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:

Options

A. `1`
B. `2`
C. `0.693`
D. `1.44`

Answer

c

Explanation

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`