\(k\,= \,\frac{{2.303}}{t}\,\,\log \,\,\frac{{{{[A]}_0}}}{{{{[A]}_t}}}\)
\(\log \frac{{{[A]}_{0}}}{{{[A]}_{t}}}=\frac{kt}{2.303}\) \(=\frac{4.5\times {{10}^{-3}}\,{{\min }^{-1}}\times \,60\,\min }{2.303}\) \(=0.11729\)
\(\log \frac{{{{[A]}_0}}}{{{{[A]}_t}}}\,\, = \,Antilog \,0.1172\,= \,1.310\)
\({[A]_t}\, = \,\frac{{{{[A]}_0}}}{{1.310}}\, = \,\frac{M}{{1.310}}\, = \,0.763\,M\)
તબક્કો \(2:1\) કલાક \((60 \) મિનિટ) પછી દરની ગણતરી
\(60\) મિનિટ પછી દર \(=k{{[A]}_{t}}\) \(=4.5\times {{10}^{-3}}\,mi{{n}^{-1}}\) \(\times 0.763\,M=\) \(3.4354\times {{10}^{-3}}M\,mi{{n}^{-1}}\)
$[X]$ $0.1\,M$, $[Y]$ $0.1\,M$ દર $\rightarrow 0.002\,Ms^{-1}$
$[X]$ $0.2\,M$, $[Y]$ $0.1\,M$ દર $\rightarrow 0.002\,Ms^{-1}$
$[X]$ $0.3\,M$, $[Y]$ $0.2\,M$ દર $\rightarrow 0.008\,Ms^{-1}$
$[X]$ $0.4\,M$, $[Y]$ $0.3\,M$ દર $\rightarrow 0.018\,Ms^{-1}$
તો દર નિયમ ......
${A}+{B} \rightarrow {M}+{N}$ $......$ ${kJ} {mol}^{-1}$ બરાબર છે. (નજીકના પૂર્ણાંકમાં)