Consider the following equilibrium system at 900°C:
H20(g) + CO(g) <==> H2(g) + CO2(g)
Initially 5.0 moles of H2O and 4.0 moles of CO were reacted. At equilibrium, it is found that 2.0 moles of H2 are present. How many moles of H2O remain in the mixture?
Would be greatly thankful if someone could help me!
I have removed the (g) designation in order to get all of this on one line.
..........H20 + CO <==> H2 + CO2
I.........5.0...4.0.....0.....0
C.........-x.....-x.....x.....x.
E.......................2.........
If 2.0 mols H2 are present at equilbrium then x must be 2.0 mols H2. That means mols H2O used must have been x = 2.0 so 5.0-2.0=3.0
You also know that mols CO2 at 4quilibrium = 2.0 and mols CO remaining = 4-2=2
Thanks very much!
To determine the number of moles of H2O that remain in the mixture at equilibrium, we need to use the stoichiometry of the balanced equation and the given information.
The balanced equation is:
H2O(g) + CO(g) <==> H2(g) + CO2(g)
We are given the initial moles of H2O as 5.0 and the initial moles of CO as 4.0. At equilibrium, we have 2.0 moles of H2.
To solve the problem, we'll first assume the number of moles of H2O that reacted is x. Then, the mole ratio tells us that x moles of CO also reacted, and thus x moles of H2, and x moles of CO2 are produced.
Using the law of conservation of mass, we can write the equation:
Initial moles of reactants - Moles reacted = Moles at equilibrium
For H2O:
5.0 moles - x moles = Moles at equilibrium
For CO:
4.0 moles - x moles = Moles at equilibrium
Since we know that the moles of H2 at equilibrium is 2.0, we can set up the equation:
x moles of H2 = 2.0 moles
Solving this equation gives us the value of x.
x = 2.0 moles
Now we substitute the value of x into the equation for H2O:
5.0 moles - 2.0 moles = Moles of H2O at equilibrium
Moles of H2O at equilibrium = 3.0 moles
Therefore, 3.0 moles of H2 remains in the mixture at equilibrium.
To find the number of moles of H2O remaining at equilibrium, we can use the principle of the equilibrium constant (Kc) and the stoichiometry of the balanced equation for the reaction.
First, we need to write the expression for the equilibrium constant (Kc) for the given reaction:
Kc = [H2][CO2] / [H2O][CO]
Since we are working with moles, we need to convert the given initial and equilibrium amounts to moles.
Given:
Initial moles of H2O = 5.0 moles
Initial moles of CO = 4.0 moles
Equilibrium moles of H2 = 2.0 moles
Now, we need to determine the concentrations of the substances involved in the equilibrium expression.
For the numerator:
[H2] = 2.0 moles / total volume of the system (unknown in this case)
[CO2] = unknown at equilibrium
For the denominator:
[H2O] = unknown amount remaining at equilibrium
[CO] = 4.0 moles / total volume of the system (unknown in this case)
Since we have the moles of CO, we still need the volume of the system to calculate the initial concentration of CO, which is required to determine the equilibrium concentration of H2O. Unfortunately, the problem does not provide information about the volume, so we cannot solve for the exact number of moles of H2O remaining.
However, we can still analyze the problem qualitatively:
- The equilibrium constant (Kc) is a constant at a given temperature, and it tells us the ratio of product concentrations to reactant concentrations at equilibrium.
- Since we are given the equilibrium concentration of H2 (2.0 moles) and the initial concentrations of H2O (5.0 moles) and CO (4.0 moles), we can set up the equilibrium expression with the given values and solve for [H2O]:
Kc = [H2][CO2] / [H2O][CO]
Using the given values:
Kc = (2.0 moles)(unknown) / (5.0 moles)(4.0 moles)
We can see that [H2O] is present in the denominator, so as [H2O] decreases, the fraction increases, and the equilibrium constant (Kc) would have to be larger than the calculated value to satisfy the given conditions.
Therefore, based on the information provided, we cannot determine the exact number of moles of H2O remaining in the mixture, but we can conclude that less than 5.0 moles of H2O will be present at equilibrium.