Half-life `(t_(1))` of the first order reaction and half-life `(t_(2))` of the second order reaction are equal. Hence ratio of the rate at the start of the start of the reaction:A. `1`B. `2`C. `0.693`D. `1.44`

Half-life `(t_(1))` of the first order reaction and half-life `(t_(2))

1.	Half-life `(t_(1))` of the first order reaction and half-life `(t_(2))` of the second order reaction are equal. Hence ratio of the rate at the start of the start of the reaction:A. `1`B. `2`C. `0.693`D. `1.44`
Answer» Correct Answer - c For I order, `K_(1)=0.693/t_(1//2)`, For II order, `K_(2)=1/(t_(1//2)a)` `K_(1)=0.693/T_(1), K_(2)=1/(T_(2)a)` If `T_(1)=T_(2)`, then `K_(1)/K_(2)=0.693a` Initially `r_(1)=K_(1)[a]^(1), r_(2)=K_(2)[a]^(2)` `:. R_(1)/r_(2)=K_(1)/(K_(2)a)=(K_(2)xx0.693xxa)/(K_(2)xxa)=0.693`

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Half-life `(t_(1))` of the first order reaction and half-life `(t_(2))` of the second order reaction are equal. Hence ratio of the rate at the start of the start of the reaction:A. `1`B. `2`C. `0.693`D. `1.44`

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