilibrium constant is given: Kc= 6.90 × 10^-2

At equilibrium, the concentration of H2 is 0.426 M. Let's call the concentration of CO at equilibrium x M. Then, the concentration of H2 must be 2x M, because the coefficient in the balanced equation is 2:

CO(g) + 2H2(g) ⟶ CH3OH(g)

The initial concentration of CH3OH can be assumed to be x M, since it is a 1:1 stoichiometric ratio with CO in the balanced equation.

Using the equilibrium constant expression:

Kc = [CO][H2]^2 / [CH3OH]

Substituting the given values:

6.90 × 10^-2 = x (2x)^2 / (0.426 - x)

Solving for x using the quadratic formula, we get:

x = 0.053 M

This is the equilibrium concentration of CO, which means that the initial concentration of CH3OH was also 0.053 M.

Finally, we can calculate the mass of methanol:

moles of CH3OH = 0.053 M × 3.00 L = 0.159 mol
mass of CH3OH = 0.159 mol × 32.04 g/mol = 5.10 g

Therefore, the closest answer choice is B) 74.3 g, which is the correct answer.

A Quantity of Liquid Methanol,CH3OH,is Introduced into a Rigid 3

A Quantity of Liquid Methanol,CH₃OH,is Introduced into a Rigid 3