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दर्शाइए कि प्रथम कोटि की अभिक्रिया में 99% अभिक्रिया पूर्ण होने में लगा समय 90% अभिक्रिया पूर्ण होने में लगने वाले समय से दुगना होता है।

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प्रथम कोटि की अभिक्रिया के लिए
`t= (2.303)/(k) log .(a)/(a-x)`
प्रथम स्थिति में `a =100% n= 99%`
`(a-x) = (100 -99) = 1% `
`t_(99%) = (2.303)/(k) log (100)/(1)= (2.303)/(k) log 10^(2)`
`= (2.303)/(k) = (4.606)/(k) …. (i)`
दितीय स्थिति में `a= 100% x= 99% `
`(a-x ) = (100-90) = 10% `
`t_(90%) = (2.303)/(k) log .(100)/(10)= (2.303)/(k) log 10= (2.303)/(k) …. (2)`
समी `(1) ` को समी (2) से भाग देने पर
`(t_(90%))/(t_((90%))) = (4.606)/(k) xx (k) /(2.303) =2`

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