The rate constants K 1 and K 2 for two different reactions are 1015 e -1000- T and 1016 e -2000- T respectively- The temperature at which K 1- K 2 is-A- 2000-2-303 KB- 2000 KC- 1000-2-303 KD- 1000 K

The correct option is C 1000/2.303 KK_1=K_2 10^16 l_n 2000/T=10^15 l_n 1000/T ln 10^16 2000/T=ln 10^15 1000/T 2.303× 16 2000/T=2.303× 15 1000/T T=1000/2.303 ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Chemistry

Zero Order Reaction

The rate cons...

Question

The rate constants $K_{1}$ and $K_{2}$ for two different reactions are $10^{15} e^{- \frac{1000}{T}}$ and $10^{16} e^{- \frac{2000}{T}}$ respectively. The temperature at which $K_{1} = K_{2}$ is:

\frac{2000}{2.303} K

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2000 K

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{1000}{2.303} K

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

1000 K

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $\frac{1000}{2.303} K$
$K_{1} = K_{2}$
$10^{16} l_{n} - \frac{2000}{T} = 10^{15} l_{n} - \frac{1000}{T}$
$l n 10^{16} - \frac{2000}{T} = l n 10^{15} - \frac{1000}{T}$
$2.303 \times 16 - \frac{2000}{T} = 2.303 \times 15 - \frac{1000}{T}$
$T = \frac{1000}{2.303} K$

Suggest Corrections

Similar questions

Q. The rate constants

k_{1}

and

k_{2}

for two different reactions are

10^{15} e^{- 200 / T}

and

10^{16} e^{- 1000 / T}

respectively.
The temperature at which

k_{1} = k_{2}

is:

Q. The ratio constants for

k_{1}

and

k_{2}

for two different reactions are

10^{16} e^{- 2000 T}

and

10^{15} e^{- 1000 / T}

respectively. Temperature at which

k_{1} = k_{2}

is:

Q. The rate constants

k_{1}

and

k_{2}

for two different reactions are

10^{16} . e^{- 2000 / T}

and

10^{15} . e^{- 1000 / T}

respectively, the temperature at which

k_{1} = k_{2}

, is:

Q. The rate constants

K_{1}

and

K_{2}

for two different group are

10^{1} 6

e^{- 2000 / 4}

and

10^{15} e^{- 1000 / T}

The temperature at which

K_{1}

and

K_{2}

is:

Q. The rate constant

k_{1}

and

k_{2}

for two different reactions are

10^{16} e^{- 2000 / T}

and

10^{15} e^{- 2000 / T}

respectively the temperature at which

k_{1} = k_{2}

is:

Join BYJU'S Learning Program

The rate constants K1 and K2 for two different reactions are 1015e−1000T and 1016e−2000T respectively. The temperature at which K1=K2 is:

The correct option is C 10002.303KK1=K2 1016ln−2000T=1015ln−1000T ln 1016−2000T=ln 1015−1000T 2.303×16−2000T=2.303×15−1000T T=10002.303K

The rate constants $K_{1}$ and $K_{2}$ for two different reactions are $10^{15} e^{- \frac{1000}{T}}$ and $10^{16} e^{- \frac{2000}{T}}$ respectively. The temperature at which $K_{1} = K_{2}$ is:

The correct option is C $\frac{1000}{2.303} K$
$K_{1} = K_{2}$
$10^{16} l_{n} - \frac{2000}{T} = 10^{15} l_{n} - \frac{1000}{T}$
$l n 10^{16} - \frac{2000}{T} = l n 10^{15} - \frac{1000}{T}$
$2.303 \times 16 - \frac{2000}{T} = 2.303 \times 15 - \frac{1000}{T}$
$T = \frac{1000}{2.303} K$