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At a certain temperature, the equilibriu...

At a certain temperature, the equilibrium constant `(K_(c ))` is `16` for the reaction:
`SO_(2)(g)+NO_(2)(g)hArrSO_(3)(g)+NO(g)`
If we take one mole of each of the equilibrium concentration of `NO` and `NO_(2)`?

Text Solution

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`SO_(2)(g)+NO_(2)(g)hArrSO_(3)(g)+NO(g)`
`{:("Initial conc".,1,1,1,1),("Equilibrium conc".,1-x,1-x,1+x,1+x):}`
Applying the law of mass action,
`K_(c)=([SO_(3)][NO])/([SO_(2)][NO_(2)])=((1+x)(1+x))/((1-x)(1-x))=16`
`(1+x)/(1-x)=4` or `1+x=4-4x`
or `5x=3`, i.e., `x=3/5=0.6`
Concentratio of `NO_(2)` at equilibrium `=(1-0.6)=0.4 "mol"`
Concentration of `NO` at equilibrium`=(1+0.6)=1.6 "mol"`
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