Rate of two reactions whose rate constants are k1 and k2 are equal at 300 K such that: Ea2-Ea1=2 R T, So calculate ln (A2/A1)

Question

Rate of two reactions whose rate constants are k1 and k2 are equal at 300 K such that: Ea2-Ea1=2 R T, So calculate ln (A2/A1) (A)  ln 4 (B) 2 (C)  log 2 (D) 2- ln 2

Tardigrade Admin · Accepted Answer

( text rate )1=( text rate )2 and rate propto k Thus, (k2/k1)=(( text rate )2/( text rate )1)=1 ln k= ln A-(Ea/R T) ln (k2/k1)= ln (A2/A1)+(Ea2-Ea1/R T) ln 1= ln (A2/A1)-(2 R T/R T) ln (A2/A1)=2

Q. Rate of two reactions whose rate constants are $k_{1}$ and $k_{2}$ are equal at $300 K$ such that :
$E_{a_{2}} - E_{a_{1}} = 2 RT,$
So calculate $ln \frac{A _{2}}{A _{1}}$

$ln 4$

$2$

$lo g 2$

$2 - ln 2$

Q. Rate of two reactions whose rate constants are k1​ and k2​ are equal at 300K such that : Ea2​​−Ea1​​=2RT, So calculate lnA1​A2​​

ln4

2

log2

2−ln2

( rate )1​=( rate )2​ and rate ∝k Thus, k1​k2​​=( rate )1​( rate )2​​=1 lnk=lnA−RTEa​​ lnk1​k2​​=lnA1​A2​​+RTEa2​​−Ea1​​​ ln1=lnA1​A2​​−RT2RT​lnA1​A2​​=2

Q. Rate of two reactions whose rate constants are $k_{1}$ and $k_{2}$ are equal at $300 K$ such that :
$E_{a_{2}} - E_{a_{1}} = 2 RT,$
So calculate $ln \frac{A _{2}}{A _{1}}$

$ln 4$

$2$

$lo g 2$

$2 - ln 2$