Embibe Experts Solutions for Chapter: Chemical Kinetics, Exercise 2: Exercise 2

In a first order reaction, the concentration of the reactant is decreased from $1.0 \mathrm{M}$ to $0.25 \mathrm{M}$ in $20$ minute. The rate constant of the reaction would be :

Catalyst increases the rate of reaction because :

For a first order reaction $\text{A}\stackrel{}{\to }\text{P}$, the temperature $\left(\mathrm{T}\right)$ dependent rate constant ($\mathrm{k}$) was found to follow the equation  $\text{log k = -(2000)}\frac{1}{\mathrm{T}}\text{ + 6.0}$. The pre-exponential factor $\mathrm{A}$ and the activation energy ${\text{E}}_{\text{a}}$, respectively, are ______.

Plots showing the variation of the rate constant $\left(\mathrm{k}\right)$ with temperature $\left(\mathrm{T}\right)$ are given below. The plot that follows Arrhenius equation is:

For the elementary reaction $\mathrm{M}\to \mathrm{N}$, the rate of disappearance of $\mathrm{M}$ increases by a factor of $8$ upon doubling the concentration of $\mathrm{M}$. The order of the reaction with respect to $\mathrm{M}$ is

For a chemical reaction, $\mathrm{A}\to \mathrm{B}$, it is observed that the rate of reaction doubles when the concentration of $\mathrm{A}$ is increased four times. The order of reaction in $\mathrm{A}$ is :

For the reaction $2{\mathrm{NO}}_2\to {\mathrm{N}}_2{\mathrm{O}}_2+{\mathrm{O}}_2,$ rate expression is as follows $-\frac{\mathrm{d}\left[{\mathrm{NO}}_2\right]}{\mathrm{dt}}=\mathrm{K}{\left[{\mathrm{NO}}_2\right]}^{\mathrm{n}},$ where $\mathrm{K}=3\times {10}^{-3} {\mathrm{mol}}^{-}{}^1 \mathrm{L} {\sec }^{-1}$. If the rate of formation of oxygen is $1.5\times {10}^{-4} \mathrm{mol} {\mathrm{L}}^{-1} {\min }^{-1}$ then the molar concentrations of ${\mathrm{NO}}_2$ in mole ${\mathrm{L}}^{-1}$ is

A radioactive element has a half-life of one day. After three days the amount of the element left will be: