Question

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:

• A. $T=431.2K$
  
• B. $T=432.2K$
• C. $T=404.2K$
• D. $T=434.2K$

Answer 1

Answer: (D)

$T=434.2K$

$K_1={10}^{16}{e}^{-2000/T}$

$K_2={10}^{15}{e}^{-1000/T}$

The temperature at which $k_1=k_2$ will be:

${10}^{16}{e}^{-2000/T}={10}^{15}{e}^{-1000/T}$

$\frac{e^{-2000/T}}{e^{-1000/T}}=\frac{{10}^{15}}{{10}^{16}}$

$e^{-1000/T}={10}^{-1}$

$lo{g}_e{e}^{-1000/T}=lo{g}_e{10}^{-1}$

$2.303\times lo{g}_{10}{e}^{-1000/T}=2.303\times lo{g}_{10}{10}^{-1}$

$\frac{-1000}{T}\times lo{g}_{10}e=-1$

On solving, we get,

$T=1000/2.303K$

$T=434.2K$