Two reactions, (i) A to Products, follow first order kinetics. The rate of reaction (i) is doubled when temperature is raised from 300 K to 310 K. The half-life for this reaction at 310 K is 30 minutes. At the same temperature, B decomposes twice as fast as A. If the energy of activation for the reaction for the reaction (ii) is half taht of reaction (i), calculate of reaction (ii) at 300 K.

Two reactions, (i) A to Products, follow first order kinetics. The rat

1.	Two reactions, (i) A to Products, follow first order kinetics. The rate of reaction (i) is doubled when temperature is raised from 300 K to 310 K. The half-life for this reaction at 310 K is 30 minutes. At the same temperature, B decomposes twice as fast as A. If the energy of activation for the reaction for the reaction (ii) is half taht of reaction (i), calculate of reaction (ii) at 300 K.
Answer» Solution :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)`