A 1.0 L reaction vessel contained 0.750 mol of CO(g) and 0.275 mol of H2O(g). After 1.0 h, equilibrium was reached. Analysis showed that 0.25 mol of CO2(g) was present. The equation for the reaction is:
CO(g) + H2O(g) ↔ CO2(g) + H2(g)
Find Keq for the reaction.
(CO) = mols/L = 0.750/1.0 = 0.750 M
(H2O) = 0.275 M
(CO2) = 0.25 M
.........................CO(g) + H2O(g) ↔ CO2(g) + H2(g)
I........................0.750.......0.275...........0
C..........................-x...........-x.................+x
E....................0.750-x.....0.275-x...........0.250
Keq = (CO2)/(CO)(H2O)
Plug the E line into Keq expression and solve for x. Then evaluate the concentrations of CO, H2O and CO2 and solve for Keq. Post your work if you get stuck.
To find Keq for the reaction, we need to use the concentrations of the substances at equilibrium.
Given:
Initial amount of CO(g) = 0.750 mol
Initial amount of H2O(g) = 0.275 mol
Amount of CO2(g) at equilibrium = 0.25 mol
To find the concentration of H2(g) at equilibrium, we can use the stoichiometry of the reaction:
CO(g) + H2O(g) ↔ CO2(g) + H2(g)
From the balanced equation, we can see that the ratio between CO(g) to H2(g) is 1:1.
Since the amount of CO(g) at equilibrium is 0.25 mol, the amount of H2(g) at equilibrium is also 0.25 mol.
Now, let's calculate the concentrations at equilibrium:
Moles of CO(g) at equilibrium = 0.25 mol
Volume of the reaction vessel = 1.0 L
Concentration of CO(g) = Moles / Volume
Concentration of CO(g) = 0.25 mol / 1.0 L
Concentration of CO(g) = 0.25 M
Moles of H2(g) at equilibrium = 0.25 mol (as calculated above)
Volume of the reaction vessel = 1.0 L
Concentration of H2(g) = Moles / Volume
Concentration of H2(g) = 0.25 mol / 1.0 L
Concentration of H2(g) = 0.25 M
Moles of CO2(g) at equilibrium = 0.25 mol
Volume of the reaction vessel = 1.0 L
Concentration of CO2(g) = Moles / Volume
Concentration of CO2(g) = 0.25 mol / 1.0 L
Concentration of CO2(g) = 0.25 M
Now that we have the concentrations of each species at equilibrium, we can use the equation for calculating Keq:
Keq = [CO2(g)] * [H2(g)] / ([CO(g)] * [H2O(g)])
Keq = (0.25 M * 0.25 M) / (0.25 M * 0.275 M)
Keq = 0.0625 / 0.06875
Keq = 0.9091
Therefore, the value of Keq for the reaction is approximately 0.9091.
To find Keq for the reaction, we need to use the equation relating the equilibrium constant (Keq) to the concentrations of the reactants and products at equilibrium.
The given equation is: CO(g) + H2O(g) ↔ CO2(g) + H2(g)
According to the stoichiometry of the reaction, the equilibrium constant expression can be written as:
Keq = ([CO2] * [H2]) / ([CO] * [H2O])
Here, [CO2], [H2], [CO], and [H2O] represent the molar concentrations of CO2, H2, CO, and H2O, respectively, at equilibrium.
To find these concentrations, we need to consider the initial amounts and the changes that occurred in the reaction.
Given:
Initial moles of CO = 0.750 mol
Initial moles of H2O = 0.275 mol
Change in moles of CO2 = 0.25 mol
Since CO2 is a product of the reaction, its moles increased by 0.25 mol. This means that the moles of CO decreased by 0.25 mol.
Now, let's calculate the moles of CO and H2O at equilibrium:
Moles of CO at equilibrium = Initial moles of CO - Change in moles of CO = 0.750 mol - 0.25 mol = 0.500 mol
Moles of H2O at equilibrium = Initial moles of H2O = 0.275 mol
Next, let's calculate the moles of CO2 and H2 at equilibrium:
Moles of CO2 at equilibrium = Change in moles of CO2 = 0.25 mol
Moles of H2 at equilibrium = Change in moles of CO2 = 0.25 mol
Now, we can substitute these values into the equilibrium constant expression to calculate Keq:
Keq = ([CO2] * [H2]) / ([CO] * [H2O])
= (0.25 mol * 0.25 mol) / (0.500 mol * 0.275 mol)
= 0.0625 mol² / 0.1375 mol²
≈ 0.4545
Therefore, the value of Keq for the reaction is approximately 0.4545.