Chemical Kinetics and Nuclear Chemistry

Subjective Answer

(A) Assume that, the element or component that should be present at the blank space in the given reaction is X:

$_{92}^{235}U + _0^1n \to _{52}^{137}A + _{40}^{97}B + \underline X $

The atomic number of uranium is 92 and atomic mass number is 235. The atomic number of element A and B is 52 and 40 respectively. The atomic mass number of elements A and B are 137 and 97 respectively.

The element with atomic number 52 and atomic mass number 137 is tellurium, Te. Therefore, element A is tellurium.

The element with atomic number 40 and atomic mass number 97 is zirconium, Zr. Therefore, element B is zirconium.

To find out X, we need to balance the atomic mass number and atomic number on both sides.

On the left side of the reaction, the sum of atomic number is 92 + 0 = 92 and that of atomic mass number is 235 + 1 = 236.

On the right side of the reaction, the sum of atomic number is 52+40 =92 and that of atomic mass number is 137+ 97 = 234.

The calculation shows that, the missing component at the blank space will have atomic number 0 and atomic mass number 2. The component with 0 atomic number will be neutron which will have atomic mass number 1. Thus, X equals to 2 neutrons.

Therefore, the complete equation with filled blank space can be written as,

$_{92}^{235}U + _0^1n \to _{52}^{137}A + _{40}^{97}B + \underline {2\,_0^1n} $

(B) Assume that, the element or component that should be present at the blank space in the given reaction is Y.

$_{34}^{82}Se \to 2_{ - 1}^0e + \underline Y $

The atomic number of selenium is 34 and atomic mass number is 82. The selenium is given 2 electrons and component Y.

To find out Y, we need to balance the atomic mass number and atomic number on both sides.

On the left side of the reaction, atomic number is 34 and that of atomic mass number is 92 as shown.

On the right side of the reaction, the electrons have atomic number as $–$1. There are two electrons, therefore, 2x ($–$1) = $–$2.

The atomic mass number of component Y must be 82.

The atomic number of component Y must be 36, as 36 + ($–$2) = 34 to balance both sides.

The element with atomic number 36 and atomic mass number 82 is Krypton and is denotes as Kr. Therefore, the complete equation with filled blank space can be written as,

$_{34}^{82}Se \to 2_{ - 1}^0e + _{36}^{82}\underline {Kr} $

Final Answer

(A) $_{92}^{235}U + _0^1n \to _{52}^{137}A + _{40}^{97}B + \underline {2\,_0^1n} $

(B) $_{34}^{82}Se \to 2_{ - 1}^0e + _{36}^{82}\underline {Kr} $

Hints:

The atomic number at the right side of equation should be equal to the sum of atomic number at the left side for both equations.

The given chemical reaction is,

2X(g) $\to$ 3Y(g) + 2Z(g)

The data given for the above chemical reaction is,

(A) From the given data, the duration of decreasing partial pressure of X from 800 to 400 and from 400 to 200, is 100 minutes for both. Therefore, the time required for decreasing of partial pressure is constant for decrement of pressure from 800 to 400 and 400 to 200.

Thus, the above chemical reaction is first order reaction.

(B) For first order reaction, rate constant can be calculated as follows:

$k = {{0.693} \over {{t_{1/2}}}}$ ..... (i)

Here, k is the rate constant and t$_{1/2}$ is the half life time.

As given, $t_{1/2}=100$ min

Substitute the respective value in equation (i), we get

$k = {{0.693} \over {100}}$

$\therefore~k=6.93\times10^{-3}$ min$^{-1}$

The rate constant for the reaction is $6.93\times10^{-3}$ min$^{-1}$.

(C) The time taken for completion of 75% reaction is 2$t_{1/2}$.

(D) For the given chemical reaction,

The partial pressure at t = 0 and after some time is given below.

P$_{total}$ = 800 $-$ 2x + 3x + 2x

= 800 + 3x

Given, 800 $-$ 2x = 700

x = 50 mm

P$_{total}$ = 800 + 3 $\times$ 50 = 950 mm

The total pressure of the reaction mixture when $p_x=700$ mm.

Final Answer:

(A) First order reaction

(B) Rate constant = 6.93 $\times$ 10$^{–3} min$^–1}$

(C) Time taken for completion of 75% reaction is 2$t_{1/2}$

(D) Total pressure when $p_x = 700$ mm is 950 mm.

Shortcut Method:

Alternatively, the rate constant for the first order reaction can be calculated as follows:

Hence, for first order $k = {{2.303} \over t}{\log _{10}}{{{p_0}} \over {{p_1}}}$

At t = 100 min ${k_{100}} = {{2.303} \over {100}}{\log _{10}}{{800} \over {400}}$

$ = {{2.303} \over {100}}{\log _{10}}2 = {{2.303 \times 0.3010} \over {100}}$ min$^{-1}$

$ = 6.93 \times {10^{ - 3}}$ min$^{-1}$

According to the given data, we see that when the concentration of A alone is doubled (0.1 $\to$ 0.2 mol L^$-$1), the rate of the reaction gets doubled (0.05 $\to$ 0.1 mol L^$-$1 s^$-$1). Therefore, the order with respect to A is 1.

When the concentration of B alone is doubled (0.1 $\to$ 0.2 mol L^$-$1), the rate of the reaction remains unaltered. Therefore, the order with respect to B is 0.

(1) The rate equation is, Rate = $k[A]{[B]^0} = k[A]$

(2) Rate $ = k[A]$ or, $k = {{rate} \over {[A]}}$ or, $k = {{0.05} \over {0.1}}$ $\therefore$ k = 0.5 s^$-$1

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