The temperature at which \(k_{1}=k_{2}\) will be
\(10^{16} e^{-2000 / T}=10^{15} e^{-1000 / T}\)
\(\Rightarrow \frac{e^{-20001 T}}{e^{-1000 T}}=\frac{10^{15}}{10^{16}}\)
\(\Rightarrow e^{\frac{-1000}{T}}=10^{-1} \Rightarrow \log _{e} e^{\frac{-1000}{T}}=\log _{e} 10^{-1}\)
\(\Rightarrow 2.303 \log _{10} e^{\frac{-1000}{T}}=2.303 \times \log _{10} 10^{-1}\)
\(\Rightarrow \frac{-1000}{T} \times \log _{10} e=-1\) \(\Rightarrow T=\frac{1000}{2.303} \mathrm{K}\)
$2NO \rightleftharpoons {N_2}O + \left[ O \right]$
${O_3} + \left[ O \right] \to 2{O_2}\,(slow)$
તો પ્રકિયાનો કમ જણાવો.
$2{N_2}{O_5}\, \to \,4N{O_2}\, + \,{O_2}$
પ્રકિયા નો દર શું હશે ?
(આપેલ : એન્ટીલોગ $antilog$ $0.125=1.333$,
$\text { antilog } 0.693=4.93 \text { ) }$