Chemical Kinetics and Nuclear Chemistry

${t_{1/2}} = 3\,$ hrs. $T=18$ hours

as $\,\,\,\,T = n \times {t_{1/2}}$

$\therefore$ $\,\,\,\,n = {{18} \over 3} = 6$

Initial mass $\left( {{C_0}} \right) = 256\,g$

$\therefore$ $\,\,\,\,{C_n} = {{{C_0}} \over {{2^n}}} = {{256} \over {{{\left( 2 \right)}^6}}} = {{256} \over {64}} = 4g.$

$r = k\left[ {{O_2}} \right]{\left[ {NO} \right]^2}.$

When the volume is reduced to $1/2,$ the conc. will double

$\therefore$ $\,\,\,\,\,$ New rate $ = k\left[ {2{O_2}} \right]{\left[ {2NO} \right]^2} = 8k\left[ {{O^2}} \right]{\left[ {NO} \right]^2}$

The new rate increases to eight times of its initial.

$Rat{e_1} = k{\left[ A \right]^n}{\left[ B \right]^m};$

$Rat{e_2} = k{\left[ {2A} \right]^n}{\left[ {{\raise0.5ex\hbox{$\scriptstyle 1$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 2$}}B} \right]^m}$

$\therefore$ $\,\,\,\,{{Rat{e_2}} \over {Rat{e_1}}} = {{k{{\left[ {2A} \right]}^n}{{\left[ {{\raise0.5ex\hbox{$\scriptstyle 1$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 2$}}B} \right]}^m}} \over {k{{\left[ A \right]}^n}{{\left[ B \right]}^m}}}$

$ = {\left[ 2 \right]^n}{\left[ {{\raise0.5ex\hbox{$\scriptstyle 1$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 2$}}} \right]^m} = {2^n}{.2^{ - m}} = {2^{n - m}}$

${}_{90}^{234}Th\buildrel { - \beta } \over \longrightarrow {}_{91}^{234}X\buildrel {\, - \beta } \over \longrightarrow {}_{92}^{234}Th\buildrel { - \alpha } \over \longrightarrow {}_{90}^{230}Th$

The Arrhenius equation reads

$k = A\,e^{-\tfrac{E_a}{RT}}$

Let’s check each term:

A is the pre-exponential (or frequency) factor, not an adsorption factor.

$E_a$ is indeed the activation energy.

R is the universal gas constant ($\approx8.314$\,J·mol⁻¹·K⁻¹), not Rydberg’s constant.

k is the rate constant, not the equilibrium constant.

Hence, the correct statement is:

Option B: $E_a$ is energy of activation.

$Rt = \log {C_o} - {{\mathop{\rm logC}\nolimits} _t}$

It is clear from the equation that if we plot a graph between $\log \,{C_t}$ and time, a straight line with a slope equal to $ - {k \over {2.303}}$ and intercept equal to $\log \,\left[ {{A_o}} \right]$ will be obtained.

For a zero order reaction.

rate $ = k{\left[ A \right]^ \circ }\,\,\,$ i.e. rate $=k$

hence unit of $k = M.{\sec ^{ - 1}}$

For a first order reaction.

rate $ = k\left[ A \right]$

$k=M.$ ${\sec ^{ - 1}}/M = {\sec ^{ - 1}}$

${}_0{n^1} \to {}_{ + 1}{p^1} + {}_{ - 1}{e^{}}$

${t_{1/2}} = 5\,\,$ years, $T=15$ years hence -

total number of half life periods $ = {{15} \over 3} = 3.$

$\therefore$ Amount left $ = {{64} \over {{{\left( 2 \right)}^3}}} = 8g$

NOTE : Order is the sum of the power of the concentrations terms in rate law expression.

Hence the order of reaction is $=1+2=3$

rate of appearance of $HI = {1 \over 2}{{d\left[ {HI} \right]} \over {dt}}$

rate of formation of ${H_2} = {{ - d\left[ {{H_2}} \right]} \over {dt}}$

rate of formation of ${I_2} = {{ - d\left[ {{I_2}} \right]} \over {dt}}$

hence ${{ - d\left[ {{H_2}} \right]} \over {dt}} = - {{ - d\left[ {{I_2}} \right]} \over {dt}} = {1 \over 2}{{d\left[ {HI} \right]} \over {dt}}$

or $\,\,\,\,$ $ - {{2d\left[ {{H_2}} \right]} \over {dt}} = - {{2d\left[ {{I_2}} \right]} \over {dt}} = {{d\left[ {HI} \right]} \over {dt}}$