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 ...

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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

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2000 K

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\frac{1000}{2.303} K

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1000 K

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Solution

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

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