- A$[X]$
- B$[X]^2$
- C$\ln\,[X]$
- ✓$\frac{1}{[X]}$
$\int_{x_{0}}^{x}-\frac{d x}{\mid X]^{2}}=\int_{0}^{t} k d t$
$=>\frac{1}{[X]}-\frac{1}{\left[X_{0}\right]}=k t$
or $\frac{1}{[X]}=k t+\frac{1}{\left[X_{0}\right]}$
Hence plot of $\frac{1}{X}$ against time will be a straight line
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when $\underset{(i)}{\mathop{\gamma =C{{H}_{3}}CO_{2}^{-},}}\,\,\,\,\,\,\underset{(ii)}{\mathop{Cl-C{{H}_{2}}-CO_{2}^{-},}}\,\,\,\,\,\,\,\underset{(iii)}{\mathop{Ph-SO_{3}^{-}}}\,$
$(A)$ $\left[ Cr \left( NH _3\right)_5 Cl ^2 Cl _2\right.$ and $\left[ Cr \left( NH _3\right)_4 Cl _2\right) Cl$
$(B)$ $\left[ Co \left( NH _3\right)_4 Cl _2\right]^{+}$and $\left[ Pt \left( NH _3\right)_2\left( H _2 O \right) Cl \right]^{+}$
$(C)$ $\left[ CoBr _2 Cl _2\right]^{2-}$ and $\left[ PtBr _2 Cl _2\right]^{2-}$
$(D)$ $\left[ Pt \left( NH _3\right)_3\left( NO _3\right) Cl\right.$ and $\left[ Pt \left( NH _3\right)_3 Cl \right] Br$
| $[R]$ (molar) | $1.0$ | $0.75$ | $0.40$ | $0.10$ |
| $\mathrm{t}$ (min.) | $0.0$ | $0.05$ | $0.12$ | $0.18$ |
The order of the reaction is
$[P]\xrightarrow{\begin{subarray}{l}
(i)\,NAN{O_2}/HCl,\,0 - {5^o}C \\
(ii)\,\beta - napthol/NaOH
\end{subarray} }Colored\,\,Solid$
$[P]\xrightarrow{{B{r_2}/{H_2}O}}{C_7}{H_6}NB{r_3}$
The compound $[P]$ is