N$$_2$$O$$_5$$ decomposes to NO$$_2$$ and O$$_2$$ and follows first order kinetics. After 50 minutes, the pressure inside the vessel increases from 50 mm Hg to 87.5 mm Hg. The pressure of the gaseous mixture after 100 minutes at constant temperature will be:

We begin with the balanced gaseous decomposition

$$2\,N_2O_5(g) \;\longrightarrow\; 4\,NO_2(g)+O_2(g)$$

The vessel initially contains only $$N_2O_5$$ at a pressure of $$P_0 = 50\;{\rm mm\;Hg}$$. Let the fraction of the original $$N_2O_5$$ that has decomposed after any time $$t$$ be denoted by $$\alpha$$ (so $$0\le \alpha \le 1$$).

Because pressure is directly proportional to the number of moles at constant volume and temperature, we translate the stoichiometry into partial pressures:

• Initial moles (or mole‐pressure units) of $$N_2O_5$$  $$= 1$$ (reference basis).
• When the fraction $$\alpha$$ has reacted, the mole distribution is

$$\begin{aligned} N_2O_5 &: 1-\alpha\\[4pt] NO_2 &: 2\alpha \quad\;(\text{because }2N_2O_5\to4NO_2\Rightarrow1\to2)\\[4pt] O_2 &: \dfrac{\alpha}{2}\quad (\text{because }2N_2O_5\to O_2\Rightarrow1\to\frac12) \end{aligned}$$

Hence the total number of mole‐units, and therefore the total pressure, become

$$P = P_0\big[(1-\alpha)+2\alpha+\tfrac{\alpha}{2}\big] = P_0\big[1+1.5\,\alpha\big].$$

Data for 50 minutes. After 50 min the pressure has risen from 50 mm Hg to 87.5 mm Hg. Substituting these numbers,

$$87.5 = 50\,(1+1.5\,\alpha_1).$$

Dividing both sides by 50,

$$1+1.5\,\alpha_1 = 1.75 \quad\Longrightarrow\quad 1.5\,\alpha_1 = 0.75 \quad\Longrightarrow\quad \alpha_1 = 0.50.$$

Thus, exactly 50 % of the $$N_2O_5$$ has decomposed in 50 minutes.

Finding the first-order rate constant. For a first-order reaction we have the integrated rate law

$$k t = -\ln(1-\alpha),$$

where $$\alpha$$ is the fraction decomposed in time $$t$$.

Using $$\alpha_1 = 0.50$$ at $$t_1 = 50\;{\rm min}$$,

$$k = \frac{-\ln(1-\alpha_1)}{t_1} = \frac{-\ln(1-0.50)}{50} = \frac{\ln 2}{50}\;{\rm min^{-1}}.$$

Fraction decomposed in 100 minutes. For $$t_2 = 100\;{\rm min}$$,

$$k t_2 = \frac{\ln 2}{50}\times 100 = 2\,\ln 2 = \ln 4.$$

Therefore,

$$1-\alpha_2 = e^{-k t_2}=e^{-\ln 4}= \frac{1}{4}\quad\Longrightarrow\quad \alpha_2 = 1-\frac14 = \frac34 = 0.75.$$

Total pressure after 100 minutes. Substituting $$\alpha_2 = 0.75$$ into $$P = P_0(1+1.5\alpha)$$,

$$P_{100} = 50\Bigl(1 + 1.5 \times 0.75\Bigr) = 50\bigl(1 + 1.125\bigr) = 50 \times 2.125 = 106.25\;{\rm mm\;Hg}.$$

The calculated pressure, 106.25 mm Hg, matches Option B.

Hence, the correct answer is Option B.