For the reaction A + B ` to` products, the following initial rates were obtained at various given initial concentration
`{:(“S.No.”,,”[A]mol/L”,,”[B] mol/L”,,”Initial rate M/s”),(1.,,” “0.1,,” “0.1,,” “0.05),(2.,,” “0.2,,” “0.1,,” “0.10),(3.,,” 0.1″,,” 0.2″,,” 0.05″):}`
Determine the half-life period.
`{:(“S.No.”,,”[A]mol/L”,,”[B] mol/L”,,”Initial rate M/s”),(1.,,” “0.1,,” “0.1,,” “0.05),(2.,,” “0.2,,” “0.1,,” “0.10),(3.,,” 0.1″,,” 0.2″,,” 0.05″):}`
Determine the half-life period.
`”rate” = k [A]^(x)[B]^(y)`
` 0.05 = k[0.1]^(x)[0.1]^(y) `……(i)
` 0.10 = k[0.2]^(x)[0.1]^(y)` …….(ii)
` 0.05 = k[0.1]^(x) [0.2]^(y)`……(iii)
`(ii) div (i)`
` (0.10)/(0.05) = (2)^(x)`
` = x = 1`
` (iii) div (i) `
` (0.05)/(0.05) = (2)^(y)`
` y = 0`
`”rate” = k[A]^(1)[B]^(0)`
It is a first order reaction.
`k = (“rate”)/([A]) = 0.5 s^(-1)`
` t _(½) = (0.693)/k = (0.693)/(0.5)`
` t _(½) = 1.386 s `
Or
` t _(½) = (0.693)/k `
` k_(2) = (0.693)/(25) ” ” 350 K`
` k_(1) = (0.693)/(50) ” ” 300 K `
` k_(2)/k_(1) = 2 `
`log. k_(2)/k_(1) = E_(a)/(2.303 R) [1/T_(1)-1/T_(2)]`
` log 2 = E_(a)/(2.303 x 8.314) [(350-300)/(350 x 300)]`
` Ea = 12.104″ kJ”//mol`.