The equlibrium constant Kc for the reaction: has the value of 4.00 at 500K. If a mixture of 1.00 mol CO and 1.00 mol H2O is allowed to reach equilibrium at this temperature in a 1.00L flask, calculate the equilibrium concentrations of each species in the flask.
You should have shown the reaction. I assume it is
CO(g) + H2O(g) ==>H2(g) + CO2(g)
If not, just ignore what follows.
Set up an ICE chart and solve.
initial:
CO = 1.00 mol/L
H2O = 1.00 mol/L
CO2 = 0
H2 = 0
change:
CO = -x
H2O = -x
H2 = x
CO2 = x
equilibrium:
CO = 1-x
H2O = 1-x
CO2 = x
H2 = x
Post your work if you get stuck.
Kc=(x)(x)/(1-x)(1-x)
What is x in this equation and is is 4.0 Kc?
To calculate the equilibrium concentrations of each species in the flask, we need to use the equation and the given value of the equilibrium constant Kc.
The balanced equation for the reaction is:
CO(g) + H2O(g) ⇌ CO2(g) + H2(g)
The Kc expression for this reaction is:
Kc = [CO2] * [H2] / [CO] * [H2O]
Given that Kc = 4.00, we can substitute the known values into the equation:
4.00 = [CO2] * [H2] / [CO] * [H2O]
We are given that the initial mixture contains 1.00 mol of CO and 1.00 mol of H2O in a 1.00 L flask. This means that the initial concentrations of CO and H2O are both 1.00 M.
Substituting these values into the equation, we have:
4.00 = [CO2] * [H2] / (1.00 * 1.00)
Simplifying the expression, we have:
4.00 = [CO2] * [H2]
Since we have one mole of CO and one mole of H2O initially, we can assume that there is no CO2 or H2 initially. Therefore, the concentration of CO2 and H2 at equilibrium will be the same.
Let's denote the equilibrium concentration of CO2 as x. Then the equilibrium concentration of H2 will also be x.
Substituting these values into the equation, we have:
4.00 = x * x
Solving this equation for x, we get:
x^2 = 4.00
x = √4.00
x = 2.00
So, at equilibrium, the concentration of CO2 and H2 will be both 2.00 M.
Therefore, the equilibrium concentrations of each species in the flask are:
[CO] = 1.00 M
[H2O] = 1.00 M
[CO2] = 2.00 M
[H2] = 2.00 M