| $[A] (mol\,L^{-1})$ | $[B] (mol\,L^{-1})$ | Initial Rate $(mol\, L^{-1}\,s^{-1} )$ |
| $0.05$ | $0.05$ | $0.045$ |
| $0.10$ | $0.05$ | $0.090$ |
| $0.20$ | $0.10$ | $0.72$ |
- ARate $= k[A]^2[B]^2$
- ✓Rate $= k[A][B]^2$
- CRate $= k[A][B]$
- DRate $= k[A]^2[B]$
| $[A] (mol\,L^{-1})$ | $[B] (mol\,L^{-1})$ | Initial Rate $(mol\, L^{-1}\,s^{-1} )$ |
| $0.05$ | $0.05$ | $0.045$ |
| $0.10$ | $0.05$ | $0.090$ |
| $0.20$ | $0.10$ | $0.72$ |
$0.045\, = \,K{(0.05)^x}{(0.05)^y}$ ...... $(1)$
$0.090\, = \,K{(0.10)^x}{(0.05)^y}$ .......$(2)$
$0.72\, = \,K{(0.20)^x}{(0.10)^y}$ ........$(3)$
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, $r\, = \,K[A]\,{[B]^2}$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$A$. Phenol $B$. $p -$ Cresol
$C$. $m-$ Nitrophenol $D$. $p-$ Nitrophenol is
Reason : The concentration of salt content in the cells increases.