Chemical Kinetics and Nuclear Chemistry

Rate = k[A]^a[B]^b

6 × 10^–3 = k(0.1)^a(0.1)^b ...(1)

2.4 × 10^–3 = k(0.1)^a(0.2)^b ...(2)

1.2 × 10^–3 = k(0.2)^a (0.1)^b ...(3)

(3) ÷ (1) $ \Rightarrow $ x = 1

(2) ÷ (3) $ \Rightarrow $ x = 2

So, order with respect to A = 1

Order with respect to B = 2

6 × 10^–3 = k(0.1)¹ (0.1)² from (1)

$ \Rightarrow $ k = 6

From experiment IV,

7.2 × 10^–2 = 6(x)¹(0.2)² $ \Rightarrow $ x = 0.3

From experiment V,

2.88 × 10^–2 = 6(0.3)¹(y)² $ \Rightarrow $ y = 0.4

K₁ = A${e^{ - {{{E_a}} \over {R \times 700}}}}$

K₂ = A${e^{ - {{\left( {{E_a} - 30} \right)} \over {R \times 500}}}}$

$ \because $Rate is same

$ \therefore $ Rate constant will also be same

K₁ = K₂

A${e^{ - {{{E_a}} \over {R \times 700}}}}$ = A${e^{ - {{\left( {{E_a} - 30} \right)} \over {R \times 500}}}}$

$ \Rightarrow $ ${{{\left( {{E_a} - 30} \right)} \over {R \times 500}}}$ = ${{{{E_a}} \over {R \times 700}}}$

$ \Rightarrow $ 5E_a = 7E_a – 210

$ \Rightarrow $ 210 = 2E_a

$ \Rightarrow $ E_a = 105 kJ/mole

$ \therefore $ Activation energy in the presence of catalyst = 105 – 30 = 75 kJ/mol

We know

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

$ \Rightarrow $ ${\log _e}k = {\log _e}A - {{{E_a}} \over {2.303RT}}$

According to Arhenius equation plot of ${\log _e}k$ versus ${1 \over T}$ is linear.

Slope = $ - {{{E_a}} \over {2.303R}}$

From plot we conclude :

Slope : c $>$ a $>$ d $>$ b

So correct order of activation energies

E_c $>$ E_a $>$ E_d $>$ E_b

The rate constant of a reaction without catalyst is

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

The rate constant in presence of catalyst is given by

$k' = A{e^{ - {{E{'_a}} \over {RT}}}}$

$ \therefore $ ${{k'} \over k} = {e^{ - {{\left( {E{'_a} - {E_a}} \right)} \over {RT}}}}$

$ \Rightarrow $ 10⁶ = ${e^{ - {{\left( {E{'_a} - {E_a}} \right)} \over {RT}}}}$

$ \Rightarrow $ ln 10⁶ = ${ - {{\left( {E{'_a} - {E_a}} \right)} \over {RT}}}$

$ \Rightarrow $ ${E{'_a} - {E_a}}$ = -RTln 10⁶

$ \Rightarrow $ ${E{'_a} - {E_a}}$ = -6RT$ \times $2.303

K_eq = ${{{k_f}} \over {{k_b}}}$ = ${{\left[ {{N_2}} \right]{{\left[ {{H_2}O} \right]}^2}} \over {{{\left[ {{H_2}} \right]}^2}{{\left[ {NO} \right]}^2}}}$

${k_f}\left[ {{H_2}} \right]{\left[ {NO} \right]^2} = {k_b}{{\left[ {{N_2}} \right]{{\left[ {{H_2}O} \right]}^2}} \over {\left[ {{H_2}} \right]}}$

At equilibrium r_f = r_b

Given r_f = ${k_f}\left[ {{H_2}} \right]{\left[ {NO} \right]^2}$

$ \therefore $ r_b = ${k_b}{{\left[ {{N_2}} \right]{{\left[ {{H_2}O} \right]}^2}} \over {\left[ {{H_2}} \right]}}$

${(C{H_3})_3}C - Br + NaOH\mathrel{\mathop{\kern0pt\longrightarrow} \limits_{(First\,order\,reaction)}^{{S_{{N^1}}}}} {(C{H_3})_3}C - OH + NaBr$

This is first order reaction and for first order reaction ${t_{1/2}} = {{0.693} \over k}$

So, half-life is independent of initial concentration.

Therefore, the plot (a) correct.

For first order reaction,

$(r) = k[{(C{H_3})_3}C - Br]$ ;

$\ln \left( {{{{P_0}} \over P}} \right) = k\,t$

or $\ln \left( {{P \over {{P_0}}}} \right) = - k\,t$

Hence, plot (b) and (d) are incorrect. For first order reaction,

Q = [P₀](1 $-$ e^kt)

Or ${{[Q]} \over {[{P_0}]}} = (1 - {e^{kt}})$

Hence, plot (c) is incorrect.

Rate of disappearance of N₂O₅ = $ - {{\Delta \left[ {{N_2}{O_5}} \right]} \over {\Delta \left[ t \right]}}$ = $ - {{\left[ {2.75 - 3} \right]} \over {30}}$ = $ + {1 \over {120}}$ mol L^-1 min^-1

Rate of formation of NO₂ = $ + {{\Delta \left[ {N{O_2}} \right]} \over {\Delta \left[ t \right]}}$

As $ - {1 \over 2}{{\Delta \left[ {{N_2}{O_5}} \right]} \over {\Delta \left[ t \right]}} = + {1 \over 4}{{\Delta \left[ {N{O_2}} \right]} \over {\Delta \left[ t \right]}}$
$ \Rightarrow $ ${{\Delta \left[ {N{O_2}} \right]} \over {\Delta \left[ t \right]}}$ = $ + {1 \over {120}} \times {4 \over 2}$ = 0.01667 = 1.667 × 10^–2 mol L^–1 min^–1

xA $ \to $ yB

${1 \over x}\left\{ { - {{d\left[ A \right]} \over {dt}}} \right\} = {1 \over y}\left\{ {{{d\left[ B \right]} \over {dt}}} \right\}$

$ \Rightarrow $ $ - {{d\left[ A \right]} \over {dt}} = {x \over y}\left\{ {{{d\left[ B \right]} \over {dt}}} \right\}$

Taking log both sides, we get

$ \Rightarrow $ ${\log _{10}}\left[ { - {{d\left[ A \right]} \over {dt}}} \right] = {\log _{10}}\left[ {{{d\left[ B \right]} \over {dt}}} \right] + {\log _{10}}\left( {{x \over y}} \right)$

Given that,

${\log _{10}}\left[ { - {{d\left[ A \right]} \over {dt}}} \right] = {\log _{10}}\left[ {{{d\left[ B \right]} \over {dt}}} \right] + 0.3010$

So, by comparing both of them we get

${\log _{10}}\left( {{x \over y}} \right)$ = 0.3010 = ${\log _{10}}\left( 2 \right)$

$ \therefore $ ${{x \over y}}$ = 2

$ \Rightarrow $ x = 2y

If x = 2 then y = 1

The reaction is of type 2A $ \to $ B

So possible reaction is

2C₂H₄ $ \to $ C₄H₈

From Arrhenius equation, we get

$\log {{{K_2}} \over {{K_1}}} = {{{E_a}} \over {2.303R}}\left( {{1 \over {{T_1}}} - {1 \over {{T_2}}}} \right)$

$ \Rightarrow $ $\log {1 \over {2.5 \times {{10}^{ - 4}}}} = {{{E_a}} \over {2.303 \times 8.314}}\left( {{1 \over {600}} - {1 \over {800}}} \right)$

$ \Rightarrow $ 3.6 = ${{{E_a}} \over {2.303 \times 8.314}} \times {{200} \over {600 \times 800}}$

$ \Rightarrow $ E_$a$ = 166 kJ/mol

${{dN} \over {dt}} = - 5{N^2}$         for t > 1 hr

'N' increases upto 1 hr and then start decreasing after 1 hr.

$ \therefore $ ${{{N_0}} \over N}$ will increase after 1 hr.

Therefore option (D) is correct.

$ \therefore $ E_$a$ = (A + B) $ \to $ C = 15

E_$a$ = (A + B) $ \to $ D = 10

So, Activation enthalpy to form C is 5 kJ mol^-1 more than that to form D.

For 1^st order reaction, we know

$\ln {{{R_0}} \over {{R_T}}} = kt$

$ \Rightarrow $ $\ln {R_0}$ - $\ln {R_t}$ = kt

$ \Rightarrow $ $\ln {R_t}$ = - kt + $\ln {R_0}$

So, $\ln {R_t}$ vs t is a straight line with negative slope.

$ \therefore $ First graph represent 1^st order reaction.

For Zeroorder reaction, we know

R₀ - R_t = kt

$ \therefore $ R_t = - kt + R₀

So, ${R_t}$ vs t is a straight line with negative slope.

$ \therefore $ Second graph represent zero order reaction.

Given, the rate of formation of B is set to be zero.

$ \therefore $ ${{d\left[ B \right]} \over {dt}} = 0$

For this reaction,

$A\mathrel{\mathop{\kern0pt\longrightarrow} \limits_{{R_1}}^{{k_1}}} B$

Rate of reaction for this reaction (R₁) = ${k_1}\left[ A \right]$

For this reaction,

$B\mathrel{\mathop{\kern0pt\longrightarrow} \limits_{{R_2}}^{{k_2}}} C$

Rate of reaction for this reaction (R₂) = ${k_2}\left[ B \right]$

Net rate of formation of B = ${{d\left[ B \right]} \over {dt}}$ = R₁ - R₂

As ${{d\left[ B \right]} \over {dt}} = 0$

$ \therefore $ R₁ - R₂ = 0

$ \Rightarrow $ ${k_1}\left[ A \right]$ - ${k_2}\left[ B \right]$ = 0

$ \Rightarrow $ $\left[ B \right] = {{{k_1}} \over {{k_2}}}\left[ A \right]$

Rate law for the reaction,

2A + B $ \to $ C

Rate law (R) = k[A]^x[B]^y

From experiment 1 :

R₁ = 0.045 = k[0.05]^x [0.05]^y ............(i)

From experiment 2 :

R₂ = 0.090 = k[0.1]^x [0.05]^y ...............(ii)

From experiment 3 :

R₃ = 0.72 = k[0.2]^x [0.1]^y ...................(iii)

Divide equation (ii) by equation (i),

${{{R_2}} \over {{R_1}}} = {{0.09} \over {0.045}} = {{k{{\left[ {0.1} \right]}^x}{{\left[ {0.05} \right]}^y}} \over {k{{\left[ {0.05} \right]}^x}{{\left[ {0.05} \right]}^y}}}$

$ \Rightarrow $ 2 = (2)^x

$ \Rightarrow $ x = 1

Divide equation (iii) by equation (ii),

${{{R_3}} \over {{R_2}}} = {{0.72} \over {0.09}} = {{k{{\left[ {0.2} \right]}^x}{{\left[ {0.1} \right]}^y}} \over {k{{\left[ {0.1} \right]}^x}{{\left[ {0.05} \right]}^y}}}$

$ \Rightarrow $ 8 = 2^x 2^y

$ \Rightarrow $ 8 = 2¹ 2^y [as x = 1]

$ \Rightarrow $ 2^y = 4

$ \Rightarrow $ y = 2

$ \therefore $ Rate law (R) = k[A]¹[B]²

From Arhenius equation,

K = Ae$^{ - {{Ea} \over {RT}}}$

taking log both sides,

lnk = lnA $-$ ${{{Ea} \over {RT}}}$

So in lnK vs ${1 \over T}$ graph

slope = $-$ ${{{Ea} \over R}}$

Rate constant at 400 K is = 10^$-$5 s^$-$1

Let rate constant at 500 K is = K

$ \therefore $   ln(10^$-$5) = ln A $-$ ${{Ea} \over {R\left( {400} \right)}}$ . . . (1)

ln K = ln A $-$ ${{Ea} \over {R\left( {500} \right)}}$ . . . .(2)

Subtracting (2) from (1) we get,

ln ${K \over {{{10}^{ - 5}}}}$ = ${{{Ea} \over R}}$ $\left[ {{1 \over {400}} - {1 \over {500}}} \right]$

$ \Rightarrow $   2.303 log ${K \over {{{10}^{ - 5}}}}$ = 4606 $\left[ {{1 \over {400}} - {1 \over {500}}} \right]$

$ \Rightarrow $   K = 10^$-$4 s^$-$1

According to unit of rate constant it is a zero order reaction, then half life of zero order reaction.

t_1/2 = ${{{a_0}} \over {2k}} = {5 \over {2 \times 0.05}}$ = 50 years

For zero order reaction,

t_1/2 = ${{{a_0}} \over {2k}}$

$ \Rightarrow $ k = ${{{a_0}} \over {2{t_{1/2}}}}$ = ${{0.2} \over {2 \times 6}}$ = 1.67 $ \times $ 10^-2 mol L^–1h^–1

For zero order reaction,

A₀ - A_t = kt

$ \Rightarrow $ 0.5 - 0.2 = 1.67 $ \times $ 10^-2 t

$ \Rightarrow $ t = ${{0.3} \over {1.67 \times {{10}^{ - 2}}}}$ = 18 h

According to Arrhenius equation,
k = A${e^{ - {{{E_a}} \over {RT}}}}$

or ln k = ln A - ${{{{E_a}} \over {RT}}}$

Comparing the above equation with straight line equation,

y = mx + c,

we get, slope or gradient (m) = –E_a
and Intercept (c) = ln A

Also given that slope or gradient (m) = -y

$ \therefore $ -y = –E_a

$ \Rightarrow $ E_a = y

So the activation energy of the reactant, E_a = y unit

Arrhenius Equation -

Arrhenius gave the quantitative dependence of rate constant on temperature by the Arrhenius equation.

- wherein

$k = A{e^{ - {E_a}/RT}}$

${\mathop{\rm lnk}\nolimits} = lnA - {{{E_a}} \over {RT}}$

k = Rate constant

as we have learned in Arrhenius equation when we increase Ea, K should decrease As T increases, power of exponential increases so k also increase.

Hence, both the plots are correct.

$r = K{\left[ A \right]^x}{\left[ B \right]^y}$

$ \Rightarrow 8 = {2^3} = {2^{x + y}}$

$ \Rightarrow x + y = 3\,...(1)$

$ \Rightarrow 2 = {2^x}$

$ \Rightarrow x = 1,y = 2$

Order w.r.t. A = 1

Order w.r.t. B = 2

To determine the time required to consume half of A, we must first determine the rate law for the reaction based on the provided kinetic data.

From the data, we can write the general rate law for the reaction as:

$ \text{Rate} = k [A]^m [B]^n $

We can now use the experimental data to find the values of the exponents $ m $ and $ n $, as well as the rate constant $ k $.

From Experiment I and II, we can see that the initial concentration of A remains constant at 0.10 M while the concentration of B changes.

$ \begin{align*} \text{Experiment I:} \quad \text{Rate} = k [0.10]^m [0.20]^n &= 6.93 \times 10^{-3} \, \text{mol L}^{-1} \text{min}^{-1} \\ \text{Experiment II:} \quad \text{Rate} = k [0.10]^m [0.25]^n &= 6.93 \times 10^{-3} \, \text{mol L}^{-1} \text{min}^{-1} \end{align*} $

By comparing the rates from Experiment I and II:

$ \frac{[0.10]^m [0.25]^n}{[0.10]^m [0.20]^n} = \frac{6.93 \times 10^{-3}}{6.93 \times 10^{-3}} $

$ \frac{[0.25]^n}{[0.20]^n} = 1 \implies \left(\frac{0.25}{0.20}\right)^n = 1 \implies n = 0 $

Thus, the reaction is zero order with respect to B.

Next, we compare Experiment I and III to determine the order with respect to A:

$ \begin{align*} \text{Experiment I:} \quad \text{Rate} = k [0.10]^m [0.20]^0 &= 6.93 \times 10^{-3} \, \text{mol L}^{-1} \text{min}^{-1} \\ \text{Experiment III:} \quad \text{Rate} = k [0.20]^m [0.30]^0 &= 1.386 \times 10^{-2} \, \text{mol L}^{-1} \text{min}^{-1} \end{align*} $

$ \frac{k [0.20]^m}{k [0.10]^m} = \frac{1.386 \times 10^{-2}}{6.93 \times 10^{-3}} $

$ \left( \frac{0.20}{0.10} \right)^m = 2 \implies (2)^m = 2 \implies m = 1 $

Therefore, the reaction is first-order with respect to A.

Now, we can write the rate law as:

$ \text{Rate} = k [A]^1 $

Let's calculate the rate constant $ k $ using data from any experiment (say Experiment I):

$ 6.93 \times 10^{-3} = k \times 0.10 $

$ k = \frac{6.93 \times 10^{-3}}{0.10} = 6.93 \times 10^{-2} \, \text{min}^{-1} $

To determine the half-life (time required to consume half of A) for a first-order reaction, we use the formula:

$ t_{1/2} = \frac{0.693}{k} $

Substituting the value of $ k $:

$ t_{1/2} = \frac{0.693}{6.93 \times 10^{-2}} = 10 \, \text{minutes} $

Therefore, the time required to consume half of A is:

Option B: 10

In decay sequence,



X₁ particle will deflect towards negatively charged plate due to presence of positive charge on $\alpha $- particles.

Hence, options (a, b, c) are correct.

Assume initial concentration of the reactant = 1 M

After 100 second, concentration becomes of the reactant

= 1 $ \times $ ${{50} \over {100}}$ = 0.5 M

After 200 second, concentration of the reactant becomes

= 1 $ \times $ ${{25} \over {100}}$ = 0.25 M.

So, in first 100 second reactant concentration becomes half of initial and in the second 100 second concentration becomes 0.5 M to 0.25 M, which is also half of 0.5 M.

So, after each 100 second period, concentration of reactant becomes half of initial concentration.

So, 100 second is the half life period and it is independent of concentration of reactant. This is characteristic of first order reaction.

For a n^th order reaction, the rate of reaction at time t ,

Rate = K [P_t] ⁿ

Here P_t = pressure at time t, k = constant.

Note :

Here instead of concentration of product, pressure of product is given.

When 5% is reacted at a rate 1 Toss S^$-$1

Then un-reacted is 95%..

As initial pressure is 363 Torr

then after 5% reaction completed the pressure will be

= 363 $ \times $ ${{95} \over {100}}$ Torr.

$\therefore\,\,\,$ 1 = K ${\left[ {363 \times {{95} \over {100}}} \right]^n}$ . . . . . . . .(1)

When 33% is reacted at a rate 0.5 , Torr S^$-$1 then un-reacted is 67%

So, after 33% reaction completion, the pressure is = $363 \times {{67} \over {100}}$ Torr.

$\therefore\,\,\,$ 0.5 = K ${\left[ {363 \times {{67} \over {100}}} \right]^n}.......$ (2)

Dividing (1) by (2), we get

${1 \over {0.5}} = {{{{\left[ {363 \times {{95} \over {100}}} \right]}^n}} \over {{{\left[ {363 \times {{67} \over {100}}} \right]}^n}}}$

$ \Rightarrow \,\,2 = {\left[ {{{95} \over {67}}} \right]^n}$

$ \Rightarrow \,\,$ 2 = ${\left[ {1.41} \right]^n}$

$ \Rightarrow \,\,$ 2 = ${\left[ {\sqrt 2 } \right]^n}$

$ \Rightarrow \,\,\,\,2 = {2^{{n \over 2}}}$

$\therefore\,\,\,$ ${n \over 2}$ = 1

$ \Rightarrow \,\,\,\,$ n = 2

This is a 2nd order reaction.

For first order reaction,

The half life, t_{${1 \over 2}$} = ${{0.693} \over k}$

Here given t_{${1 \over 2}$} = 10 days

$\therefore\,\,\,$ k = ${{0.693} \over {10}}$ = 0.0693 days^$-$1

Now, the time required for ${1 \over 4}$th conversion of A is ,

t = ${{2.303} \over k}$ log₁₀ $\left( {{a \over {a - x}}} \right)$

= ${{2.303} \over {0.0693}}$ log $\left( {{1 \over {1 - {1 \over 4}}}} \right)$

= ${{2.303} \over {0.0693}}$ log $\left( {{4 \over 3}} \right)$

= 4.1 days.

	N2O5	$\rightleftharpoons$	2NO2	+ ${1 \over 2}$O2
At t = 0	50		0	0
At t = 50 min	50 $-$ P1		2P1	${{{P_1}} \over 2}$

Total pressure after 50 min,

= 50 $-$ P₁ + 2P₁ + ${{{P_1}} \over 2}$ = 87.5

$\therefore\,\,\,$ 50 + ${{3{P_1}} \over 2}$ = 87.5

$ \Rightarrow $ $\,\,\,$ P₁ = 25

So, 50 min is the Half $-$ life period of the reaction.

Then 100 min is two half life

	N2O5	$\rightleftharpoons$	2NO2	+ ${1 \over 2}$O2
At t = 100 min	50 $-$ P2		2P2	${{{P_2}} \over 2}$

$\therefore\,\,\,$ 50 $-$ P₂ = ${{25} \over 2}$

P₂ = 37.5

$\therefore\,\,\,$ Total pressure at t = 100 min

= 50 $-$ P₂ + 2P₂ + ${{{P_2}} \over 2}$

= 50 + ${3 \over 2}$ $ \times $ 37.5

= 106.25 mm of Hg

Total pressure of gaseous mixture at beginning $(T=0)=\mathrm{P}_0$

Total pressure of gaseous mixture after time $t$ $(\mathrm{T}=t)=\mathrm{P}_t$

Initial concentration of $A=[A]_0$

Temperature of the reaction mixture (T) $=300 \mathrm{~K}$

For the first order reaction,


Total pressure after time $(\mathrm{T}=t)=\mathrm{P}_0-\mathrm{P}+2 \mathrm{P} +\mathrm{P}$

$ \begin{aligned} \mathrm{P}_t=\mathrm{P}_0+2 \mathrm{P} \\\\ \mathrm{P} =\frac{\mathrm{P}_t-\mathrm{P}_0}{2} \end{aligned} $

According to integrated rate law for the first order reaction:

$ \begin{aligned} &t =\frac{2.303}{k} \log \frac{\mathrm{P}_0}{\mathrm{P}_t} ~~~~...(i)\\\\ &t =\frac{2.303}{k} \log \frac{\mathrm{P}_0}{\mathrm{P}_0-\mathrm{P}} \\\\ &t =\frac{2.303}{k} \log \frac{\mathrm{P}_0}{\mathrm{P}_0-\frac{\left(\mathrm{P}_t-\mathrm{P}_0\right)}{2}} \\\\ &t =\frac{2.303}{k} \log \frac{2 \mathrm{P}_0}{2 \mathrm{P}_0-\mathrm{P}_t+\mathrm{P}_0} \\\\ &t =\frac{2.303}{k} \log \frac{2 \mathrm{P}_0}{3 \mathrm{P}_0-\mathrm{P}_t} \\\\ &t =\frac{1}{k} \ln \frac{2 \mathrm{P}_0}{3 \mathrm{P}_0-\mathrm{P}_t} \\\\ &\ln \left(3 \mathrm{P}_0-\mathrm{P}_t\right) =\ln \left(2 \mathrm{P}_0-\mathrm{P}_t\right)-k t ~~~~...(ii) \end{aligned} $

Comparing equation (ii) with equation of straight option line

$ \begin{aligned} & \ln \left(3 \mathrm{P}_0-\mathrm{P}_t\right) =\ln 2 \mathrm{P}_0-k t \\\\ &y =c+m x \end{aligned} $

Plot of $\ln \left(3 \mathrm{P}_0-\mathrm{P}_t\right)$ verses time $(t)$ is represented as:


Option (A) is correct.

(ii) For the first order reaction,


$ \mathrm{P}_t=\frac{\mathrm{P}_0}{3}+\frac{4 \mathrm{P}_0}{3}+\frac{2 \mathrm{P}_0}{3}=\frac{7 \mathrm{P}_0}{3} $

Substituting the value of $P_t$ in $e q$ (i)

$ \begin{aligned} t_{1 / 3} & =\frac{2.303}{k} \log \frac{\mathrm{P}_0}{\frac{7 \mathrm{P}_0}{3}} \\\\ t_{1 / 3} & =\frac{2.303}{k} \log \frac{3}{7} \end{aligned} $

$\mathrm{P}_{t_3}$ is independent of concentration of (A).

Option (B) is not correct.

(iii) At any time $t$, rate constant $(k)$ is independent of concentration of [A].

For example:

At $t_{y_3}$ :

$ k=\frac{2.303}{t_{1 / 3}} \times \log \frac{3}{7} $

Hence, Option (D) is correct.

According to Arrhenius equation,

$\log {{{k_2}} \over {{k_1}}} = {{{E_a}} \over {2.303R}}\left( {{{{T_2} - {T_1}} \over {{T_1}{T_2}}}} \right)$

Substituting the given values in the equation, we get

$\log 4 = {{{E_a}} \over {2.303 \times 8.314\,J\,{K^{ - 1}}mo{l^{ - 1}}}}\left( {{{310 - 300} \over {300 \times 310}}} \right)$

${E_a} = {{0.602 \times 2.303 \times 8.314 \times 300 \times 310} \over {10}}$

= 107197.12 J/mol^$-$1 or 107.2 kJ/mol^$-$1

For reaction A, T₁ = 300 K, T₂ = 310 K, k₂ = 2 k₁

$\log {{{k_2}} \over {{k_1}}} = {{{E_{{a_1}}}} \over {2.303R}}\left[ {{1 \over {{T_1}}} - {1 \over {{T_2}}}} \right]$

$ \therefore $ $\log {{2{k_1}} \over {{k_1}}} = \log 2 = {{{E_{{a_1}}}} \over {2.303R}}\left[ {{1 \over {300}} - {1 \over {310}}} \right]$ ...(1)

For reaction B, T₁ = 300 K, T₂ = ?, k₂ = 2k₁, E_a2 = 2E_a1

$\log {{2{k_1}} \over {{k_1}}} = \log 2 = {{2{E_{{a_1}}}} \over {2.303R}}\left[ {{1 \over {300}} - {1 \over {{T_2}}}} \right]$

From eq. (i) and (ii), we get

${{2{E_{{a_1}}}} \over {2.303R}}\left[ {{1 \over {300}} - {1 \over {{T_2}}}} \right] = {{{E_{{a_1}}}} \over {2.303R}}\left[ {{1 \over {300}} - {1 \over {310}}} \right]$

$ \Rightarrow $ $2\left[ {{1 \over {300}} - {1 \over {{T_2}}}} \right] = \left[ {{1 \over {300}} - {1 \over {310}}} \right]$

$ \Rightarrow $ T₂ = ${{{300 \times 310} \over {610}}}$ $ \times $ 2 = 304.92 K

$ \therefore $ Increased temperature = (304.92 – 300) = 4.92 K

We know, from arrhenius equation,

k = A.${e^{{{ - {E_a}} \over {RT}}}}$

$ \therefore $ k₁ = A.${e^{{{ - {E_{{a_1}}}} \over {RT}}}}$ ......(1)

k₂ = A.${e^{{{ - {E_{{a_2}}}} \over {RT}}}}$ ......(2)

On dividing equation (2) by (1), we get

${{{k_2}} \over {{k_1}}} = {e^{{{\left( {{E_{{a_1}}} - {E_{{a_2}}}} \right)} \over {RT}}}}$

$ \Rightarrow $ $\ln \left( {{{{k_2}} \over {{k_1}}}} \right) = {{\left( {{E_{{a_1}}} - {E_{{a_2}}}} \right)} \over {RT}}$ = ${{10,000} \over {8.314 \times 300}}$ = 4

Arrhenius equation is

$k = A{e^{ - {E_a}/RT}}$

where, A = Frequency factor

Taking into account orientation factor,

$k = P{Z_{AB}}{e^{ - {E_a}/RT}}$

where, P = steric factor, Z_AB = collision frequency

The value of steric factor lies between 0 and 1 predicted by Arrhenius equation. Thus, the experimentally determined value of frequency factor is higher than that predicted by Arrhenius equation.

The activation energy of the reaction does not depend upon the value of the steric factor.

If P is very small, then catalyst is required to carry out the reaction at measurable rate.

Rate constant only depends on temperature only and it is independed of concentration of reactants.

We have

${O_3}(g) + C{l^ \bullet }(g) \to {O_2}(g) + Cl{O^ \bullet }(g)$ ...... (1)

$Cl{O^ \bullet }(g) + {O^ \bullet }(g) \to {O_2}(g) + C{l^ \bullet }(g)$ ...... (2)

On adding Eq. (1) and Eq. (2), we get the required equation for overall reaction

${O_3}(g) + {O^ \bullet }(g) \to 2{O_2}(g)$

Therefore,

${k_{overall}} = {k_1} \times {k_2}$

$ = 5.2 \times {10^9} \times 2.6 \times {10^{10}}$

$ = 1.4 \times {10^{20}}$ L mol^$-$1 s^$-$1

${H_2}{O_2}\left( {aq} \right) \to {H_2}O\left( {aq} \right) + {1 \over 2}{O_2}\left( g \right)$

For a first order reaction

$k = {{2.303} \over t}\log {a \over {\left( {a - x} \right)}}$

Given $a = 0.5,\left( {a - x} \right) = 0.125,\,t = 50\,\,$ min

$\therefore$ $\,\,\,\,\,k = {{2.303} \over {50}}\log {{0.5} \over {0.125}}$

$ = 2.78 \times {10^{ - 2}}\,{\min ^{ - 1}}$

$r = k\left[ {{H_2}{O_2}} \right] = 2.78 \times {10^{ - 2}} \times 0.05$

$ = 1.386 \times {10^{ - 3}}\,\,mol\,{\min ^{ - 1}}$

Now

$ - {{d\left[ {{H_2}{O_2}} \right]} \over {dt}} = {{d\left[ {{H_2}O} \right]} \over {dt}} = {{2d\left[ {{O_2}} \right]} \over {dt}}$

$\therefore$ $\,\,\,\,\,{{2d\left[ {{O_2}} \right]} \over {dt}} - {{d\left[ {{H_2}{O_2}} \right]} \over {dt}}$

$\therefore$ $\,\,\,\,\,$ ${{d\left[ {{O_2}} \right]} \over {dt}} = {1 \over 2} \times {{d\left[ {{H_2}{O_2}} \right]} \over {dt}}$

$ = {{1.386 \times {{10}^{ - 3}}} \over 2} = 6.93 \times {10^{ - 4}}\,mol\,{\min ^{ - 1}}$

Plot of Z (number of protons) verses N (number of neutrons) is represented as :


(a) For a stable nucleus $\mathrm{N} / \mathrm{Z}\left(\frac{\text { no. of neutrons }}{\text { no. of protons }}\right)$ should be equal to 1.

This happens for elements till atomic number $(Z=20)$. For such nucleus number of proton are equal to number of neutrons.

(b) For heavier nucleides with atomic number $(\mathrm{Z})$ greater than $20(Z>20)$, the number of neutron are more than proton as seen by curved (green) appearance in the plot.

(c) For nucleids with $\mathrm{N} / \mathrm{Z}<1$, the number of neutrons are less than the protons making the nucleus unstable. This is due to increase in coulombic repulsions between positively charged protons.

Such nuclei become stable by K-electron capture and positron emission.

K-electron capture :

$ { }_1^1 p+{ }_{-1}^0 e \rightarrow{ }_0^1 n $

The K-electron capture, proton is converted to the neutron as increases the $\mathrm{N} / \mathrm{Z}$ ratio.

$\beta^{ \pm}$decay (positron emission) :

$ { }_1^1 p \rightarrow{ }\underset{neutron}{_0^1 n}+\underset{positron}{e^{+}}+\underset{electron \,nutrino}{v_e} $

Proton decays to a neutron, positron and electron nutrino. This decreases the number of protons and increases the number of neutrons, thereby increasing $\mathrm{N} / \mathrm{Z}$ ratio.

Options (B) and (D) are correct.

The Arrhenius equation is a mathematical model of the temperature dependence of reaction rates. It is expressed as follows:

$ k=\mathrm{Ae}^{-\mathrm{E}_a / \mathrm{RT}} $

where :

- k is the rate constant

- A is the pre-exponential factor (also known as the frequency factor)

- E_a is the activation energy

- R is the universal gas constant

- T is the temperature (in Kelvin)

- e is Euler's number, a mathematical constant approximately equal to 2.71828

The options provided correlate to the Arrhenius equation as follows:

Option A : Incorrect. A high activation energy usually implies a slower reaction, not a fast one. The activation energy is the energy barrier that needs to be overcome for a reaction to occur. The higher the activation energy, the fewer the number of molecules that have enough energy to surpass this barrier and react, resulting in a slower reaction rate.


Option B : Correct. Temperature $\mathrm{T}_2$ is greater than $\mathrm{T}_1$; hence, the fraction of molecules (shaded in green) having energy greater than activation energy increases with increase in temperature result, rate of reaction increases.

Higher temperature, more number of collisions (between extents) having energy greater than activation energy happen increases speed of reaction.



Option C : Correct. A reaction at higher activation energy makes the reaction rate slow. An increase in temperature causes reactant molecules to cross the high energy barrier; i.e., $E_a$ and form activated complex.

Option D : Correct. The pre-exponential factor (A) takes into account the probability of proper orientation of reactant molecules to form the product and the frequency at which reactant molecules collide to form the product.

$ \mathrm{A}=\mathrm{Z}_{\mathrm{AB}} \times \mathrm{P} $

$Z_{A B}$ : Collision frequency or frequency with which A collides with B.

P : Probability of reactant molecules oriented in right orientation with respect to each other to form the product.

Option (D), where pre-exponential factor affects the rate reaction irrespective of the energy they possess.

Reactions of higher order $\left( { > 3} \right)$ are very rate due to very less chances of many molecules to undergo effective collisions.

The equilibrium yield of ammonia gets lowered with the increase of temperature. However, at higher temperature the initial rate of forward reaction would be greater than at lower temperature that is why the percentage yield of NH₃ would be more initially.

Let rate of reaction $ = {{d\left[ C \right]} \over t} = k{\left[ A \right]^x}{\left[ B \right]^y}$

Now from the given data

$1.2 \times {10^{ - 3}} = k{\left[ {0.1} \right]^x}{\left[ {0.1} \right]^y}\,\,\,\,\,...\left( i \right)$

$1.2 \times {10^{ - 3}} = k{\left[ {0.1} \right]^x}{\left[ {0.2} \right]^y}\,\,\,\,\,...\left( {ii} \right)$

$2.4 \times {10^{ - 3}} = k{\left[ {0.2} \right]^x}{\left[ {0.1} \right]^y}\,\,\,\,...\left( {iii} \right)$

Dividing equation $(i)$ by $(ii)$

$ \Rightarrow \,\,\,\,{{1.2 \times {{10}^{ - 3}}} \over {1.2 \times {{10}^{ - 3}}}} = {{k{{\left[ {0.1} \right]}^x}{{\left[ {0.1} \right]}^y}} \over {k{{\left[ {0.1} \right]}^x}{{\left[ {0.2} \right]}^y}}}$

We find, $y=0$

Now dividing equation $(i)$ by $(iii)$

$ \Rightarrow \,\,\,\,{{1.2 \times {{10}^{ - 3}}} \over {2.4 \times {{10}^{ - 3}}}} = {{k{{\left[ {0.1} \right]}^x}{{\left[ {0.1} \right]}^y}} \over {k{{\left[ {0.2} \right]}^x}{{\left[ {0.1} \right]}^y}}}$

We find, $x=1$

Hence ${{d\left[ C \right]} \over {dt}} = k{\left[ A \right]^1}{\left[ B \right]^0}$

The rate law for an elementary reaction where the concentration of reactant M impacts the rate of the reaction is expressed as:

$\text{Rate} = k[M]^n$

where k is the rate constant, M is the concentration of the reactant, and n is the order of the reaction with respect to M.

In the scenario described, when the concentration of M is doubled, the rate of disappearance of M increases by a factor of 8. Let’s denote the initial concentration of M by [M] and the doubled concentration by 2[M]. According to the rate law, the initial rate would be:

$\text{Rate}_1 = k[M]^n$

Upon doubling M:

$\text{Rate}_2 = k(2[M])^n = k2^n[M]^n$

It is given that this rate is 8 times the initial rate:

$ \text{Rate}_2 = 8 \times \text{Rate}_1 $

$ k2^n[M]^n = 8k[M]^n $

Cancelling out the common terms and simplifying the equation, we have:

$ 2^n = 8 $

To find n, solve for n in:

$ 2^n = 2^3 $

Hence, n = 3, indicating that the order of the reaction with respect to M is 3. The correct option is therefore:

Option B: 3

Using Graham's law of diffusion, we have that for identical conditions, rate of diffusion varies inversely with square root of mass of the gases. So,

${x \over {24 - x}} = {\left( {{{40} \over {10}}} \right)^{1/2}} \Rightarrow x = 16$

Collision frequency of X is greater than Y.

Activation energy can be calculated from the equation

${{\log \,{k_2}} \over {\log \,{k_1}}} = {{{E_a}} \over {2.303R}}\left( {{1 \over {{T_1}}} - {1 \over {{T_2}}}} \right)$

given ${{{k_2}} \over {{k_1}}} = 2\,\,{T_2} = 310\,K\,\,{T_1} = 300\,K$

$ = \log 2 = {{ - {E_a}} \over {2.303 \times 8.314}}\left( {{1 \over {310}} - {1 \over {300}}} \right)$

${E_a} = 53598.6J/mol$

$ = 53.6kJ/mol.$