Chemical Kinetics

The drug follows first-order kinetics. Using the formula $k=\frac{2.303}{t}\log \frac{[A_0]}{[A]}$ , we find $k=0.1156$ month⁻¹. The half-life is $t_{1\mathrm{/}2}=\frac{0.693}{k}=6$ months.

Correct answer: A (6 months).

When temperature changes from $T_1=27^\circ\!C=300\,K$ to $T_2=57^\circ\!C=330\,K,$ the rate quadruples:

$$\ln\left(\frac{k_2}{k_1}\right) = \ln(4)\approx1.386 = \frac{E_a}{R}\Bigl(\frac{1}{T_1}-\frac{1}{T_2}\Bigr).$$

So,

$$E_a = \frac{1.386\,R}{\left(\tfrac{1}{300}-\tfrac{1}{330}\right)} =1.386\times8.314\times\frac{1}{\tfrac{1}{300}-\tfrac{1}{330}} \approx 38\,\text{kJ mol}^{-1}.$$

For the elementary reaction A + B → C + D, rate law is Rate = k[A][B]. When volume becomes 1/3rd, concentration increases 3 times. Since rate ∝ [A][B], new rate = 9 times the initial rate. Final answer: Rate increases 9 times.

Rate $=k_2\left[A\right]\left[B_2\right]$

$K=\frac{[A{]}^2}{\left[A_2\right]}$

Rearrange to express [A]:

$\left[A\right]=[K{]}^{1/2}{\left[A_2\right]}^{1/2}$

Substituting this into the rate law:

$$\begin{gathered} \text{ Rate }=k_2\left(K^{1/2}{\left[A_2\right]}^{1/2}\right)\left[B_2\right] \\ \text{ Rate }=k{\left[A_2\right]}^{1/2}\left[B_2\right] \end{gathered}$$

where $k=k_2{K}^{1/2}$ is a new rate constant. From the rate law:

$\text{ Rate }=k{\left[A_2\right]}^{1/2}{\left[B_2\right]}^1$

Order with respect to $A_2=0.5$
Order with respect to $B_2=1$
The overall order of the reaction:

$0.5+1=1.5$

Using the Arrhenius equation, knowing k at two temperatures allows calculating the activation energy Eₐ from the slope of a plot of ln(k) vs 1/T.

Radioactive decay follows first-order kinetics, where the decay constant k is independent of temperature

Using $k=\frac{0.693}{600}$and Arrhenius equation,

$$\ln k=\ln A-\frac{E_a}{RT}$$

Substituting values and solving, we get T ≈ 309 K. Closest answer: (D) 311.15 K.

The order of the reaction is determined by the rate law expression of the slowest step of the reaction, which is the rate-determining step

The slowest step in the given question is $\mathrm{A}+\mathrm{B} \to \mathrm{D}$

Rate expression for the slowest step $\mathrm{Rate} =\mathrm{K}\left[\mathrm{A}\right]\left[\mathrm{B}\right], \mathrm{hence} \mathrm{the} \mathrm{order} \mathrm{of} \mathrm{reaction} =2$

For first-order reactions, the rate law is:

$$\begin{gathered} k =\frac{1}{12}\ln \left(\frac{16}{4}\right) \\ k \approx 0.1155{\text{ month }}^{-1} \end{gathered}$$

Calculate Time for 50 % Decomposition

$$\begin{gathered} t=\frac{1}{0.1155}\ln \left(\frac{16}{8}\right) \\ t\approx 6\text{ months } \end{gathered}$$

The rate law for the reaction is given by: $R_1=k\left[A{]}^1\right[B{]}^1$
Now, if the volume is reduced to one-third, the concentration of each reactant increases by a factor of 3 . Therefore, the new rate $R_2$ is:

$$\begin{gathered} R_2=k\left[3A{]}^1\right[3B{]}^1 \\ R_2=9R_1 \end{gathered}$$