Chemical Kinetics and Nuclear Chemistry

For a first order reaction, rate = $k[A]$ . If A₁ and A₂ be the initial and final concentrations of the reactant respectively then,

${r_1} = k[{A_1}]$ or, $0.04 = k[{A_1}]$ .............. [1]

and ${r_2} = k[{A_2}]$ or, $0.03 = k[{A_2}]$ ....... [2]

$\therefore$ ${{[{A_1}]} \over {[{A_2}]}} = {{0.04} \over {0.03}} = {4 \over 3}$ (Dividing equation [1] by equation [2])

At 10 min, $10 = {{2.303} \over k}\log {{[{A_0}]} \over {[{A_1}]}}$, at 20 min, $20 = {{2.303} \over k}\log {{[{A_0}]} \over {[{A_2}]}}$

$\therefore$ $20 - 10 = 10 = {{2.303} \over k}\log {{[{A_1}]} \over {[{A_2}]}}$

or, $k = {{2.303} \over {10}}\log \left( {{4 \over 3}} \right)$ or, $k = 2.855 \times {10^{ - 2}}$ min^$-$1

$\therefore$ The half-life for the first order reaction,

$ = {{0.693} \over k} = {{0.693} \over {2.855 \times {{10}^{ - 2}}}} = 24.27$ min

Given, initial moles of A = 10; initial moles of B = 12. Let, the number of moles of A = n, when polymerization is arrested. Moles of solute added = 0.525.

$\therefore$ Total number of moles = (n + 12 + 0.525) = (n + 12.525)

$\therefore$ Mole fraction of $A({x_A}) = {n \over {n + 12.525}}$ and

Mole fraction of $B({x_B}) = {{12} \over {n + 12.525}}$

The total vapour pressure of the solution = $P_A^0{X_A} + P_B^0{X_B}$

or, $400 = 300 \times {n \over {n + 12.525}} + 500 \times {{12} \over {n + 12.525}}$ or, n = 9.9

For a first order reaction, $k = {{2.303} \over t}\log {{{{[A]}_0}} \over {[A]}}$

or, $k = {{2.303} \over {100}}\log {a \over {a - x}} = {{2.303} \over {100}}\log {{10} \over {9.9}}$ [$\because$ a $-$ x = n = 9.9]

or, $k = 1.004 \times {10^{ - 4}}$ min^$-$1

From Arrhenius equation, $\log k = \log A - {{{E_a}} \over {2.303RT}}$

$\therefore$ $\log {k_{500}} = \log A - {{{E_{{a_1}}}} \over {2.303R{T_1}}}$, $\log {k_{400}} = \log A - {{{E_{{a_2}}}} \over {2.303R{T_2}}}$

Given, ${k_{500}} = {k_{400}}$ $\therefore$ $\log {k_{500}} = \log {k_{400}}$

or, ${{{E_{{a_1}}}} \over {{T_1}}} = {{{E_{{a_2}}}} \over {{T_2}}}$ or, ${{{E_{{a_1}}}} \over {500}} = {{{E_{{a_2}}}} \over {400}}$ or, ${{{E_{{a_1}}}} \over {{E_{{a_2}}}}} = {{500} \over {400}} = {5 \over 4}$ ...... [1]

According to given data, ${E_{{a_1}}} - {E_{{a_2}}} = 20$

$\therefore$ Substituting in equation [1], we get ${{{E_{{a_1}}}} \over {{E_{{a_1}}} - 20}} = {5 \over 4}$

or, $4{E_{{a_1}}} - 5{E_{{a_1}}} - 100$ or, ${E_{{a_1}}} = 100$ kJ mol^$-$1

$\therefore$ ${E_{{a_2}}} = 100 - 20 = 80$ kJ mol^$-$1

The unit of rate constant = min^$-$1 implies that the reaction is of first order.

For a first order reaction, $k = {{2.303} \over t}\log {a \over {a - x}}$

or, $k = {{2.303} \over t}\log {{{{[A]}_0}} \over {[A]}}$ or, $4.5 \times {10^{ - 3}} = {{2.303} \over {60}}\log {1 \over {[A]}}$

Hence, after 1 hour concentration of A, [ A ] = 0.764 (M)

$\therefore$ Reaction-rate after 1 hour = $k[A] = 4.5 \times {10^{ - 3}} \times 0.746$

= 3.438 $\times$ 10^$-$3 mol L^$-$1min^$-$1

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

$\therefore$ $\log k = \log A - {{{E_a}} \over {2.303RT}}$ or, $\log \left( {{{{k_2}} \over {{k_1}}}} \right) = {{{E_a}} \over {2.303R}}\left[ {{1 \over {{T_1}}} - {1 \over {{T_2}}}} \right]$

Given : $\log \left( {{{4.5 \times {{10}^7}} \over {1.5 \times {{10}^7}}}} \right) = {{{E_a}} \over {2.303 \times 8.314}}\left[ {{1 \over {323}} - {1 \over {373}}} \right]$

or, ${E_a} = 2.2 \times {10^4}$ J mol^$-$1

From Arrhenius equation, $k = A{e^{ - {E_a}/RT}}$

or, $\log k = \log A - {{{E_a}} \over {2.303RT}}$

or, $\log (4.5 \times {10^7}) = \log A - {{2.2 \times {{10}^4}} \over {2.303 \times 8.314 \times 373}}$

or, $A = 5.42 \times {10^{10}}$ s^$-$1.

Given, NH$_4^ + $ + H₂O $\rightleftharpoons$ NH₃ + H₃O⁺

${K_a} = {{[N{H_3}][{H_3}{O^ + }]} \over {[NH_4^ + ][{H_2}O]}}$ (K_a = 5.6 $\times$ 10^$-$10) ..... [1]

$NH_4^ + + \mathop O\limits^ - H$ $\mathrel{\mathop{\kern0pt\rightleftharpoons} \limits_{{k_b}}^{{k_f}}} $ $N{H_3} + {H_2}O$ (k_f = 3.4 $\times$ 10¹⁰ Lmol^$-$1 s^$-$1) ...... [2]

${K_{eq}} = {{{k_f}} \over {{k_b}}} = {{[N{H_3}][{H_3}O]} \over {[NH_4^ + ][O{H^ - }]}}$

$ = {{[N{H_3}][{H_3}{O^ + }]} \over {[NH_4^ + ][{H_2}O]}} \times {{{{[{H_2}O]}^2}} \over {[{H_3}{O^ + }][O{H^ - }]}} = {{{K_a}} \over {{K_w}}}$

$\therefore$ ${k_b} = {{{K_w}} \over {{K_a}}} \times {k_f} = {{1.0 \times {{10}^{ - 4}}} \over {5.6 \times {{10}^{ - 10}}}} \times 3.4 \times {10^{10}} = 6.07 \times {10^5}$

For the first order reaction half-life = ${{0.693} \over k}$

or, $360 = {{0.693} \over k}$ $\therefore$ ${k_{380^\circ C}} = {{0.693} \over {360}}$

We know, $\log {{{k_2}} \over {{k_1}}} = {{{E_a}} \over {2.303R}}\left[ {{1 \over {{T_1}}} - {1 \over {{T_2}}}} \right]$

$\therefore$ $\log {{{k_{450^\circ C}}} \over {{{0.693} \over {360}}}} = {{200 \times 1000} \over {2.303 \times 8.314}}\left[ {{1 \over {653}} - {1 \over {723}}} \right]$

or, ${k_{450^\circ C}} = 6.18 \times {10^{ - 2}}$ min^$-$1

For 75% decomposition at 723K (450$^\circ$C),

${k_{450^\circ C}} = {{2.303} \over t}\log {a \over {a - x}}$

or, $6.81 \times {10^{ - 2}} = {{2.303} \over t}\log {{100} \over {(100 - 75)}}$ or, t = 20.358 min

From Arrhenius equation, $\log k = \log A - {{{E_a}} \over {2.303RT}}$

or, $\log {{{k_2}} \over {{k_1}}} = {{{E_a}} \over {2.303}}\left[ {{1 \over {{T_1}}} - {1 \over {{T_2}}}} \right]$

$\therefore$ $\log {{{k_2}} \over {{k_1}}} = {{70} \over {2.303 \times 8.314}}\left[ {{1 \over {298}} - {1 \over {313}}} \right]$ or, ${{{k_2}} \over {{k_1}}} = 3.872$ ...... [1]

For a first order reaction, $k = {{2.303} \over t}\log {a \over {a - x}}$

Given x = 0.25 a,

$\therefore$ a $-$ x = (a $-$ 0.25 a) = 0.75 a at t = 20 min

$\therefore$ ${k_1} = {{2.303} \over {20}}\log {a \over {0.75\,a}} = 0.014386$ min^$-$1 ...... [2]

Substituting for k₁ in equation 1, we have

${k_2} = 3.872 \times 0.014386 = 0.05571$ min^$-$1

For a first order reaction, ${k_2} = {{2.303} \over t}\log {a \over {a - x}}$

or, $0.05571 = {{2.303} \over {20}}\log {a \over {a - x}}$ [x = decrease in concentration of reactant]

or, $\log {a \over {a - x}} = {{0.05571 \times 20} \over {2.303}}$

or, $\log {a \over {a - x}} = 0.48381$

or, x = 0.6717 a

Hence, percentage decomposition $ = {{0.6717\,a} \over a} \times 100 = 67.17\% $

Given reaction : 2N₂O₅(g) $\to$ 4NO₂(g) + O₂(g)

$\therefore$ Total no. of moles at time, $t = a - x + 2x + {x \over 2} = a + {3 \over 2}x$

Total no. of moles after complete decomposition

$ = 2a + {a \over 2} = {5 \over 2}a$

At a given volume and temperature, P $\propto$ n [Assuming ideal behaviour of gas mixture]

According to the given data, ${5 \over 2}$a $\propto$ 584.5 mm Hg ...... [1]

and $a + {3 \over 2}x \propto 284.5$ mm Hg ............... [2]

$\therefore$ a $\propto$ 233.8 mm Hg and ${3 \over 2}x \propto 284.5-a$

or, ${3 \over 2}x \propto 284.5 - 233.8 \propto 50.7$ mm Hg

$\therefore$ x $\propto$ 33.8 mm Hg

Therefore, (a $-$ x) $\propto$ (233.8 $-$ 33.8) mm Hg

i.e., (a $-$ x) $\propto$ 200 mm Hg

$\therefore$ $k = {{2.303} \over t}\log {a \over {a - x}} = {{2.303} \over {30}}\log {{233.8} \over {200}}$

or, k = 5.21 $\times$ 10^$-$3 min^$-$1