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Step-by-Step Solution
Step 1: Write the Balanced Decomposition Reaction
The given decomposition reaction is:
$N_{2}O_{5} \rightarrow 2\,NO_{2} + \tfrac{1}{2}\,O_{2}$
Step 2: Identify Initial Pressure and Final Pressure After 50 Minutes
Initial total pressure inside the vessel = 50 mmHg.
After 50 minutes, total pressure increases to 87.5 mmHg.
Step 3: Define the Amount of $N_{2}O_{5}$ Decomposed After 50 Minutes
Let the partial pressure of $N_{2}O_{5}$ that decomposed in 50 minutes be $P_{1}$. Thus:
Remaining $N_{2}O_{5}$ has partial pressure $(50 - P_{1})$.
Formed $NO_{2}$ has partial pressure $2\,P_{1}$ (because 2 moles of $NO_{2}$ are produced per mole of $N_{2}O_{5}$ decomposed).
Formed $O_{2}$ has partial pressure $\tfrac{P_{1}}{2}$ (because $\tfrac{1}{2}$ mole of $O_{2}$ is produced per mole of $N_{2}O_{5}$ decomposed).
Step 4: Write the Total Pressure Expression After 50 Minutes
The total pressure at 50 minutes is given by summing all partial pressures:
$(50 - P_{1}) + 2P_{1} + \tfrac{P_{1}}{2} = 87.5
Combine like terms:
$50 + \left(2P_{1} - P_{1}\right) + \frac{P_{1}}{2} = 87.5$
$50 + \frac{3P_{1}}{2} = 87.5
Step 5: Solve for $P_{1}$
$\frac{3P_{1}}{2} = 87.5 - 50 = 37.5
$P_{1} = \frac{37.5 \times 2}{3} = 25
Step 6: Infer the Half-Life from the Change in Pressure
Since the partial pressure of $N_{2}O_{5}$ has been reduced by half (decomposed from 50 mmHg to 25 mmHg of reacted portion) in 50 minutes, the half-life ($t_{1/2}$) of the reaction is 50 minutes.
Step 7: Determine Conditions After 100 Minutes (Two Half-Lives)
After two half-lives (i.e., 100 minutes), the amount of $N_{2}O_{5}$ that remains is:
$ \text{Remaining fraction} = \left(\frac{1}{2}\right)^{2} = \frac{1}{4}
In terms of partial pressures, we can use a stoichiometric approach similar to Step 3, but now the decomposition is repeated for the second half-life:
Let $P_{2}$ be the decomposed amount after 100 minutes.
The partial pressure of $N_{2}O_{5}$ then is $(50 - P_{2})$.
Partial pressure of $NO_{2}$ is $2P_{2}$.
Partial pressure of $O_{2}$ is $\tfrac{P_{2}}{2}$.
However, because one more half-life has passed, another half of the remaining $N_{2}O_{5}$ would decompose. From the stoichiometric step or by direct half-life logic, we find:
$50 - P_{2} = \frac{50}{2} = 25 \quad \Rightarrow \quad P_{2} = 25
But since 50 mmHg was the initial total, after the first half-life, 25 mmHg of $N_{2}O_{5}$ had reacted. In the second half-life, half of the remaining $N_{2}O_{5}$ (which was 25 mmHg leftover) will decompose, meaning an additional 12.5 mmHg of $N_{2}O_{5}$ will react. So total $N_{2}O_{5}$ decomposed is $25 + 12.5 = 37.5$ mmHg. Thus, directly:
$P_{2} = 37.5
Step 8: Calculate the Total Pressure After 100 Minutes
Total pressure = Unreacted $N_{2}O_{5}$ + $NO_{2}$ formed + $O_{2}$ formed
$= (50 - P_{2}) + 2P_{2} + \tfrac{P_{2}}{2}
$= 50 - 37.5 + 2 \times 37.5 + \tfrac{37.5}{2}
$= 50 + \frac{3 \times 37.5}{2}
$= 50 + \frac{3 \times 37.5}{2} = 50 + \frac{112.5}{2} = 50 + 56.25 = 106.25 \text{ mmHg}
Final Answer: 106.25 mmHg
