In a reaction if the concentration of reactant A is tripled, the rate of reaction becomes twenty seven times. What is the order of the reaction ?
In a reaction if the concentration of reactant A is tripled, the rate of reaction becomes twenty seven times. What is the order of the reaction ?
Reaction order $=3$
For any reaction $\mathrm{R} \rightarrow \mathrm{P}$ and order $=n$ so rate $r_{1}=k[\mathrm{R}]^{n}$
If the concentration becomes 3 times. i.e. $3 \mathrm{R}$, then rate, $r_{2}=k[3 \mathrm{R}]^{n}$
So, $\frac{r_{2}}{r_{1}}=\frac{k[3 \mathrm{R}]^{n}}{k[\mathrm{R}]^{n}}=\frac{27}{\mathrm{I}}$
$\therefore \frac{{ }^{27} r_{1}}{r_{1}}=27$
$\therefore 27=3^{n}$
$\therefore 3=3^{n}$
So, $n=3$
Similar Questions
The rate of reaction $A + 2B \to 3C$ becomes $72\, times$ when concentration of $A$ is tripled and concentration of $B$ is doubled then the order of reaction with respect to $A$ and $B$ respectively is
The rate of reaction $A + 2B \to 3C$ becomes $72\, times$ when concentration of $A$ is tripled and concentration of $B$ is doubled then the order of reaction with respect to $A$ and $B$ respectively is
Consider the following gas-phase reaction.
$2HI(g) \longrightarrow H_2(g) + I_2(g)$
and the following experimental data obtained at $555\, K$, What is the order of the reaction with respect to $HI$ $(g)$ ?
$[HI]$, $M$
rate, $Ms^{-1}$
$0.0500$
$8.80 \times {10^{ - 10}}$
$0.1000$
$3.52 \times {10^{ - 9}}$
$0.1500$
$7.92 \times {10^{ - 9}}$
Consider the following gas-phase reaction.
$2HI(g) \longrightarrow H_2(g) + I_2(g)$
and the following experimental data obtained at $555\, K$, What is the order of the reaction with respect to $HI$ $(g)$ ?
| $[HI]$, $M$ | rate, $Ms^{-1}$ |
| $0.0500$ | $8.80 \times {10^{ - 10}}$ |
| $0.1000$ | $3.52 \times {10^{ - 9}}$ |
| $0.1500$ | $7.92 \times {10^{ - 9}}$ |
The reaction between $A$ and $B$ is first order with respect to $A$ and zero order with respect to $B$. Fill in the blanks in the following table:
Experiment
$[ A ] / mol\, ^{-1}$
$[ B ] / mol\, ^{-1}$
Initial rate $/$ $mol$ $L^{-1}$ $min$ $^{-1}$
$I$
$0.1$
$0.1$
$2.0 \times 10^{-2}$
$II$
-
$0.2$
$4.0 \times 10^{-2}$
$III$
$0.4$
$0.4$
-
$IV$
-
$0.2$
$2.0 \times 10^{-2}$
The reaction between $A$ and $B$ is first order with respect to $A$ and zero order with respect to $B$. Fill in the blanks in the following table:
| Experiment | $[ A ] / mol\, ^{-1}$ | $[ B ] / mol\, ^{-1}$ | Initial rate $/$ $mol$ $L^{-1}$ $min$ $^{-1}$ |
| $I$ | $0.1$ | $0.1$ | $2.0 \times 10^{-2}$ |
| $II$ | - | $0.2$ | $4.0 \times 10^{-2}$ |
| $III$ | $0.4$ | $0.4$ | - |
| $IV$ | - | $0.2$ | $2.0 \times 10^{-2}$ |
Assertion : The kinetics of the reaction -
$mA + nB + pC \to m' X + n 'Y + p 'Z$
obey the rate expression as $\frac{{dX}}{{dt}} = k{[A]^m}{[B]^n}$.
Reason : The rate of the reaction does not depend upon the concentration of $C$.
Assertion : The kinetics of the reaction -
$mA + nB + pC \to m' X + n 'Y + p 'Z$
obey the rate expression as $\frac{{dX}}{{dt}} = k{[A]^m}{[B]^n}$.
Reason : The rate of the reaction does not depend upon the concentration of $C$.
- [AIIMS 2011]
Consider following two reaction,
$A \to {\text{Product ;}}\,\, - \frac{{d[A]}}{{dt}} = {k_1}{[A]^o}$
$B \to {\text{Product ;}}\,\, - \frac{{d[B]}}{{dt}} = {k_2}{[B]}$
Units of $k_1$ and $k_2$ are expressed in terms of molarity $(M)$ and time $(sec^{-1})$ as
Consider following two reaction,
$A \to {\text{Product ;}}\,\, - \frac{{d[A]}}{{dt}} = {k_1}{[A]^o}$
$B \to {\text{Product ;}}\,\, - \frac{{d[B]}}{{dt}} = {k_2}{[B]}$
Units of $k_1$ and $k_2$ are expressed in terms of molarity $(M)$ and time $(sec^{-1})$ as