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)`

Options

A. the energy below which colloiding molecules will not react
B. the total energy of the reacting molecules at a temperature, T
C. The fraction of molecules with energy greater than the activation energy of the reaction
D. `0.03 xx 10^(-2) S^(-1)`

Answer

C

Explanation

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)`