At a certain temperature, the equilibrium
constant Kc is 0.154 for the reaction
2 SO2(g) + O2(g) *) 2 SO3(g)
What concentration of SO3 would be in
equilibrium with 0.250moles of SO2 and 0.853
moles of O2 in a 1.00 liter container at this
temperature? Note: These latter moles are
the equilibrium values.
Answer in units of M
Since these mols are in 1.00L, mols = M.
.........2SO2 + O2 ==> 2SO3
E........0.250..0.853...x
Kc = (SO3)^2/(SO2)^2(O2)
Substitute and solve for x.
1.70M
To find the concentration of SO3 in equilibrium, we can use the equilibrium constant expression. The equilibrium constant expression for the given reaction is:
Kc = [SO3]^2 / ([SO2]^2 * [O2])
Where [SO3], [SO2], and [O2] represent the concentrations of SO3, SO2, and O2, respectively.
Given:
[SO2] = 0.250 moles
[O2] = 0.853 moles
Volume (V) = 1.00 liter
To find the concentrations, we need to convert the moles to molar concentration (Molarity). Molarity is defined as moles of solute per liter of solution.
Concentration of SO2: [SO2] = moles of SO2 / volume of the container
[SO2] = 0.250 moles / 1.00 L = 0.250 M
Concentration of O2: [O2] = moles of O2 / volume of the container
[O2] = 0.853 moles / 1.00 L = 0.853 M
Now, we can substitute these values into the equilibrium constant expression:
Kc = [SO3]^2 / ([SO2]^2 * [O2])
Solving for [SO3]:
[SO3]^2 = Kc * ([SO2]^2 * [O2])
[SO3]^2 = 0.154 * (0.250^2 * 0.853)
Taking the square root of both sides:
[SO3] = sqrt(0.154 * (0.250^2 * 0.853))
Calculating this expression, we get:
[SO3] ≈ 0.116 M
Therefore, the concentration of SO3 in equilibrium with 0.250 moles of SO2 and 0.853 moles of O2 in a 1.00 liter container at this temperature is approximately 0.116 M.