1) the value of Kb after the decomposition of sulfur trioxide (S03). 2SO3 (g)--2SO2 (g) +O2 (g) is 1.79 x10^-5 at 350 degree celcius. if SO3 initially at 0.200 atm pressure comes to equilibrium in an otherwise empty tank at 350 degree celcius, determine the partial pressure of O2 at equilibrium.
To determine the partial pressure of O2 at equilibrium, we need to use the equilibrium constant expression (Kb) and the given initial pressure of SO3.
Given:
Equilibrium constant (Kb) = 1.79 x 10^-5
Initial pressure of SO3 (P_SO3) = 0.200 atm.
The balanced equation for the decomposition of SO3 is:
2SO3 (g) --> 2SO2 (g) + O2 (g)
The equilibrium constant expression for this reaction is:
Kb = (P_SO2)^2 * P_O2
Since the stoichiometric coefficient of O2 in the balanced equation is 1, the equilibrium expression simplifies to:
Kb = P_SO2^2 * P_O2
We need to rearrange this expression to solve for the partial pressure of O2 at equilibrium:
P_O2 = Kb / (P_SO2)^2
Given that the initial pressure of SO3 is 0.200 atm, we need to find the partial pressure of SO2.
Since the stoichiometric coefficient of SO2 in the balanced equation is also 2, the pressure of SO2 at equilibrium (P_SO2) can be calculated as:
P_SO2 = (Initial pressure of SO3) / 2
= 0.200 atm / 2
= 0.100 atm
Now, substitute the values into the rearranged equation to find the partial pressure of O2 at equilibrium:
P_O2 = Kb / (P_SO2)^2
= (1.79 x 10^-5) / (0.100 atm)^2
= 1.79 x 10^-5 / (0.01 atm)
= 1.79 x 10^-3 atm
Therefore, the partial pressure of O2 at equilibrium is 1.79 x 10^-3 atm.