At a certain temperature, the equilibrium constant Kc is 0.154 for the reaction 2SO2(g) + O2(g)<-> 2SO3(g)
What concentration of SO3 would be in
equilibrium with 0.250 moles of SO2 and 0.885 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
Substitute the equilibrium value into Kc and solve for the one unknown.
if 96g so2 is added to 2 moles of oxygen at stp,calculate the mass of so3 that is formed
To find the concentration of SO3 in equilibrium, we need to make use of the equilibrium constant expression and solve for the concentration of SO3.
The equilibrium constant expression for the given reaction is:
Kc = [SO3]^2 / ([SO2]^2 * [O2])
Given that Kc = 0.154, [SO2] = 0.250 M, [O2] = 0.885 M, and the volume of the container is 1.00 L, we can substitute these values into the expression and solve for [SO3].
0.154 = [SO3]^2 / (0.250^2 * 0.885)
Rearranging the equation and solving for [SO3]:
[SO3]^2 = 0.154 * (0.250^2 * 0.885) = 0.021581875
Taking the square root of both sides:
[SO3] = √0.021581875 ≈ 0.146 M
Therefore, the concentration of SO3 in equilibrium with 0.250 moles of SO2 and 0.885 moles of O2 in a 1.00 liter container at this temperature is approximately 0.146 M.