EXERCISE-III (Analytical Questions)

The rate of a first order reaction is $1.5\times {10}^{-2} \mathrm{mol} {\mathrm{L}}^{-1} {\min }^{-1}$ at $0.5 \mathrm{M}$ concentration of the reactant. The half life of the reaction is :
 

The rate equation for the reaction : $2\mathrm{A}+\mathrm{B}\to \mathrm{C}$ is found to be rate$=\mathrm{k}\left[\mathrm{A}\right]\left[\mathrm{B}\right]$. The correct statement in relation to this reaction is that the
 

The rate of a chemical reaction doubles for every $10 º\mathrm{C}$ rise of temperature. If the temperature is raised by $50 º\mathrm{C}$, the rate of the reaction increases by about.

For a first order reaction, $\mathrm{A}\to$Products, the concentration of $\mathrm{A}$ changes from $0.1 \mathrm{M}$ to $0.025 \mathrm{M}$ in $40$ minutes The rate of reaction when the concentration of $\mathrm{A}$ is $0.01 \mathrm{M}$ is
 

A reactant (A) forms two products :

$\begin{array}{cc}\text{A}\stackrel{{\text{k}}_1}{⟶}\text{B}& \text{Activation energy}{\text{E}}_{{\text{a}}_1}\\ \text{A}\stackrel{{\text{k}}_2}{⟶}\text{C}& \text{Activation energy}{\text{E}}_{{\text{a}}_2}\end{array}$

If ${\text{E}}_{{\text{a}}_2}=2{\text{E}}_{{\text{a}}_1}$, then k₁ and k₂ will be related as :

The rate of a reaction doubles when its temperature changes from $300 \mathrm{K}$ to $310 \mathrm{K}$. Activation energy of such a reaction will be $(\mathrm{R}=8.314 {\mathrm{JK}}^{-1} {\mathrm{mol}}^{-1}$ and $\log 2=0.301)$

The half life of a radio isotope is four hours. If the initial mass of the isotope was 200 g, the mass remaining after 24 hours undecayed is

Assertion: rate of reaction doubles when the concentration of the reactant is doubles if it is a first order reaction.

Reason: rate constant also doubles.