Academic Experts Solutions for Chapter: Chemical Kinetics & Nuclear Chemistry, Exercise 1: Exercise

For an elementary reaction of reversible nature, net rate is: $\left(\frac{\mathrm{dx}}{\mathrm{dt}}\right)={\mathrm{k}}_1{\left[\mathrm{A}\right]}^2{\left[\mathrm{B}\right]}^1-{\mathrm{k}}_2\left[\mathrm{C}\right]$, hence, given reaction is:

A substance '$\mathrm{A}$' decomposes in solution following the first order kinetics. Flask $\mathrm{I}$ contains $1 \mathrm{L}$ of $1 \mathrm{M}$ solution of $\mathrm{A}$ and flask $\mathrm{II}$ contains $100 \mathrm{ml}$ of $0.6 \mathrm{M}$ solution. After $8 \mathrm{hr}$, the concentration of $\mathrm{A}$ in flask $\mathrm{I}$ becomes $0.25 \mathrm{M}$. What will be time for concentration of $\mathrm{A}$ in flask $\mathrm{II}$ to become $0.3 \mathrm{M}$? 

The rate constant for the reaction, $2 {\mathrm{N}}_2{\mathrm{O}}_5⟶4{\mathrm{NO}}_2+{\mathrm{O}}_2$, is $3\times {10}^{-5} {\mathrm{s}}^{-1}$. If the rate is $2.4\times {10}^{-5} \mathrm{mol} {\mathrm{L}}^{-1} {\mathrm{s}}^{-1}$, then determine the concentration of ${\mathrm{N}}_2{\mathrm{O}}_5($ in ${\mathrm{molL}}^{-1})$.

In the biologically-catalysed oxidation of ethanol, the concentration of ethanol decreases in a first order reaction from $800 \mathrm{mol}e {\mathrm{dm}}^{-3}$ to $50 \mathrm{mol}e {\mathrm{dm}}^{-3}$ in $2\times {10}^4 \mathrm{s}$. Determine the rate constant $\left({\mathrm{s}}^{-1}\right)$ of the reaction. If answer is $\mathrm{x}\times {10}^{-4} {\sec }^{-1}$ then x is.

Under the same reaction conditions, initial concentration of $1.386 \mathrm{mol} {\mathrm{dm}}^{-3}$ of a substance becomes half in $40$ $\mathrm{seconds}$ and $20$ $\mathrm{seconds}$ through first order and zero order kinetics, respectively. Determine the ratio $\left(\frac{{\mathrm{k}}_1}{{\mathrm{k}}_0}\right)$ of the rate constant for first order $\left({\mathrm{k}}_1\right)$ and zero order $\left({\mathrm{k}}_0\right)$ of the reaction.

For a first order reaction $\mathrm{A}\to \mathrm{P}$, the temperature $\left(\mathrm{T}\right)$ dependent rate constant $\left(\mathrm{k}\right)$ was found to follow the equation $\mathrm{logk}=-\left(2000\right)\frac{1}{ \mathrm{T}}+6.0$. The activation energy ${\mathrm{E}}_{\mathrm{a}}$ is _____$\mathrm{kJ} {\mathrm{mol}}^{-1}$.$(38.3/3.83)$

Bombardment of aluminum by $\mathrm{\alpha }$ -particle leads to its artificial disintegration in two ways, $\left(\mathrm{i}\right)$ and $\left(\mathrm{ii}\right)$ as shown. Determine the products $\mathrm{X},\mathrm{Y}$ and $\mathrm{Z}$.

${}_{92}{}^{238}\mathrm{U}$ is radioactive and it emits $\mathrm{\alpha }$ and $\mathrm{\beta }$ particles to form ${}_{82}{}^{206} \mathrm{Pb}$. Calculate the number of $\mathrm{\alpha }$ and $\mathrm{\beta }$ particles emitted in this conversion. An ore of ${}_{92}{}^{238}\mathrm{U}$ is found to contain ${}_{92}{}^{238}\mathrm{U}$ and ${}_{82}{}^{206} \mathrm{Pb}$ in the weight ratio of $1: 0.9$, the half life period of ${}_{92}{}^{238}\mathrm{U}$ is $4.5\times {10}^9$ years. Calculate the age of the ore. If answer is $=4.64\times {10}^y \mathrm{years}$ then y is.