\(\log \,\frac{{{K_2}}}{{{K_1}}}\,\, = \,\,\frac{{{E_a}}}{{2.303R}}\,\left( {\frac{{{T_2}\, - \,\,{T_1}}}{{{T_1}{T_2}}}} \right)\)
આપેલ : તાપમાને આચળાંક \(\, = \,\,\frac{{{K_2}}}{{{K_1}}}\,\, = \,\,2\)
\({T_1}\,\, = \,\,25\,\, + \,\,273\,\, = \,\,298,\)
\(\,{T_2}\,\, = \,\,35\,\, + \,\,273\,\, = \,308,\,\,R\,\, = \,\,8.314\)
\(\log \,2\,\, = \,\,\frac{{{E_a}}}{{2.303\,\, \times \,\,8.314}}\,\, \times \,\,\left( {\frac{{10}}{{298\,\, \times \,\,308}}} \right)\)
\({E_a}\,\, = \,\,52.31\,\,KJ\)
$-\frac{d[{{N}_{2}}{{O}_{5}}]}{dt}={{K}_{1}}[{{N}_{2}}{{O}_{5}}]$ ,
$\frac{d[N{{O}_{2}}]}{dt}={{k}_{2}}[{{N}_{2}}{{O}_{5}}]$ ,
$\frac{d[{{O}_{2}}]}{dt}={{K}_{3}}[{{N}_{2}}{{O}_{5}}]$
તો $K_1$, $K_2$ અને $K_3 $ વચ્ચેનો સંબંધ શું થાય?
$T$ (in, $K$) $- 769$ , $1/T$ (in, $K^{-1}$ ) $- 1.3\times 10^{-3},$
$\log_{10}K - 2.9\,T$ (in, $K$) $- 667$, $1/T$ (in, $K^{-1}) - 1.5\times 10^{-3}$, $\log_{10}\,K - 1.1$
(લો: $\log 2=0.30 ; \log 2.5=0.40)$