નીચે આપેલ પ્રક્કિયાવિધી દ્વારા થઈ રહી છે.
$NO + Br _2 \Leftrightarrow NOBr _2 \text { (fast) }$
$NOBr _2+ NO \rightarrow 2 NOBr$(ધીમી)
પ્રક્રિયાનો સમગ્ર ક્રમ $........$
$\left. r = K \left[ NOBr _2\right] NO \right] \quad--- \text { (i) }$
$Keq =\frac{\left[ NOBr _2\right]}{[ NO ]\left[ Br _2\right]}----- \text { (ii) }$
$\text { From (i) and (ii) }$
$r = K \cdot Keq \cdot[ NO ]\left[ Br _2\right][ NO ]$
$r = K ^{\prime}[ NO ]^2\left[ Br _2\right]$
$\text { Overall order }=3$
| No | $[NH_4^+]$ | $[NO_2^-]$ | rate of reaction |
| $1.$ | $0.24\, M$ | $0.10\, M$ | $7.2 \times {10^{ - 6}}$ |
| $2.$ | $0.12\, M$ | $0.10\, M$ | $3.6 \times {10^{ - 6}}$ |
| $3.$ | $0.12\, M$ | $0.15\, M$ | $5.4 \times {10^{ - 6}}$ |
$[X]$ $0.1\,M$, $[Y]$ $0.1\,M$ 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 ......
