Sulfur dioxide and oxygen react to form sulfur trioxide as follows
2SO_2(g) + O_2(g) ⇌ 2SO_3(g)
1 mole of sulfur dioxide is mixed with 0.5 moles of oxygen gas in a 1 L container. There is 0.9 mol of SO_3 at equilibrium. Calculate the following.
a) Equilibrium concentrations of sulfur dioxide and oxygen.
b) Equilibrium constant for this reaction.
Anyone can check if a) = [SO_2) = 0.1M
[O_2] = 0.05M
and b) = 1620
Thanks.
I agree.
To calculate the equilibrium concentrations of sulfur dioxide (SO2) and oxygen (O2), we first need to determine the initial concentrations of each reactant.
Given:
Initial moles of SO2 = 1 mol
Initial moles of O2 = 0.5 mol
Equilibrium moles of SO3 = 0.9 mol
Since the reaction is in a 1 L container, the initial and equilibrium concentrations will have the same values as moles and volumes are directly proportional. Therefore, the initial and equilibrium concentrations of SO2 and O2 are as follows:
Initial concentration of SO2 = 1 mol/L
Initial concentration of O2 = 0.5 mol/L
Equilibrium concentration of SO3 = 0.9 mol/L
Now, let's calculate the equilibrium constant (K) for this reaction. The equilibrium constant expression for this reaction can be written as:
K = [SO3]^2 / ([SO2]^2 * [O2])
Using the given equilibrium concentrations, we can substitute the values and calculate K:
K = (0.9)^2 / ((1)^2 * (0.5))
K = 0.81 / 0.5
K = 1.62
Hence, the equilibrium constant (K) for this reaction is 1.62.