at a certain temperature, .700 mol of SO3 is place in a 4.50-L container.
2SO3(g)--2SO2(g)+O2(g)
At equilibrium, .140 mol of O2 is present. Calculate Kc
To find the equilibrium constant, Kc, we need to use the given information and apply the principles of the law of mass action.
The balanced equation is:
2SO3(g) ↔ 2SO2(g) + O2(g)
We are given:
Initial amount of SO3 = 0.700 mol
Amount of O2 at equilibrium = 0.140 mol
Volume of the container = 4.50 L
First, let's calculate the concentrations of SO3, SO2, and O2 at equilibrium.
The concentration (in mol/L) of SO3 at equilibrium can be calculated using the formula:
[SO3] = (moles of SO3 at equilibrium) / (volume of the container)
= (0.700 mol) / (4.50 L)
= 0.1556 mol/L
Since 2 moles of SO3 produce 1 mole of O2, the concentration of O2 at equilibrium can be calculated as follows:
[O2] = (moles of O2 at equilibrium) / (volume of the container)
= (0.140 mol) / (4.50 L)
= 0.0311 mol/L
Similarly, the concentration of SO2 at equilibrium can be calculated using the stoichiometry ratio of the balanced equation:
[SO2] = 2 * [SO3]
= 2 * (0.1556 mol/L)
= 0.3112 mol/L
Now, let's calculate the equilibrium constant, Kc.
Kc = ([SO2]^2 * [O2]) / ([SO3]^2)
= (0.3112 mol/L)^2 * (0.0311 mol/L) / (0.1556 mol/L)^2
= 0.00335
Therefore, the equilibrium constant, Kc, for the given equation is approximately 0.00335.
To calculate the equilibrium constant (Kc), we need to know the concentrations of all the species at equilibrium. In this case, we are given the initial number of moles of SO3 and the number of moles of O2 at equilibrium.
The balanced equation for the given reaction is:
2SO3(g) -> 2SO2(g) + O2(g)
We are told that initially, there are 0.700 mol of SO3. Since this is a 4.50-L container, the initial concentration of SO3 is:
[SO3] = (0.700 mol) / (4.50 L) = 0.156 M
At equilibrium, we are told that there are 0.140 mol of O2. We need to find the concentrations of SO2 and O2 at equilibrium.
Since the coefficients of SO2 and O2 in the balanced equation are both 2, the change in the number of moles of SO3 is -2 * x, where x is the extent of reaction. This means that at equilibrium, the number of moles of SO3 is 0.700 - 2x.
According to the stoichiometry of the reaction, the number of moles of SO2 formed is also 2 * x, and the number of moles of O2 formed is x.
So, at equilibrium, we have:
[SO3] = (0.700 - 2x) / (4.50 L)
[SO2] = (2x) / (4.50 L) = (x) / (2.25 L)
[O2] = (x) / (4.50 L)
Now, we can use the given concentration values of [O2] = 0.140 M to determine the value of x, which will allow us to calculate the equilibrium concentrations of SO3 and SO2.
0.140 M = (x) / (4.50 L)
x = 0.140 M * 4.50 L
x = 0.630
Now, we can substitute the value of x back into the expressions for [SO3], [SO2], and [O2] to get the equilibrium concentrations:
[SO3] = (0.700 - 2 * 0.630) / (4.50 L)
[SO2] = (2 * 0.630) / (4.50 L)
[O2] = (0.630) / (4.50 L)
Simplifying these equations will give us the equilibrium concentrations in terms of M (molarity). We can then use these concentrations to calculate Kc.
Finally, we can plug in the values of the equilibrium concentrations into the equilibrium expression:
Kc = ([SO2]^2 * [O2]) / [SO3]^2
By substituting the values we found for [SO3], [SO2], and [O2] into the equation, we can calculate Kc.