For the following reaction
Co(g)+H2o(g)=Co2(g)+H2(g)
If the initial concentrations of Co and H2o are 1.00 mol in a 50 L vessel, what is the number of mole of each species at equilibrium, if the equilibrium constant Kc= 0.85 at 1000 C?
please can any one explain how to do this ?
the two reactants react in equal amounts to give two products in equal amounts
let x = concentration of each product
Kc = 0.85 = [x]^2 / [1.00 - x]^2
.85 - 1.7 x + .85 x^2 = x^2
.15 x^2 + 1.7 x - .85 = 0
use quadratic formula to find x
moles = concentration * vessel size
should I use 1mole/50L =0.02 M instead of using 1 ?
Yes, you should use 1/50, calculate as shown above but x will be concentration and not mols. Then convert x = concentration in mols/L to mols.
To solve this problem, we need to apply the concept of equilibrium and the given equilibrium constant (Kc).
First, let's write the balanced chemical equation for the reaction:
Co(g) + H2O(g) → CO2(g) + H2(g)
Now, let's assume that at equilibrium, the number of moles of Co(g), H2O(g), CO2(g), and H2(g) are represented by x, y, z, and w, respectively.
According to the balanced equation, the stoichiometric coefficients for Co(g) and H2O(g) are both 1. Therefore, the change in moles for Co(g) and H2O(g) will be -x and -y, respectively.
The change in moles for CO2(g) and H2(g) will be +z and +w, respectively.
At equilibrium, the concentrations of Co(g), H2O(g), CO2(g), and H2(g) are given by:
[Co(g)] = (1-x) mol/L
[H2O(g)] = (1-y) mol/L
[CO2(g)] = z mol/L
[H2(g)] = w mol/L
Now, we can substitute these concentrations into the expression for the equilibrium constant (Kc):
Kc = [CO2(g)]/[Co(g)] * [H2(g)]/[H2O(g)]
= (z/(1-x)) * (w/(1-y))
= 0.85
Next, we can substitute the given initial concentrations into the equation above:
Kc = (z/(1-1)) * (w/(1-1))
= z * w
= 0.85
Since the value of Kc is given as 0.85, we can set the equation 0.85 = z * w.
Now, we need to solve this equation to find the values of z and w. Unfortunately, we don't have enough information to solve it directly.
However, we can make an approximation by assuming that the change in moles (x and y) are small compared to the initial number of moles (1). In this case, we can neglect the -x and -y terms when compared to 1.
Now, let's solve the simplified equation 0.85 = z * w.
Since the value of Kc is less than 1, we can conclude that the forward reaction (formation of CO2 and H2) is favored.
In this simplified case, the number of moles of CO2 and H2 will be approximately equal to the initial number of moles of Co and H2O, respectively.
Therefore, at equilibrium:
[Co(g)] ≈ 0
[H2O(g)] ≈ 0
[CO2(g)] ≈ 1.00 mol
[H2(g)] ≈ 1.00 mol
Thus, the number of moles of each species at equilibrium is approximately:
[Co(g)] ≈ 0 mol
[H2O(g)] ≈ 0 mol
[CO2(g)] ≈ 1.00 mol
[H2(g)] ≈ 1.00 mol
Keep in mind that this is an approximation based on the assumption that the change in moles is small. If more accurate calculations are required, additional information about the reaction conditions or the relationship between the initial and equilibrium concentrations would be needed.