| $[A] (mol\,L^{-1})$ | $[B] (mol\,L^{-1})$ | પ્રક્રિયાનો પ્રારંભિક વેગ $(mol\, L^{-1}\,s^{-1} )$ |
| $0.05$ | $0.05$ | $0.045$ |
| $0.10$ | $0.05$ | $0.090$ |
| $0.20$ | $0.10$ | $0.72$ |
\(0.045\, = \,K{(0.05)^x}{(0.05)^y}\) ...... \((1)\)
\(0.090\, = \,K{(0.10)^x}{(0.05)^y}\) .......\((2)\)
\(0.72\, = \,K{(0.20)^x}{(0.10)^y}\) ........\((3)\)
Diving \((1)\) by \((2)\) We get
\(\frac{{0.045}}{{0.090}}\, = \,{\left( {\frac{{0.05}}{{0.10}}} \right)^x}\, \Rightarrow \,x = 1\)
Diving \((2)\) by \((3)\)
\(\frac{{0.090}}{{0.720}}\, = \,{\left( {\frac{{0.10}}{{0.20}}} \right)^x}\,{\left( {\frac{{0.05}}{{0.10}}} \right)^y}\, \Rightarrow \,y = 2\,\)
Hence, \(r\, = \,K[A]\,{[B]^2}\)
|
$[R] (molar)$ |
$1.0$ |
$0.76$ |
$0.40$ |
$0.10$ |
|
$t (min.)$ |
$0.0$ |
$0.05$ |
$0.12$ |
$0.18$ |
તો પ્રક્રિયાનો ક્રમ $...$ થશે.
$T$ (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, $K^{-1}) - 1.5\times 10^{-3}$, $\log_{10}\,K - 1.1$
${\log _{10}}\,\left[ { - \frac{{d\left[ A \right]}}{{dt}}} \right] = {\log _{10}}\,\left[ {\frac{{d\left[ B \right]}}{{dt}}} \right] + 0.3010$