For reactions textA arrow textB and textP arrow textQ Arrhenius constants are 108 and 1010 respectively. If textE textA arrow textB text= text600 textcal/mole and textE textP arrow textQ text= text1200 textcal/mole , then find the temperature at which their rate constants are same. (Given: textR text= text2 textcal/mole/K )

Question

For reactions textA arrow textB and textP arrow textQ Arrhenius constants are 108 and 1010 respectively. If textE textA arrow textB text= text600 textcal/mole and textE textP arrow textQ text= text1200 textcal/mole , then find the temperature at which their rate constants are same. (Given: textR text= text2 textcal/mole/K ) (A) 600 K (B) 300× 4.606 K (C) (300/4.606)K (D) (4.606/600)K

Tardigrade Admin · Accepted Answer

A1e- E a1 / R T=A2e- E a2 / R T (A2/A1)=e(E a2 - E a1) / R T 102=Exp (600/R T) R=2 c a l / K-mol 2ln10=(600/2 T) T= (300/2 × 2.303) K

Q. For reactions $A \to B$ and $P \to Q$ Arrhenius constants are $1 0^{8}$ and $1 0^{10}$ respectively. If $E_{A \to B} = 600 cal/mole$ and $E_{P \to Q} = 1200 cal/mole$ , then find the temperature at which their rate constants are same.

(Given: $R = 2 cal/mole/K$ )

$600 K$

$300 \times 4.606 K$

$\frac{300}{4.606} K$

$\frac{4.606}{600} K$

$A_{1} e^{- E a_{1} / RT} = A_{2} e^{- E a_{2} / RT}$

$\frac{A _{2}}{A _{1}} = e^{(E a_{2} - E a_{1}) / RT}$

$1 0^{2} = E x p {\frac{600}{RT}}$ $R = 2 c a l / K - m o l$

$2 l n 10 = \frac{600}{2 T}$

$T = {\frac{300}{2 \times 2.303}} K$

Q. For reactions A→B and P→Q Arrhenius constants are 108 and 1010 respectively. If EA→B​=600cal/mole and EP→Q​=1200cal/mole , then find the temperature at which their rate constants are same. (Given: R=2cal/mole/K )

600K

300×4.606K

4.606300​K

6004.606​K