InterviewSolution
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InterviewSolution
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The equilibrium composition for the reaction is `{:(PCl_(3),+,Cl_(2),hArr,PCl_(5)),(0.20,,0.10,,0.40 mol L^(-1)):}` What will be the equilibrium concentration of `PCl_(5)` on adding `0.10 mol` of `Cl_(2)` at the same temperature? |
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Answer» Correct Answer - A::D `{:(PCl_(3),+,Cl_(2),hArr,PCl_(5)),(0.2,,0.1,,0.4):}` `K_(c ) 0.4/(0.2xx0.1)xx20` After adding `0.1` mole of `Cl_(2)`, New initial concentration of `Cl_(2)=0.1+0.1 =0.2 "mole"` New initial concentration of `PCl_(3)` and `PCl_(5)` remains the same. Now, suppose `x` mol of `PCl_(3)` reacts, the new equilibrium concentration. will be `[PCl_(3)]=0.2-x, [Cl_(2)]=0.2-x, [PCl_(5)]=0.4+x` `K_(c )=([PCl_(5)])/([PCl_(3)][Cl_(2)])=((0.4+x))/((0.2-x)(0.2-x))` `:. 20=((0.4+x))/((0.2-x)(0.2-x))` solve for `x` `x=0.45 "mol" L^(-1)` |
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