Why can we not determine the order of a reaction by taking into consideration the balanced chemical equation ?
Why can we not determine the order of a reaction by taking into consideration the balanced chemical equation ?
Balanced chemical equation often leads to incorrect order or rate law. For example, the following reaction appears to be a tenth order reaction.
$(i)$ $\mathrm{CHCl}_{3}+\mathrm{Cl}_{2} \rightarrow \mathrm{CCl}_{4}+\mathrm{HCl}$
Experimental rate equation : $k\left[\mathrm{CHCl}_{3}\right]\left[\mathrm{Cl}_{2}\right]^{\frac{1}{2}}$
$(ii)$ $\mathrm{KClO}_{3}+6 \mathrm{FeSO}_{4}+3 \mathrm{H}_{2} \mathrm{SO}_{4} \rightarrow \mathrm{KCl}+3 \mathrm{H}_{2} \mathrm{O}+3 \mathrm{Fe}_{2}\left(\mathrm{SO}_{4}\right)_{3}$
However, this is actually a second order reaction. Actually the reaction is complex and occurs in several steps. The order of such reaction mechanism order is determined experimentally and gives the actual dependence of observed rate of reaction on the concentration of reactants.
Similar Questions
The rate of the reaction $CC{l_3}CHO + NO \to CHC{l_3} + NO + CO$ is given by Rate $ = K\,[CC{l_3}CHO]\,[NO]$. If concentration is expressed in moles/litre, the units of K are
The rate of the reaction $CC{l_3}CHO + NO \to CHC{l_3} + NO + CO$ is given by Rate $ = K\,[CC{l_3}CHO]\,[NO]$. If concentration is expressed in moles/litre, the units of K are
The reaction $2NO + Br_2 \rightarrow 2NOBr,$ follows the mechanism given below
$(I)$ $NO + Br_2 \rightleftharpoons NOBr_2 $ ........ Fast
$(II)$ $NOBr_2 + NO \rightarrow 2NOBr$ ......... Slow
The overall order of this reaction is
The reaction $2NO + Br_2 \rightarrow 2NOBr,$ follows the mechanism given below
$(I)$ $NO + Br_2 \rightleftharpoons NOBr_2 $ ........ Fast
$(II)$ $NOBr_2 + NO \rightarrow 2NOBr$ ......... Slow
The overall order of this reaction is
Which of the following reaction will have fractional order for $A_2$ or $B_2$ ?
Which of the following reaction will have fractional order for $A_2$ or $B_2$ ?
Hydrolysis of methyl acetate in aq. solution has been studied by titrating the liberated acetic acid against solidum hydroxide. The conc. of the ester at different time is given below :
Time $(t)$ $\min$
$0$
$30$
$60$
$90$
Con. of ester
$(C)$
$0.850$
$0.800$
$0.754$
$0.710$
Show that it follows a pseudo first order reaction as the conc. of $H_2O$ remain nearly constant $(54.2\,mol\,L^{-1})$ during the course of the reaction. What is the value of $k'$ in this reaction ?
Hydrolysis of methyl acetate in aq. solution has been studied by titrating the liberated acetic acid against solidum hydroxide. The conc. of the ester at different time is given below :
| Time $(t)$ $\min$ | $0$ | $30$ | $60$ | $90$ |
| Con. of ester $(C)$ |
$0.850$ | $0.800$ | $0.754$ |
$0.710$ |
Show that it follows a pseudo first order reaction as the conc. of $H_2O$ remain nearly constant $(54.2\,mol\,L^{-1})$ during the course of the reaction. What is the value of $k'$ in this reaction ?
The following data was obtained for chemical reaction given below at $975\, \mathrm{~K}$.
$2 \mathrm{NO}_{(\mathrm{g})}+2 \mathrm{H}_{2(\mathrm{~g})} \rightarrow \mathrm{N}_{2(\mathrm{~g})}+2 \mathrm{H}_{2} \mathrm{O}_{(\mathrm{g})}$
$[NO]$
$\mathrm{mol} \mathrm{L}^{-1}$
${H}_{2}$
$\mathrm{mol} \mathrm{L}^{-1}$
Rate
$\mathrm{mol}L^{-1}$ $s^{-1}$
$(A)$
$8 \times 10^{-5}$
$8 \times 10^{-5}$
$7 \times 10^{-9}$
$(B)$
$24 \times 10^{-5}$
$8 \times 10^{-5}$
$2.1 \times 10^{-8}$
$(C)$
$24 \times 10^{-5}$
$32 \times 10^{-5}$
$8.4 \times 10^{-8}$
The order of the reaction with respect to $\mathrm{NO}$ is ..... .
The following data was obtained for chemical reaction given below at $975\, \mathrm{~K}$.
$2 \mathrm{NO}_{(\mathrm{g})}+2 \mathrm{H}_{2(\mathrm{~g})} \rightarrow \mathrm{N}_{2(\mathrm{~g})}+2 \mathrm{H}_{2} \mathrm{O}_{(\mathrm{g})}$
|
$[NO]$ $\mathrm{mol} \mathrm{L}^{-1}$ |
${H}_{2}$ $\mathrm{mol} \mathrm{L}^{-1}$ |
Rate $\mathrm{mol}L^{-1}$ $s^{-1}$ |
|
| $(A)$ | $8 \times 10^{-5}$ | $8 \times 10^{-5}$ | $7 \times 10^{-9}$ |
| $(B)$ | $24 \times 10^{-5}$ | $8 \times 10^{-5}$ | $2.1 \times 10^{-8}$ |
| $(C)$ | $24 \times 10^{-5}$ | $32 \times 10^{-5}$ | $8.4 \times 10^{-8}$ |
The order of the reaction with respect to $\mathrm{NO}$ is ..... .
- [JEE MAIN 2021]