1.

(i) A first order reaction takes 8 hours for 90% completion. Calculate the time required for 80% completion . (log 5 = 0 .6989, log 10 =1 )

Answer»

Solution :For a first ORDER reaction,
`K=(2.303)/tlog(([A_0])/([A]))""..........(1)`
Let `[A_0]=100M `
When `t=t_(90%),[A]=10M` (GIVEN that `t_(90%)=8`hours)
`t = t = _(80%),[A]=20M`
`t_(80%)=(2.303)/(t_(90%))log(100/10)`
`k=(2.303)/(8"hours")log10`
`k=(2.303)/(8 "hours")(1)`
Substitute the value of k in equation (2)
`t_(80%)=(2.303)/(2.303//8"hours")log (5)`
`t_(80%)=2.303//8"hours"xx0.6989`
`t_(80%)=5.59` hours


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