for a reaction 2so2 +o2=2so3,1.5 moles of so2 and 1 mole of o2 are taken in 2L vessel.at equilbrium the concentration of so3 was found to be 0.35mol/L. the equilibrium constant for the reaction would be??
Dear aspirant
Hope you are doing well
Now let me go through your question
For the given rexn
2SO2+O22SO3
Initial moles ; 1.5 1 0
At eqm : 1.5−2x 1−x 2x (let us suppose)
Now given that conc. of SO3 at eqm =0.35 mol/L
⇒2x=0.70⇒x=0.35 ( the container is of 2L)
Now, KC=[SO2]2[O2][SO3]2
=(1.5−2x)2(1−x)(2x)2
⇒KC=(21.5−0.7)2×(21−0.35)[20.7]2
=0.8×0.8×0.650.7×0.7×2= 2.35Lmol−1
hope it will help you
Thank you