1. Given reaction

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

This is a first-order decomposition at constant volume.

Initial total pressure:

$P_0 = 0.1\ \text{atm}$

At time $t=115\,\text{s}$, total pressure:

$P_t = 0.28\ \text{atm}$

2. Relate total pressure to partial pressure of A

Let initial pressure of $A$ be $P_0=0.1$ atm.

Suppose fraction decomposed at time $t$ is $\alpha$.

Then:

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

So total pressure at time $t$ is

$P_t = P_0(1-\alpha) + 2P_0\alpha + P_0\alpha$

$P_t = P_0(1+2\alpha)$

Substitute values:

$0.28 = 0.1(1+2\alpha)$

$1+2\alpha = 2.8$

$2\alpha = 1.8$

$\alpha = 0.9$

Thus, pressure of $A$ remaining is

$P_A = P_0(1-\alpha) = 0.1(0.1) = 0.01\ \text{atm}$

3. Use first-order rate law

For a first-order reaction,

$k = \frac{2.303}{t}\log\frac{P_{A,0}}{P_{A,t}}$

Here,

$P_{A,0}=0.1, \qquad P_{A,t}=0.01, \qquad t=115\,\text{s}$

So,

$k = \frac{2.303}{115}\log\left(\frac{0.1}{0.01}\right)$

$k = \frac{2.303}{115}\log(10)$

Since $\log 10 = 1$,

$k = \frac{2.303}{115} = 0.0200\ \text{s}^{-1}$

$k = 2.0 \times 10^{-2}\ \text{s}^{-1}$

4. Nearest integer for $\underline{\hspace{2cm}} \times 10^{-2}\,\text{s}^{-1}$

The blank is:

$2$

5. Comparison with stored answer

Stored correct answer = $2$

Our derived answer = $2$

So they agree.