1. Set up the reaction and pressure relation

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

Suppose initially only $\mathrm{A}$ is present and its initial pressure is $P_0$.

If a fraction $\alpha$ of $\mathrm{A}$ has decomposed at time $t$, then:

• Pressure of $\mathrm{A}$ left $= P_0(1-\alpha)$
• Pressure of $\mathrm{B}$ formed $= 2\alpha P_0$
• Pressure of $\mathrm{C}$ formed $= \alpha P_0$

So total pressure at time $t$ is $P_t = P_0(1-\alpha) + 2\alpha P_0 + \alpha P_0 = P_0(1+2\alpha)$

2. Use the pressure at complete decomposition

At complete decomposition, $\alpha = 1$, so $P_\infty = P_0(1+2) = 3P_0$

Given: $P_\infty = 300\ \text{torr}$

Hence, $3P_0 = 300 \Rightarrow P_0 = 100\ \text{torr}$

3. Use the pressure after 23 s

Given: $P_t = 200\ \text{torr at } t=23\ \text{s}$

Now, $200 = 100(1+2\alpha)$ $2 = 1 + 2\alpha$ $2\alpha = 1$ $\alpha = \frac{1}{2}$

So after 23 s, half of $\mathrm{A}$ has decomposed.

Thus pressure of $\mathrm{A}$ remaining is $P_A = P_0(1-\alpha) = 100\left(1-\frac12\right)=50\ \text{torr}$

4. Apply first-order kinetics

For a first-order reaction, $k = \frac{2.303}{t}\log\frac{[A]_0}{[A]_t}$

Since concentration is proportional to partial pressure for gases, $k = \frac{2.303}{23}\log\frac{100}{50}$ $k = \frac{2.303}{23}\log 2$

Given: $\log 2 = 0.301$

Therefore, $k = \frac{2.303 \times 0.301}{23}$

Now, $2.303 \times 0.301 \approx 0.693$

So, $k \approx \frac{0.693}{23} \approx 0.0301\ \text{s}^{-1}$

5. Convert into the asked form

We need $k = (\text{integer}) \times 10^{-2}\ \text{s}^{-1}$

Since $0.0301 = 3.01 \times 10^{-2}\ \text{s}^{-1}$

Nearest integer $= 3$.

Final Answer

$\boxed{3}$