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For the decomposition of azoisopropane to hexane and nitrogen at 543 K, following data are obtained : {:("t (sec)",,,,"P (mm of Hg)"),(0,,,,""35.0),(360,,,,""54.0),(720,,,,""63.0):} Calculate the rate constant. |
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Answer» Solution :`("CH"_(3))_(2)"CH N"="N CH"("CH"_(3))_(2)(g)to"N"_(2)(g)+"C"_(6)"H"_(16)(g)` `{:("Initial pressure",,," ""P"_(0),,,,0,,,0),("After time t",,,"P"_(0)-"p",,,,"p",,,"p"):}` Total pressure after time `t(P_(t))=(P_(0)-p)+p+p=P_(0)+p" or "p=P_(t)-P_(0)` `apropP_(0)" and "(a-x)propP_(0)-p` or substituting the value of p, `a-xprop P_(0)-(P_(t)-P_(0)),i.e.,(a-x)prop P_(0)-P_(t)` As DECOMPOSITION of AZOISOPROPANE is a first order REACTION, `K=(2.303)/(t)log""(a)/(a-x)=(2.303)/(t)log""(P_(0))/(2P_(0)-P_(t))` When `t=360" sec,"k=(2.303)/(360" s")log""(350)/(2xx35.0xx54.0)=(2.303)/(360" s")log""(35)/(16)=(2.303)/(360" s")(0.3400)=2.175xx10^(-3)s^(-1)` When `t=720" sec,"k=(2.303)/(720" s")log""(35.0)/(2xx35.0-63.0)=(2.303)/(720" s")log5=(2.303)/(720)(0.6990)=2.235xx10^(-3)s^(-1)` `:." Average value of "k=(2.175+2.235)/(2)xx10^(-3)s^(-3)=2.20xx10^(-3)s^(-1).` |
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