| $[A] (mol\,L^{-1})$ | $[B] (mol\,L^{-1})$ | 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, $r\, = \,K[A]\,{[B]^2}$
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$H_2O(l) \rightarrow H^+(aq) + OH^-(aq)\,;\,\,\Delta H = 57.32\,kJ$
$H_2(g)+ \frac{1}{2} O_2(g) \rightarrow H_2O(l)\,;\,\, \Delta H=-286.20\,kJ$
The value of enthalpy of formation of $OH^-$ ion at $25\,^oC$ is : .............. $\mathrm{kJ}$

$(I)$ ${H_2}(g) + \frac{1}{2}{O_2}(g) \to {H_2}O(l);$
$\Delta {H^o_{298\,K}} = - 285.9\,kJ\,mo{l^{ - 1}}$
$(II)$ ${H_2}(g) + \frac{1}{2}{O_2}(g) \to {H_2}O(g);$
$\Delta {H^o_{298\,K}} = - 241.8\,kJ\,mo{l^{ - 1}}$
The molar enthalpy of vapourisation of water will be.....$kJ\,mol^{-1}$