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COCl(2(g)) Leftrightarrow CO(g)+Cl(2(g))...

`COCl_(2(g)) Leftrightarrow CO_(g)+Cl_(2(g))`
`K_(p)=1`
`K_p= 1` is attained by starting with `COCl_2` at an initial pressure of 1 atm. Isothermally, when volume of vessel is halved, total pressure at new equilibrium becomes

A

3.4144 atm

B

0.866 atm

C

0.471 atm

D

3.0 atm

Text Solution

Verified by Experts

The correct Answer is:
A
`underset(1)(COCl_(2(g)) Leftrightarrow underset(0)(CO_(g))+underset(0)(Cl_(2(g))`
`p_("Total")=1-p+p+p =(1+p)`
`p_(CO)=p/(1+p) xx 1, p_(Cl_2)=p/(1+p) xx 1, p_(COCl_2)=(1-p)/(1+p) xx1 `
`K_(p)=(p_(CO))(p_(Cl_2))/(p_(COCl_2))=((p)/(1+p))^2/((1-p)/(1+p))=p^2/(1-p^2)`
`1=p^2/(1-p^2) rArr 1-p^2 =p^2 rArr 2p^2=1 rArr p^2=1/2`
`p=1/sqrt2 =1/(1.414) =0.7072`
When volume of vessel is halved, the pressure will become double.
`2(1+p)=2(1+0.7072)=2 xx (1.7072)=3.4144atm 4`
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