A reaction is second order with respect to a reactant. How is the rate of reaction affected if the concentration of the reactant is
$(i)$ doubled
$(ii)$ reduced to half $?$
A reaction is second order with respect to a reactant. How is the rate of reaction affected if the concentration of the reactant is
$(i)$ doubled
$(ii)$ reduced to half $?$
Let the concentration of the reactant be $[ A ]=a$
Rate of reaction, $R=k[A]^{2}$
$=k a^{2}$
$(i)$ If the concentration of the reactant is doubled, i.e. $[ A ]=2 a$, then the rate of the reaction would be
$R ^{\prime}=k(2 a)^{2}$
$=4 ka ^{2}$
$=4 R$
Therefore, the rate of the reaction would increase by $4$ times.
$(ii)$ If the concentration of the reactant is reduced to half, i.e. $[ A ]=\frac{1}{2} a$
the reaction would be
$R ^{\prime \prime}=k\left(\frac{1}{2} a\right)^{2}$
$=\frac{1}{4} k a$
$=\frac{1}{4} R$
Therefore, the rate of the reaction would be reduced to ${\frac{1}{4}^{th}}$
Similar Questions
The rate of the reaction becomes twice when the concentration of reactant becomes $8$ times then the order of the reaction is
The rate of the reaction becomes twice when the concentration of reactant becomes $8$ times then the order of the reaction is
In a reaction between $A$ and $B$, the initial rate of reaction $\left(r_{0}\right)$ was measured for different initial concentrations of $A$ and $B$ as given below:
$A/mol\,\,{L^{ - 1}}$
$0.20$
$0.20$
$0.40$
$B/mol\,\,{L^{ - 1}}$
$0.30$
$0.10$
$0.05$
${r_0}/mol\,\,{L^{ - 1}}\,\,{s^{ - 1}}$
$5.07 \times 10^{-5}$
$5.07 \times 10^{-5}$
$1.43 \times 10^{-4}$
What is the order of the reaction with respect to $A$ and $B$?
In a reaction between $A$ and $B$, the initial rate of reaction $\left(r_{0}\right)$ was measured for different initial concentrations of $A$ and $B$ as given below:
| $A/mol\,\,{L^{ - 1}}$ | $0.20$ | $0.20$ | $0.40$ |
| $B/mol\,\,{L^{ - 1}}$ | $0.30$ | $0.10$ | $0.05$ |
| ${r_0}/mol\,\,{L^{ - 1}}\,\,{s^{ - 1}}$ | $5.07 \times 10^{-5}$ | $5.07 \times 10^{-5}$ | $1.43 \times 10^{-4}$ |
What is the order of the reaction with respect to $A$ and $B$?
For the reaction
$2 \mathrm{H}_{2}(\mathrm{g})+2 \mathrm{NO}(\mathrm{g}) \rightarrow \mathrm{N}_{2}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})$
the observed rate expression is, rate $=\mathrm{k}_{\mathrm{f}}[\mathrm{NO}]^{2}\left[\mathrm{H}_{2}\right] .$ The rate expression of the reverse reaction is
For the reaction
$2 \mathrm{H}_{2}(\mathrm{g})+2 \mathrm{NO}(\mathrm{g}) \rightarrow \mathrm{N}_{2}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})$
the observed rate expression is, rate $=\mathrm{k}_{\mathrm{f}}[\mathrm{NO}]^{2}\left[\mathrm{H}_{2}\right] .$ The rate expression of the reverse reaction is
- [JEE MAIN 2020]
The formation of gas at the surface of tungsten due to adsorption is the reaction of order
The formation of gas at the surface of tungsten due to adsorption is the reaction of order
- [AIEEE 2002]
Write general reaction. Write rate law of general reaction.
Write general reaction. Write rate law of general reaction.