The rate constant for a first order decomposition reaction is given by: log K =10-(1000/ T ). Then, what will be activation energy in kcal / mol ?

Question

The rate constant for a first order decomposition reaction is given by: log K =10-(1000/ T ). Then, what will be activation energy in kcal / mol ? (A) 4.60 (B) 17.6 (C) 3.2 (D) 9.8

Tardigrade Admin · Accepted Answer

log K = log A -( E a /2.303 RT ) ldots(1) log k =10-(1000/ T ) (given) ldots(2) Compairing equ. (1) ( 2) ( E a /2.303 R )=1000 E a =4.60 k cal mol -1

Q. The rate constant for a first order decomposition reaction is given by : $lo g K = 10 - \frac{1000}{T}$ . Then, what will be activation energy in $k c a l / m o l$ ?

4.60

17.6

3.2

9.8

$lo g K = lo g A - \frac{E _{a}}{2.303 RT} \dots$ (1)
$lo g k = 10 - \frac{1000}{T}$ (given) $\dots (2)$
Compairing equ. (1) & ( 2 $)$
$\frac{E _{a}}{2.303 R} = 1000$
$E_{a} = 4.60 k c a l m o l^{- 1}$

Q. The rate constant for a first order decomposition reaction is given by :logK=10−T1000​. Then, what will be activation energy in kcal/mol ?

4.60

17.6

3.2

9.8

logK=logA−2.303RTEa​​…(1) logk=10−T1000​ (given) …(2) Compairing equ. (1) & ( 2) 2.303REa​​=1000 Ea​=4.60kcalmol−1

Q. The rate constant for a first order decomposition reaction is given by : $lo g K = 10 - \frac{1000}{T}$ . Then, what will be activation energy in $k c a l / m o l$ ?

$lo g K = lo g A - \frac{E _{a}}{2.303 RT} \dots$ (1)
$lo g k = 10 - \frac{1000}{T}$ (given) $\dots (2)$
Compairing equ. (1) & ( 2 $)$
$\frac{E _{a}}{2.303 R} = 1000$
$E_{a} = 4.60 k c a l m o l^{- 1}$