The time required for 10% completion of a first order reaction at 298 K is equal to that required for its 25% completion at 308 k. If the value of A is 4xx10^(10)s^(-1). Calculate the rate constant, k at 318 k. 2.89xx10^(-2)s^(-1) 3.26xx10^(-2) s^(-1) 1.03xx10^(-2)s^(-1) 0.03xx10^(-2)s^(-1)

The time required for 10% completion of a first order reaction at 298

1.	The time required for 10% completion of a first order reaction at 298 K is equal to that required for its 25% completion at 308 k. If the value of A is 4xx10^(10)s^(-1). Calculate the rate constant, k at 318 k. 2.89xx10^(-2)s^(-1) 3.26xx10^(-2) s^(-1) 1.03xx10^(-2)s^(-1) 0.03xx10^(-2)s^(-1)
Answer» the energy below which colloiding molecules will not react the total energy of the reacting molecules at a temperature, T The fraction of molecules with energy greater than the activation energy of the reaction `0.03 xx 10^(-2) S^(-1)` Solution :For first order reaction, `k=(2.303)/(t)"LOG"[A]_(0)/([A])` At 298 K,`K_(1)=(2.303)/(t)"log"(100)/(90)` At 308 k,`k_(2)=(2.303)/(t)"log"(100)/(75)` On dividing Eq.(ii) by Eq. (i), we get `k_(2)/(k_(1))=(log""100/75)/(log""100/90)=2.73` According to ARRHENIUS theory, `log""k_(2)/k_(1)=E_(a)/(2.303R)xx(T_(2)-T_(1))/(T_(1)T_(2))` `log2.73=E_(a)/(2.303 R)[(308-298)/(298xx308)]` `E_(a)=76651 J" mol"^(-1)` "Now, according to Arrhenius equation," `logk=logA-E_(a)/(2.303 RT)` `logk=log(4xx10^(10))-(76651 J" mol"^(-1))/(2.303xx(8.314 J" mol"^(-1)k^(-1))xx(318 k))` `=-1.9868` k=Antilog `(-19868)=1.035xx10^(-2)s^(-1)`