If the average rate of reaction for 2 N₂O₅(g) $\rightarrow$ 4 NO₂(g) + O₂(g) is 1.30 *10^-5 M/s over five minutes of the reaction, what is the partial pressure of NO₂ after the first 5.00 minutes if the reaction takes place in a 10.0 L vessel maintained at 55.0$\degree$C?

Question

Quizplus · Accepted Answer

The average rate of reaction given is $1.30 \times 10^{-5} \, \text{M/s}$. Since the reaction involves the decomposition of $2 \, \text{N}_2\text{O}_5$ into $4 \, \text{NO}_2 + \text{O}_2$, the change in concentration of $N_2O_5$ over 5 minutes (300 seconds) is $1.30 \times 10^{-5} \, \text{M/s} \times 300 \, \text{s} = 3.9 \times 10^{-3} \, \text{M}$. Given the stoichiometry of the reaction, the decrease in $N_2O_5$ concentration directly translates to the amount of $N_2O_5$ decomposed. Since the reaction volume is 10.0 L, the moles of $N_2O_5$ decomposed are $3.9 \times 10^{-3} \, \text{M} \times 10.0 \, \text{L} = 3.9 \times 10^{-2} \, \text{mol}$. Using the ideal gas law $PV = nRT$ and converting the temperature to Kelvin ($55.0^\circ C = 328.15 \, K$), we can solve for the pressure $P$ given $n = 3.9 \times 10^{-2} \, \text{mol}$, $V = 10.0 \, \text{L}$, $R = 0.0821 \, \text{L atm/mol K}$, and $T = 328.15 \, K$. The calculation yields a pressure close to $0.420 \, \text{atm}$, indicating that option A is correct.

If the Average Rate of Reaction for 2 N2O5(g) →\rightarrow→ 4 NO2(g) + O2(g) Is 1

If the Average Rate of Reaction for 2 N₂O₅(g) $\rightarrow$ 4 NO₂(g) + O₂(g) Is 1