1.

At `250^@C`, `K_c` for `PCl_5(g)hArr PCl_3(g)+Cl_2(g)` is `0.04`. How many moles of `PCl_5` must be added to a `3-L` flask to obtain `0.15M Cl_2` at equilibrium?A. `2.1 mol`B. `1.7mol`C. `0.9 mol`D. `3.5 mol`

Answer» Correct Answer - A
Molarity `(M)=n/V_L`
or `n=M*V_L`
Thus, the number of moles of `Cl_2` at equilibrium,
`n_(Cl_2)=(0.15M)(3L)`
`=0.45mol`
`{:(,PCl_(5)hArrPCl_(3)+Cl_(2)),("Initial moles"," a 0 0"),("Change"," -0.45 +0.45 +0.45"),("Equilibrium moles", bar(" a-0.45 0.45 0.45 ")),("Equilibrium concentration"," "(a-0.45)/(3)" 0.15 0.15"):}`
Equilibrium constant expression
`K_c=(C_(PCl_3)C_(Cl_2))/(C_(PCl_5))`
Substituting, we get
`0.04=((0.15)(0.15))/(((a-0.45)/(3)))`
`0.04((a-0.45)/(3))=0.0225`
`a=2.1`


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