The rate constants k1 and k2 for two different reactions are 1016 .e-2000/T and 1015 .e-1000/T,respectively. The temperature at which k1= k2 is

Question

The rate constants k1 and k2 for two different reactions are 1016 .e-2000/T and 1015 .e-1000/T,respectively. The temperature at which k1= k2 is (A) 1000 K (B) (2000/2.303) K (C) 2000 K (D) (1000/2.303) K

Tardigrade Admin · Accepted Answer

k 1=1016 e -2000 T k 2=1015 e -2000 T By taking log on both of them Longrightarrow log k 1= log 1016-(2000/2.303) t Longrightarrow log k 2= log 1015-(1000/2.303) t Now, k 1= k 2, Hence from equations T =(1000/2.303) k

Q. The rate constants $k_{1}$ and $k_{2}$ for two different reactions are $1 0^{16} . e^{- 2000/ T}$ and $1 0^{15} . e^{- 1000/ T}$ ,respectively. The temperature at which $k_{1} = k_{2}$ is

1000 K

$\frac{2000}{2.303} K$

2000 K

$\frac{1000}{2.303} K$

$k 1 = 1 0^{16} e^{- 2000 T}$
$k 2 = 1 0^{15} e^{- 2000 T}$
By taking log on both of them
$⟹ lo g k 1 = lo g 1 0^{16} - \frac{2000}{2.303} t$
$⟹ lo g k 2 = lo g 1 0^{15} - \frac{1000}{2.303} t$
Now, $k 1 = k 2$ , Hence from equations $T = \frac{1000}{2.303} k$

Q. The rate constants k1​ and k2​ for two different reactions are 1016.e−2000/T and 1015.e−1000/T,respectively. The temperature at which k1​=k2​ is

1000 K

2.3032000​K

2000 K

2.3031000​K

k1=1016e−2000T k2=1015e−2000T By taking log on both of them ⟹logk1=log1016−2.3032000​t ⟹logk2=log1015−2.3031000​t Now, k1=k2, Hence from equations T=2.3031000​k

Q. The rate constants $k_{1}$ and $k_{2}$ for two different reactions are $1 0^{16} . e^{- 2000/ T}$ and $1 0^{15} . e^{- 1000/ T}$ ,respectively. The temperature at which $k_{1} = k_{2}$ is

$\frac{2000}{2.303} K$

$\frac{1000}{2.303} K$

$k 1 = 1 0^{16} e^{- 2000 T}$
$k 2 = 1 0^{15} e^{- 2000 T}$
By taking log on both of them
$⟹ lo g k 1 = lo g 1 0^{16} - \frac{2000}{2.303} t$
$⟹ lo g k 2 = lo g 1 0^{15} - \frac{1000}{2.303} t$
Now, $k 1 = k 2$ , Hence from equations $T = \frac{1000}{2.303} k$