For a reaction, $$ {N}_2{O}_5(g) \rightarrow 2{NO}_2(g) + \frac{1}{2} {O}_2(g) $$ in a constant volume container, no products were present initially. The final pressure of the system when 50% of the reaction gets completed is:

Question

Name the following:
(a) The most electronegative element of Period 2
(b) The largest atom of Period 3

Answer 1

To determine the final pressure of the system when 50% of the reaction is completed, we analyze the given reaction:

${N}_2{O}_5(g) \rightarrow 2{NO}_2(g) + \frac{1}{2} {O}_2(g)$

Initially, the container holds only ${N}_2{O}_5$ at an initial pressure P₀.
Assuming initial moles of ${N}_2{O}_5$ are 1, with no initial products.
At 50% reaction completion, half of ${N}_2{O}_5$ decomposes, leaving ${N}_2{O}_5$ moles at 0.5.
Based on reaction stoichiometry, the decomposition of $0.5$ moles of ${N}_2{O}_5$ yields the following product moles:
- ${NO}_2: 2 \times 0.5 = 1$
- ${O}_2: \frac{1}{2} \times 0.5 = 0.25$
The total moles in the system at 50% reaction completion is the sum of remaining reactant and formed products: 0.5 (remaining ${N}_2{O}_5$) + 1 (formed ${NO}_2$) + 0.25 (formed ${O}_2$) = 1.75 moles.
The change in total moles from the initial state (1 mole of ${N}_2{O}_5$) to the final state (1.75 moles) causes a pressure change, given constant volume and temperature.
The final pressure is calculated as:
- $\frac{\text{Final Moles}}{\text{Initial Moles}} \times P_0 = \frac{1.75}{1} \times P_0 = 1.75 \times P_0$
This is equivalent to the final pressure being $\frac{7}{4}$ times the initial pressure, as $1.75 = \frac{7}{4}$.

Therefore, the final pressure of the system is $\frac{7}{4}$ times the initial pressure.

$ \frac{7}{4} $ times of initial pressure

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