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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??


Meghanath Reddy M.C 5th Jul, 2021
Answer (1)
Ujjwal Kumawat 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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