InterviewSolution
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InterviewSolution
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(a) Explain thefollowing terms : (i) Rate of a reaction (ii) Activation energy of a reaction (b) Thedecompositon of phosphine , PH3, proceeds according to the following equation : 4 PH_(3) (g) to P_(4)(g) + 6 H_(2) (g) It is found that the reaction follows the following rate equation :Rate = k [PH_(3)]. The half-life ofPH_(3)" is "37.9" s at "120^(@)C. (i) How much time is required for 3//4^("th") "of " PH_(3) todecompose ? (ii) What fraction of the original sample of PH_(3) remains behind after 1 minute ? |
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Answer» Solution :(a) (i) Rate of a reaction : Therate of a reaction can be defined as thechange in concentration of a reactant or proudct in unit time. (II) Activation energy of a reaction : The MINIMUM extra amount of energy absorbed by the reactant molecules so that their energy becomes equal to threshold VALUE is called activation energy. It is denoted by EA. (b) (i)`t_(1//2) = 37.9 ` s Letinitial concentration = a ` :.x = 3/4a` Using the formula `t=(2.303)/klog.a/(a-x)` ` t=((2.303)/(0.693))/(t_(1//2))log. a/(a-3/4a)""[:'k=(0.693)/t_(1//2)]` `t = 2.303 xxt_(1//2)/(0.693)log.q/(1/4a)` ` t = (2.303xxt_(1//2))/(0.693) xx log 4 =(2.303 xxt_(1//2)xx2log 2)/(0.693)` `t = (2.303 xx37.9 xx0.3010)/(0.693)""[:. t_(1//2) = 37.9s]` ` = (52.544)/(0.693) = 75. 82 sec`. (ii) Heret = 1 min = 60s ` andt_(1//2) = 37.9s` Using the formula ` t = (2.303)/ klog. ([A]_(0))/([A]_(t))` ` and k = ( 0.693)/(t_(1//2))` ` :. "" t = (2.303)/((0.693)/t_(1//2))log.[A]_(0)/[A]_(t)` `t=2.303 xx t_(1//2)/(0.693) xx log.[A]_(0)/[A]_(t)` ` 60 = (2.303 xx37.9)/(0.693) log . [A]_(0)/[A]_(t) ` `:. ""log.[A]_(0)/[A]_(t)=(60 xx 0.693)/(2.303xx37.9) = 0.4768` `[A]_(0)/[A]_(t) = anti-log(0.4768) = 2.997` ` [A]_(t)/[A]_(0) = 1/(2.997) = 0.3337 = 33.37 %` |
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