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:

$$\mathrm{A(g)} \rightarrow 2\mathrm{B(g)} + \mathrm{C(g)}$$

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.

At the start: Total pressure due to A only.

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.

The total pressure change thus directly relates to the extent of reaction, i.e., how much A has decomposed into B and C.

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.

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:

$$\ln\left(\frac{[A]_0}{[A]}\right) = kt$$

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.

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.

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.

Hence, at 23 sec, $$[A] = \frac{1}{2}[A]_0$$. Plugging this into the first-order reaction formula:

$$\ln\left(\frac{[A]_0}{\frac{1}{2}[A]_0}\right) = kt$$

$$\ln(2) = k \times 23$$

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:

$$0.693 = k \times 23$$

$$k = \frac{0.693}{23} = 0.0301 \, \mathrm{s}^{-1}$$

When expressed in the format requested, $$k = 3.01 \times 10^{-2} \, \mathrm{s}^{-1}$$. Rounding to the nearest integer, $$k = 3 \times 10^{-2} \, \mathrm{s}^{-1}$$.