Calculate the value of the equilibrium constant, Kc, for the reaction below, if 0.208 moles of sulfur dioxide gas, 0.208 moles of oxygen gas and 0.425 moles of sulfur trioxide gas are present in a 1.50-liter reaction vessel at equilibrium.
2SO2(g) + O2 (g) 2SO3 (g)

Kc = 3.32 x 10^-2

Kc = 2.01 x 10^1

Kc = 2.98 x 10^1

Kc = 1.47 x 10^1

Question

Calculate the value of the equilibrium constant, Kc, for the reaction below, if 0.208 moles of sulfur dioxide gas, 0.208 moles of oxygen gas and 0.425 moles of sulfur trioxide gas are present in a 1.50-liter reaction vessel at equilibrium.
2SO2(g) + O2 (g)  2SO3 (g)

 
		
Kc = 3.32 x 10^-2
 		
Kc = 2.01 x 10^1
 		
Kc = 2.98 x 10^1
 		
Kc = 1.47 x 10^1

anonymous · Answer

Kc = (products)^moles/(reactants)^moles

and i got 20.1

anonymous · Answer

so it is Kc = 2.01 x 10^1
 		?

anonymous · Answer

yes