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)`?

At a certain temperature, the equilibrium constant `(K_(c ))` is `16`

1.	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)`?
Answer» `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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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)`?

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