The reaction $$2N_2O_5(g) \rightarrow 4NO_2(g) + O_2(g)$$ follows first order kinetics. The pressure of a vessel containing only $$N_2O_5$$ was found to increase from 50 mm Hg to 87.5 mm Hg in 30 min. The pressure exerted by the gases after 60 min. will be (Assume temperature remains constant):

The reaction is $$2N_2O_5(g) \rightarrow 4NO_2(g) + O_2(g)$$, which follows first-order kinetics. The initial pressure of $$N_2O_5$$ is 50 mm Hg, and after 30 minutes, the total pressure increases to 87.5 mm Hg. We need to find the total pressure after 60 minutes.

Since the reaction is first-order with respect to $$N_2O_5$$, the rate depends on its concentration. At constant volume and temperature, concentration is proportional to partial pressure, so we can use partial pressures in the first-order rate equation.

Let $$P_0 = 50$$ mm Hg be the initial pressure of $$N_2O_5$$. As the reaction proceeds, $$N_2O_5$$ decomposes, and products are formed. For the reaction $$2N_2O_5 \rightarrow 4NO_2 + O_2$$, when 2 moles of $$N_2O_5$$ decompose, they produce 5 moles of products (4 moles of $$NO_2$$ and 1 mole of $$O_2$$), resulting in a net increase of 3 moles (since 5 moles produced minus 2 moles decomposed).

Define $$y$$ as the pressure decrease corresponding to the decomposition of $$N_2O_5$$ based on the stoichiometry. If the initial pressure of $$N_2O_5$$ is $$P_0$$, then:

• The pressure of $$N_2O_5$$ remaining is $$P_0 - 2y$$.
• The pressure of $$NO_2$$ produced is $$4y$$.
• The pressure of $$O_2$$ produced is $$y$$.

The total pressure $$P_t$$ at any time is:

$$P_t = (P_0 - 2y) + 4y + y = P_0 + 3y$$

Solving for $$y$$:

$$y = \frac{P_t - P_0}{3}$$

The partial pressure of $$N_2O_5$$ at time $$t$$ is:

$$P_{N_2O_5} = P_0 - 2y = P_0 - 2\left(\frac{P_t - P_0}{3}\right) = \frac{3P_0 - 2P_t + 2P_0}{3} = \frac{5P_0 - 2P_t}{3}$$

For a first-order reaction:

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

Here, $$[A]_0 = P_0 = 50$$ mm Hg and $$[A] = P_{N_2O_5} = \frac{5P_0 - 2P_t}{3}$$, so:

$$\ln\left(\frac{50}{\frac{5 \times 50 - 2P_t}{3}}\right) = kt$$

Simplify the argument:

$$\ln\left(\frac{50 \times 3}{250 - 2P_t}\right) = kt \implies \ln\left(\frac{150}{250 - 2P_t}\right) = kt$$

At $$t = 30$$ min, $$P_t = 87.5$$ mm Hg:

$$\ln\left(\frac{150}{250 - 2 \times 87.5}\right) = k \times 30$$

Calculate inside the denominator:

$$2 \times 87.5 = 175, \quad 250 - 175 = 75$$

So:

$$\ln\left(\frac{150}{75}\right) = \ln(2) = k \times 30$$

Solving for $$k$$:

$$k = \frac{\ln 2}{30} \text{ min}^{-1}$$

Now, find the total pressure $$P$$ at $$t = 60$$ min:

$$\ln\left(\frac{150}{250 - 2P}\right) = k \times 60$$

Substitute $$k = \frac{\ln 2}{30}$$:

$$\ln\left(\frac{150}{250 - 2P}\right) = \frac{\ln 2}{30} \times 60 = 2 \ln 2 = \ln(4)$$

Since $$\ln a = \ln b$$ implies $$a = b$$:

$$\frac{150}{250 - 2P} = 4$$

Solve for $$P$$:

$$150 = 4(250 - 2P)$$ $$150 = 1000 - 8P$$ $$8P = 1000 - 150 = 850$$ $$P = \frac{850}{8} = 106.25 \text{ mm Hg}$$

Hence, the pressure after 60 minutes is 106.25 mm Hg, which corresponds to Option A.

So, the answer is 106.25 mm Hg.