1.2 mol of CH3OH(g) are injected into a 2.0 L container and the following equilibrium becomes established.

2H2(g) + CO(g) <=====> CH3OH(g) + 92 kJ
If at equilibrium 1.0 mol of CH3OH is still in the container the Ke must be which of the following?
a. 25 d. 12.5
b. 125 e. 0.032
c. 0.0080

Doing practice problems. Thanks!

Write the equation.

Set up an ICE Chart.
Substitute in Keq expression and solve for Keq.

25

utv

It's 125

To determine the value of the equilibrium constant (Kₑ), we need to use the given information about the moles of reactants and products at equilibrium. Kₑ is defined as the ratio of the product concentrations to the reactant concentrations, with each concentration raised to the power of its stoichiometric coefficient.

In this case, we have the equilibrium equation: 2H2(g) + CO(g) ⇌ CH3OH(g) + 92 kJ

According to the equation, the stoichiometric coefficient of CH3OH is 1, which means the concentration of CH3OH is directly related to its number of moles.

Given that at equilibrium there is 1.0 mol of CH3OH in the container, the concentration of CH3OH can be calculated by dividing the number of moles by the volume of the container.

Concentration of CH3OH = (1.0 mol) / (2.0 L) = 0.5 M

Now that we have the concentration of CH3OH, we need to determine the concentrations of H2 and CO. However, without additional information about the moles of H2 and CO at equilibrium, we cannot directly calculate their concentrations. Additionally, the equilibrium constant can only be determined from concentrations, not from the number of moles.

Therefore, based on the information given, it is not possible to determine the value of the equilibrium constant (Kₑ). None of the provided answer choices (a, b, c, d, and e) is correct.

It's important to note that to accurately calculate the equilibrium constant, we need either the ratio of reactant and product concentrations or their initial concentrations.