Consider the following first order gas phase reaction at constant temperature $$ \mathrm{A}(\mathrm{g}) \rightarrow 2 \mathrm{B}(\mathrm{~g})+\mathrm{C}(\mathrm{g})$$

If the total pressure of the gases is found to be 200 torr after 23 $\mathrm{sec}$. and 300 torr upon the complete decomposition of A after a very long time, then the rate constant of the given reaction is ________ $\times 10^{-2} \mathrm{~s}^{-1}$ (nearest integer)

[Given : $\log _{10}(2)=0.301$]

<p>To find the rate constant of the first-order reaction indicated, let us first understand the reaction dynamics based on the total pressure change over time. The reaction is:</p> <p>$\mathrm{A(g)} \rightarrow 2\mathrm{B(g)} + \mathrm{C(g)}$</p> <p>Initially, only A is present. As the reaction progresses, A decreases, and B and C are formed. For each mole of A that reacts, a total of 3 moles of gas are produced (2 moles of B and 1 mole of C), leading to an increase in the total pressure of the system if the volume remains constant.</p> <p>At the start: Total pressure due to A only.</p> <p>Finally, when A has completely decomposed: Total pressure is due to 3 times the initial amount of A, as no A is left and for every mole of A decomposed, 3 moles of gas (2B + 1C) are produced.</p> <p>The total pressure change thus directly relates to the extent of reaction, i.e., how much A has decomposed into B and C.</p> <p>The total pressure at 23 sec is 200 torr, and the final total pressure is 300 torr after a very long time, indicating that the pressure increases by 100 torr due to the complete decomposition of A.</p> <p>To use this information, we need to understand the relationship between pressure change and concentration in a closed system, especially for a first-order reaction. For a first-order reaction:</p> <p>$\ln\left(\frac{[A]_0}{[A]}\right) = kt$</p> <p>Where: $[A]_0$ is the initial concentration (or in this context, partial pressure), $[A]$ is the concentration (or partial pressure) at time t, $k$ is the rate constant, and $t$ is the time.</p> <p>Let's denote the initial total pressure due to A as $P_0$. At complete decomposition, the pressure increase is from the conversion of A to 2B + C which increases the total pressure to 300 torr, indicating that initially, $P_0$ must have been 100 torr since the pressure increase is due to the tripling effect of the reaction, completing to 300 torr.</p> <p>At 23 sec, the total pressure is at 200 torr. This means 100 torr is due to the unreacted A, and the additional 100 torr is from the formation of 2B + C. Given the nature of the reaction (1:3 stoichiometry of A to the total gas), we can deduce that at 23 seconds, half of A has reacted because the pressure due to the products matches the pressure due to the unreacted A.</p> <p>Hence, at 23 sec, $[A] = \frac{1}{2}[A]_0$. Plugging this into the first-order reaction formula:</p> <p>$\ln\left(\frac{[A]_0}{\frac{1}{2}[A]_0}\right) = kt$</p> <p>$\ln(2) = k \times 23$</p> <p>Given that $\log_{10}(2) = 0.301$, and knowing that $\ln(2) = \log_{e}(2)$ (where $\log_{e}(2) \approx 0.693$, for conversion from base 10 to e we can use the natural logarithm of 2 directly), we have:</p> <p>$0.693 = k \times 23$</p> <p>$k = \frac{0.693}{23} = 0.0301 \, \mathrm{s}^{-1}$</p> <p>When expressed in the format requested, $k = 3.01 \times 10^{-2} \, \mathrm{s}^{-1}$. Rounding to the nearest integer, $k = 3 \times 10^{-2} \, \mathrm{s}^{-1}$.</p>