Calculate the equilibrium constant 2SO2(g]+O2(g]⇌2SO3(g]

given equilibrium constant of SO2,O2,SO3 are 0.3m,0.14,1.45 respectively wt happens to the equilibrium constant a) if so2 is added b)if o2 is removed

To calculate the equilibrium constant for the given reaction, we can use the ratio of the concentrations of the products to the concentrations of the reactants. The equilibrium constant expression for the reaction is:

K = [SO3]^2 / ([SO2]^2 * [O2])

a) If SO2 is added, the concentration of SO2 will increase, while the concentrations of SO3 and O2 remain constant. Let's assume that the new concentration of SO2 is x (initial concentration + added amount).

The new equilibrium constant can be calculated as:

K' = [(1.45)^2] / [(0.3 + x)^2 * (0.14)]

b) If O2 is removed, the concentration of O2 will decrease, while the concentrations of SO2 and SO3 remain constant. Let's assume that the new concentration of O2 is y (initial concentration - removed amount).

The new equilibrium constant can be calculated as:

K'' = [(1.45)^2] / [(0.3)^2 * (0.14 - y)]

Hope this helps! Let me know if you have any further questions.

K = (SO3)^2/(SO2)^2(O2)

You are given the concentrations of each (I assume that m stands for M). Just substitute and you get K.
What happens to the equilibrium constant if SO2 is added or O2 removed? Nothing happens. The CONCENTRATIONS will change but the K will not.