Calculate the equilibrium constant for the reaction 2 SO2 (g) + O2 (g) 2 SO3 (g), if there are 0.208 moles of SO2, 0.208 moles of O2, and 0.625 moles of SO3, in a 1.50-L vessel at equilibrium.

Question

Calculate the equilibrium constant for the reaction 2 SO2 (g) + O2 (g) <--> 2 SO3 (g), if there are 0.208 moles of SO2, 0.208 moles of O2, and 0.625 moles of SO3, in a 1.50-L vessel at equilibrium.

aaronq · Answer

set up an equilibrium equation, mang

chmvijay · Answer

write equlibrium equation for that reaction

chmvijay · Answer

it is product of concenteration of product  / prodouct of the concenteration of the reactant
now u write down

chmvijay · Answer

still u cant did that
????hello say anything

anonymous · Answer

Is it 6.47 * 10^1?

aaronq · Answer

idk show your work

anonymous · Answer

I had divided each of the moles by the 1.50-Liter to find the molarity to plug into the equation. Am I correct in doing so?