Question

The rate constant k1 and k2 for two different reactions are

Answer 1

Answer:

$k_{1} =10^{16} e_{-2000/T}$

$k_{2} =10^{15 } e^{-1000/T}$

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} }$

$e^{\frac{-1000}{T} } =10^{-1}$

$log_{e} e^{2} \frac{-1000}{T} =log_{e} 10^{-1}$

$2.303log_{10} e^{\frac{-1000}{T} } =2.303*log_{10} 10^{-1}$

$\frac{-1000}{T} *log_{10} e=-1$

$T=\frac{1000}{2.303} K$

Answer 2

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