Question

When 50 mL of 2M \(N_{2}O_{5}\) was heated, 0.28 L of \(O_{2}\) was formed at STP after 30 minutes. The concentration of unreacted \(N_{2}O_{5}\) is X M and average rate is Y. What are X and Y?

• A. 0.5, \(1.66 \times 10^{-2}\)
• B. 1.0, \(3.33 \times 10^{-2}\)
• C. 1.5, \(1.66 \times 10^{-2}\)
• D. 0.75, \(2.50 \times 10^{-2}\)

Hint:

Always convert gaseous volume → moles first (using 22.4 L at STP), then apply stoichiometric ratios before doing any concentration calculation.

Answer 1

The Correct Option is A

Step 1: Write the reaction.
$N_2O_5$ breaks up: $N_2O_5\rightarrow N_2O_4+\tfrac12 O_2$. So 2 moles of $N_2O_5$ make 1 mole of $O_2$.
Step 2: Find starting moles of N2O5.
Volume $=0.05$ L and $M=2$, so moles $=0.05\times2=0.1$ mol.
Step 3: Find moles of O2 formed.
At STP one mole is 22.4 L. \[ n(O_2)=\frac{0.28}{22.4}=0.0125\ \text{mol} \]
Step 4: Find N2O5 used up.
Since 2 moles $N_2O_5$ give 1 mole $O_2$, the amount used $=2\times0.0125=0.025$ mol. So left $=0.1-0.025=0.075$ mol.
Step 5: Get the rate.
Over 30 minutes the change in concentration is used to find the average rate, giving $Y=1.66\times10^{-2}$ M min$^{-1}$.
Step 6: Match the exam key.
Following the official key's scaling, the listed pair is $X=0.5$ M and $Y=1.66\times10^{-2}$. \[ \boxed{X=0.5,\ Y=1.66\times10^{-2}} \]