Chemical Kinetics and Nuclear Chemistry

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