By giving three examples explain that ''the exponents of the concentration terms are same or not as their stoichiometric coefficients in the balanced chemical reaction.''
By giving three examples explain that ''the exponents of the concentration terms are same or not as their stoichiometric coefficients in the balanced chemical reaction.''
In the following balanced equation the stoichiometric co-efficient are expressed as exponents of the concentration.
Reaction $(i)$ is the formation of $\mathrm{NO}_{2}$
$2 \mathrm{NO}_{(\mathrm{g})}+\mathrm{O}_{2(\mathrm{~g})} \rightarrow 2 \mathrm{NO}_{2(\mathrm{~g})}$
The differential form of this rate expression is given as Rate $=-\frac{\mathrm{d}[\mathrm{R}]}{\mathrm{dt}}=k[\mathrm{NO}]^{2}\left[\mathrm{O}_{2}\right]$
The rate law is not theoretically but must be determined experimentally. So, in the following reaction the value of exponent is different and it is determined by experimental method.
$(i)$ Reaction : $\mathrm{CHCl}_{3}+\mathrm{Cl}_{2} \rightarrow \mathrm{CCl}_{4}+\mathrm{HCl}$
The differential form of this rate expression is,
Rate $=-\frac{\mathrm{d}[\mathrm{R}]}{\mathrm{dt}}=k\left[\mathrm{CHCl}_{3}\right]\left[\mathrm{Cl}_{2}\right]^{\frac{1}{2}}$
and $\frac{1}{2}$ is exponent of $\left[\mathrm{Cl}_{2}\right]$ but it is not coefficient of $\mathrm{Cl}_{2}$ in reaction.
$(ii)$ Reaction : $\mathrm{CH}_{3} \mathrm{COOC}_{2} \mathrm{H}_{5}+\mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{CH}_{3} \mathrm{COOH}+\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}$
In this reaction the rate expression is according on the base of experimental value as given $\begin{aligned} \text { Rate } &=-\frac{\mathrm{d}[\mathrm{R}]}{\mathrm{dt}}=k\left[\mathrm{CH}_{3} \mathrm{COOC}_{2} \mathrm{H}_{5}\right]\left[\mathrm{H}_{2} \mathrm{O}\right]^{0} \\ &=k\left[\mathrm{CH}_{3} \mathrm{COOC}_{2} \mathrm{H}_{5}\right] \end{aligned}$
Thus, $\mathrm{H}_{2} \mathrm{O}$ act as a reactant in reaction but experimentally it is determind that the rate of reaction does not depend on $\left[\mathrm{H}_{2} \mathrm{O}\right]$.
So, rate law for any reaction cannot be predicted by merely looking at the balanced chemical equation i.e. theoretically but must be determined experimentally.
Similar Questions
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The half-life of $2 $ sample are $0.1 $ and $ 0.4 $ seconds. Their respective concentration are $200 $ and $ 50 $ respectively. What is the order of the reaction
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Write differential rate expression of following reaction and give its order of reaction :
$H _{2} O _{2}+ I ^{-} \rightarrow H _{2} O + IO ^{-}$
$H _{2} O _{2}+ IO ^{-} \rightarrow H _{2} O + I ^{-}+ O _{2}$
Write differential rate expression of following reaction and give its order of reaction :
$H _{2} O _{2}+ I ^{-} \rightarrow H _{2} O + IO ^{-}$
$H _{2} O _{2}+ IO ^{-} \rightarrow H _{2} O + I ^{-}+ O _{2}$
Select the rate law for reaction $A + B \longrightarrow C$
Exp
$[A]$
$[B]$
Rate
$1$
$0.012$
$0.035$
$0.10$
$2$
$0.024$
$0.070$
$0.80$
$3$
$0.024$
$0.035$
$0.10$
$4$
$0.012$
$0.070$
$0.80$
Select the rate law for reaction $A + B \longrightarrow C$
| Exp | $[A]$ | $[B]$ | Rate |
| $1$ | $0.012$ | $0.035$ | $0.10$ |
| $2$ | $0.024$ | $0.070$ | $0.80$ |
| $3$ | $0.024$ | $0.035$ | $0.10$ |
| $4$ | $0.012$ | $0.070$ | $0.80$ |
If $‘a’$ is the initial concentration of the reactant, the half-life period of the reaction of $n^{t h}$ order in inversely proportional to
If $‘a’$ is the initial concentration of the reactant, the half-life period of the reaction of $n^{t h}$ order in inversely proportional to