दिखाइए कि एक प्रथम क्रम के अभिक्रिया के 99.9% सम्पन्न होने में लगा समय अभिक्रिया के आधे सम्पन्न होने में लगे समय का 10 गुना होता है।

Explanation

For first order reaction
`1^(st)` case , `a = 100, x = 99.9 , a - x =100 - 99.9 = 0.1 `
`:.t_(99.9) = (2.303)/k log"" (a)/((a-x)) = (2.303)/k log "" (100)/(0.1) = (2.303)/k log 10^3`
`=(2.303 xx 3)/k` …(1)
`2^(nd)`case `a = 100, x = 50 , a - x = 100 - 50 = 50`
`t_(50) - (2.303)/k log"" (100)/(50) = 2.303/k log 2`
` = (2.303 xx0.3010)/k` ...(ii)
eq (i) by eq. (ii) we get
`t_(99.9)/(t_(50)) = (2.303xx3)/k xx k /(2.03 xx 0.3010) = 9.97 = 10`
` = 10 xx t_50`