When 0.5 mole of CO2 and 0.5 mole of H2 were forced into a litre reaction container, and equilibrium was established:
CO2(g) + H2(g) <--> H2O (g) + CO (g)
Under the conditions of the experiment, K=2.00
a) Find the equilibrium concentration of each reactant and product.
K = [H20][CO]/[CO2][H2]
2.00 = X2/(0.5-X)^2
2(0.5-X) = X
1-2X = X
1-3X = 0
-3X = -1
X = 1/3
^ is this right?
b) how would the equilibrium concentrations differ is 0.50 mole of H2O and 0.50 moles of CO had been introduced into the reaction vessel instead of the CO2 and H2?
please help :)
Is this K = 2.00 Kp or Kc? I went with Kc.
a is not right. You have made two errors.
1. You took the square root of the left side but you didn't take the square root of 2. I get x = 0.293 mols.
2. You calculated mols but not M
3. You solved for x but not 0.5-x (but since this is in 1L mols = M.
For b, equilibrium is the same whether approached from the left or right as long as moles on the left = moles on the right.
For part a, your calculations are correct. The equilibrium concentration of H2O and CO will be 1/3 mole each, while the equilibrium concentration of CO2 and H2 will be (0.5 - 1/3) = 1/6 mole each.
For part b, let's consider the reaction:
H2O(g) + CO(g) ⇌ CO2(g) + H2(g)
The equilibrium expression for this reaction is:
K' = [CO2][H2] / [H2O][CO]
Here, we are given that K' = 2.00. Now, using the values given (0.50 moles H2O and 0.50 moles CO), we can substitute them into the equation:
2.00 = (0.50)(0.50) / (x)(x)
2.00x^2 = 0.25
x^2 = 0.125
x = √(0.125)
x ≈ 0.354
Therefore, the equilibrium concentrations of CO2 and H2 will be approximately 0.354 mole, and the equilibrium concentrations of H2O and CO will be approximately 0.50 mole each.
So, if 0.50 mole of H2O and 0.50 mole of CO had been introduced into the reaction vessel instead of 0.5 mole each of CO2 and H2, the equilibrium concentrations of the reactants and products would be different.
a) To find the equilibrium concentrations of each reactant and product, we can use the equilibrium constant expression and the given values of moles at the start:
The equilibrium constant expression for the reaction is:
K = [H2O][CO] / [CO2][H2]
Let's start by substituting the given values: [CO2] = 0.5 mol, [H2] = 0.5 mol, and since the initial concentrations of H2O and CO are not given, we'll use variables x and y for their concentrations at equilibrium:
K = (y)(x) / (0.5 - x)(0.5 - x)
Now, using the given equilibrium constant K = 2.00, we can set up the equation and solve for x:
2.00 = (y)(x) / (0.5 - x)(0.5 - x)
Simplifying further, we get:
2(0.5 - x) = y
1 - 2x = y
Now, we can substitute y with 1 - 2x in the equilibrium constant expression:
K = (1 - 2x)(x) / (0.5 - x)(0.5 - x)
Next, we need to square the denominator to continue:
K = (1 - 2x)(x) / (0.5 - x)^2
Now, you have arrived at the same expression in your question. The steps you've shown are correct so far.
Continuing with the calculation, let's solve for x:
2(0.5 - x) = x
1 - 2x = x
1 - 3x = 0
-3x = -1
x = 1/3
Therefore, the equilibrium concentration of each reactant and product is:
[CO2] = 0.5 - x = 0.5 - 1/3 = 1/6 mol/L
[H2] = 0.5 - x = 0.5 - 1/3 = 1/6 mol/L
[H2O] = x = 1/3 mol/L
[CO] = x = 1/3 mol/L
So, the equilibrium concentrations are [CO2] = [H2] = 1/6 mol/L, and [H2O] = [CO] = 1/3 mol/L.
b) If 0.50 mole of H2O and 0.50 mole of CO were introduced instead of CO2 and H2, the stoichiometry of the reaction would change. The new balanced equation for the reaction would be:
CO(g) + H2O(g) <--> CO2(g) + H2(g)
The equilibrium expression will remain the same:
K = [H2O][CO] / [CO2][H2]
Given that the initial moles of H2O = 0.50 mol and CO = 0.50 mol, we can apply the same steps as in part a) to determine the equilibrium concentrations of each species.
Substituting the given values and solving the equation, you will find the equilibrium concentrations for [CO2], [H2], [H2O], and [CO] under these new conditions.