Two reaction, `(I)A rarr` Products and `(II) B rarr` Products, follow first order kinetics. The rate of reaction `(I)` is doubled when the temperature is raised form `300 K` to `310K`. The half life for this reaction at `310K` is `30 min`. At the same temperature `B` decomposes twice as fast as `A`. If the energy of activation for reaction `(II)` is twice that of reaction `(I)`, (a) calculate the rate of constant of reaction `(II)` at `300 K`.

Explanation

Calculation of activation energy of reaction (i)
`T_(1)=300" K", T_(2)=310" K",k_(1)=k,k_(2)=2k`
`:." "log""(k_(2))/(k_(1))=(E_(a))/(2.303" R")((T_(2)-T_(1))/(T_(1)T_(2))), i.e., log2=(E_(a))/(2.303xx8.314)xx(10)/(300xx310)" or "E_(a)=53.60" kJ mol"^(-1)`
Calculation of rate constant of reaction (i) at 310 K
`k=(0.693)/(t_(1//2))=(0.693)/(30" min")=2.31xx10^(-2)min^(-1)`
Rate constant of reaction (ii) at 310 `K=2xx2.31xx10^(-2)min^(-1)=4.62xx10^(-2)min^(-1)`
Energy of activation of reaction (ii) `=(53.60" kJ mol"^(-1))/(2)=26.80" kJ mol"^(-1)`
Aim. To calculate k for reaction (ii) at 300 K
`log""(k_(300" K"))/(k_(300" K"))=(E_(a))/(2.303"R")((T_(2)-T_(1))/(T_(1)T_(2)))`
`:." "log""(4.62xx10^(-2))/(k_(300" K"))=(26.80)/(2.303xx8.314xx10^(-3))xx(10)/(300xx310)=0.0151`
or `" "(4.62xx10^(-2))/(k_(300" K"))="Antilog"0.0151=1.035`
or `" "k_(300" K")=(4.62xx10^(-2))/(1.035)=4.46xx10^(-2)min^(-1)`