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When PCl(5) is heated in a closed vessel...

When `PCl_(5)` is heated in a closed vessel at 575 K, the total pressure at equilibrium is found to be 1 atm and partial pressure of `Cl_(2)` is found to the 0.324 atm . Calculate the equilibrium constant `(K_(p))` for the decomposition reaction.

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The correct Answer is:
`0.298`
` PCl_(5) hArr PCl_(3) + Cl_(2)`
`" At eqm., " p_(Cl_(2)) = p_(pCl_(3))= 0*324 " atm . Hence " p_(PCl_(5)) = 1- (0*324 + 0*324) = 0*352 " atm"`
`K_(p) = (p_(PCl_(3)) xxp_(Cl_(2)))/(p_(PCl_(5)))=(0*324 xx 0*324)/(0*352)= 0*298 " atm "`
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