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1 mol of PCl_(5) , kept in a closed container of volume of 1 dm^(3) and was allowed to attain equilibrium at 423 K . Calculate the equilibrium composition of reaction mixture . (The K_(C) value for PCl_(5) dissociation at 423 K is 2 )

Answer» <html><body><p></p>Solution :`PCl_(5) hArr PCl_(3) + Cl_(2)` <br/> Given that `[PCl_(5)]_("initial") =1` mol , `V = 1 dm^(3) , K_(C) = 2` <br/> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/FM_CHE_XI_V02_C08_E01_055_S01.png" width="80%"/> <br/> `K_(C) = ([PCl_(3)][Cl_(2)])/([PCl_(5)])` <br/> `2 = ( x xx x)/((1 - x))` <br/> `2 - 2x = x^(2)` <br/> `x^(2) + 2x - 2 = 0` <br/> Solution for a quadratic equation <br/> `ax^(2) + bx + c = 0` are `x = (- <a href="https://interviewquestions.tuteehub.com/tag/b-387190" style="font-weight:bold;" target="_blank" title="Click to know more about B">B</a> pm sqrt(b^(2) - 4ac))/(2a)` <br/> `a = 1 . b = 2 . c = -2` <br/> `x = ( - 2 pm sqrt(4 - <a href="https://interviewquestions.tuteehub.com/tag/4xx-1883349" style="font-weight:bold;" target="_blank" title="Click to know more about 4XX">4XX</a> 1 xx (-2)))/(2 xx 1) = (-2 pm sqrt(12))/(2) = (-2 pm sqrt(4 xx 3))/(2)` <br/> `x = (-2 pm 2 sqrt3)/(2) = (- 2 + 2 sqrt3)/(2) , ( - 2 - 2 sqrt3)/(3)` <br/> `x = -1+ sqrt3 , - 1 - sqrt3` (<a href="https://interviewquestions.tuteehub.com/tag/since-644476" style="font-weight:bold;" target="_blank" title="Click to know more about SINCE">SINCE</a> x is +ve ) , `-1 - sqrt3` not possible <br/> `= - 1 + 1.732 = 0.732` <br/> `therefore` Equilibrium concentration of `[PCl_(5)]_(<a href="https://interviewquestions.tuteehub.com/tag/eq-446394" style="font-weight:bold;" target="_blank" title="Click to know more about EQ">EQ</a>) = (1- x)/(1) = 1 - 0.732 = 0. <a href="https://interviewquestions.tuteehub.com/tag/268-1832096" style="font-weight:bold;" target="_blank" title="Click to know more about 268">268</a>` M <br/> `[PCl_(3)]_(eq) = (x)/(1) = (0.732)/(1) = 0.732` <br/> `[Cl_(2)]_(eq) = (x)/(1) = (0.732)/(1) = 0.732`</body></html>


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