1. Write the decomposition reaction

For decomposition of dinitrogen pentoxide:

$2N_2O_5(g) \rightarrow 4NO_2(g) + O_2(g)$

So, if initially we start with only $N_2O_5$, then for every $2$ moles decomposed, total moles become $5$ moles.

That means total moles increase during reaction, so pressure increases at constant temperature and volume.

2. Relate total pressure to extent of decomposition

Let initial moles of $N_2O_5$ be $a$.

If fraction decomposed after time $t$ is $\alpha$, then:

• moles of $N_2O_5$ left $= a(1-\alpha)$
• moles of $NO_2$ formed $= 2a\alpha$
• moles of $O_2$ formed $= \dfrac{a\alpha}{2}$

Hence total moles at time $t$:

$n_t = a(1-\alpha) + 2a\alpha + \frac{a\alpha}{2} = a\left(1 + \frac{3\alpha}{2}\right)$

Since $P \propto n$ at constant $T,V$,

$P_t = P_0\left(1 + \frac{3\alpha}{2}\right)$

Given initial pressure:

$P_0 = 50\ \text{mmHg}$

After $50$ min, pressure is $87.5$ mmHg:

$87.5 = 50\left(1 + \frac{3\alpha}{2}\right)$

$\frac{87.5}{50} = 1 + \frac{3\alpha}{2}$

$1.75 = 1 + \frac{3\alpha}{2}$

$\frac{3\alpha}{2} = 0.75$

$\alpha = 0.5$

So after $50$ min, 50% decomposition has occurred.

3. Use first-order kinetics

For a first-order reaction,

$t_{1/2} = \frac{0.693}{k}$

Since 50% decomposes in 50 min, that means:

$t_{1/2} = 50\ \text{min}$

Therefore after another 50 min (i.e. at $t=100$ min), the remaining $N_2O_5$ again reduces by half.

So after $100$ min, amount left is:

$a \to \frac{a}{2} \to \frac{a}{4}$

Hence fraction decomposed after 100 min is:

$\alpha = 1 - \frac{1}{4} = \frac{3}{4}$

4. Calculate pressure after 100 min

Using

$P_t = P_0\left(1 + \frac{3\alpha}{2}\right)$

with $\alpha = \dfrac{3}{4}$:

$P_{100} = 50\left(1 + \frac{3}{2}\cdot\frac{3}{4}\right)$

$P_{100} = 50\left(1 + \frac{9}{8}\right)$

$P_{100} = 50\cdot\frac{17}{8}$

$P_{100} = 106.25\ \text{mmHg}$

5. Check options

• A: $175.0$ mmHg ✗
• B: $116.25$ mmHg ✗
• C: $136.25$ mmHg ✗
• D: $106.25$ mmHg ✓

So the correct answer is Option D.