sulfur trioxide gas dissociates into sulfer dioxide gas and oxygen gas at 1250 degrees C. In an experiment, 3.6 moles of sulfur trioxide were placed into an evacuated 3.0-L flask. THe concentration of sulfur dioxide gas measured at equilibrium was found to be 0.20 M. What is the equilibrium constant, Kc, for the reaction?
SO3 ==> SO2 + O2
(SO3) = 3.6 mols/3.0 L = 1.2 M initially.
(SO2) = 0.2 M at equilibrium.
(O2) = 0.2 M at equilibrium.
(SO3) = 1.2-0.2 = 1.0 M
Plug into Kc and calculate.
1.6 x 10^-2
Why did the sulfur trioxide go to therapy? Because it had a dissociation problem!
But let's get back to business. The equilibrium constant, Kc, can be calculated using the concentrations of the reactants and products at equilibrium. In this case, the concentration of sulfur dioxide gas is 0.20 M, and the concentration of sulfur trioxide gas is 1.0 M. The concentration of oxygen gas is also 0.20 M.
So, the equilibrium constant, Kc, is calculated as follows:
Kc = ([SO2] * [O2]) / [SO3]
= (0.20 * 0.20) / 1.0
= 0.04 / 1.0
= 0.04
Therefore, the equilibrium constant, Kc, for the reaction is 0.04.
To calculate the equilibrium constant (Kc), you need to set up the expression for the equilibrium constant using the concentrations of the reactants and products.
For the given reaction:
SO3 ==> SO2 + O2
The equilibrium constant expression is:
Kc = [SO2][O2] / [SO3]
Given:
[SO2] = 0.2 M
[O2] = 0.2 M
[SO3] = 1.0 M
Substituting the given values into the expression:
Kc = (0.2)(0.2) / 1.0
Simplifying:
Kc = 0.04 / 1.0
Therefore, the value of the equilibrium constant (Kc) for the given reaction is 0.04.
To calculate the equilibrium constant, Kc, for the reaction, you need to use the concentrations of the reactants and products at equilibrium.
The balanced equation for the reaction is:
SO3 ⇒ SO2 + O2
First, calculate the concentration of SO3 at equilibrium. It is given that 3.6 moles of sulfur trioxide were placed into a 3.0-L flask. So, the initial concentration of SO3 was:
(SO3) = 3.6 moles / 3.0 L = 1.2 M
Given that the concentration of SO2 at equilibrium is 0.20 M and the concentration of O2 at equilibrium is also 0.20 M, we can calculate the concentration of SO3 at equilibrium using the stoichiometry of the reaction.
Since the mole ratio of SO3 to SO2 is 1:1, the concentration of SO3 at equilibrium would be:
(SO3) = (SO3) initial - (SO2) equilibrium
(SO3) = 1.2 M - 0.20 M = 1.0 M
Now, we can substitute the values into the equilibrium expression and solve for Kc:
Kc = ((SO2)(O2)) / (SO3)
= (0.20)^1 * (0.20)^1 / (1.0)^1
= 0.04 / 1.0
= 0.04
Therefore, the equilibrium constant, Kc, for the reaction is 0.04.