Chemical Kinetics and Nuclear Chemistry

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

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

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

Option C

Nuclear fusion

Explanation :

A hydrogen bomb, also known as a thermonuclear bomb, uses the principle of nuclear fusion. In a fusion reaction, two lighter atomic nuclei combine to form a heavier nucleus, and a substantial amount of energy is released in the process. In the case of a hydrogen bomb, isotopes of hydrogen (such as deuterium and tritium) fuse together to form helium, releasing a large amount of energy.

It's worth noting that a hydrogen bomb usually involves a two-stage process. The first stage is a fission bomb (like those used in Hiroshima and Nagasaki) that creates the conditions necessary for the fusion reaction in the second stage. Despite this, the majority of the energy in a hydrogen bomb comes from fusion, which is why it is categorized as a fusion weapon.