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??
Answer (1)
5th Jul, 2021
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
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