1.

The rate of a first order reaction is 0.04" mil L"^(-1)s^(-1) at 10 minutes and 0.03" mol L"^(-1)s^(-1) at 20 minutes after initiation. Find the half life of the reaction.

Answer»

Solution :For a FIRST order reaction, `ATO"PRODUCTS, for concentration of the reactant at two different TIMES,"`
`k=(2.303)/(t_(2)-t_(1))log""([A]_(1))/([A]_(2))`
But as rate `=k[A]," therefore, "(("rate")_(1))/((rate)_(2))=([A]_(1))/([A]_(2))`
Hence, `k=(2.303)/(t_(2)-t_(1))log""(("rate")_(1))/(("rate")_(2))=(2.303)/((20-10)min)log""(0.04)/(0.03)=2.88xx10^(-2)min^(-1)`
`:.t_(1//2)=(0.693)/(k)=(0.693)/(2.88xx10^(-2)min^(-1))=24.06" min "=1443.6 s.`


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