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Chemistry · 2022 · 2 marks
CBSE 2022 · Region 2 · Set 2 · Q3
In a reaction 2 N_2 O_5( g ) → 4 NO_2( g )+ O_2( g ), the concentration of N_2 O_5 decreases from 0.5 mol L^-1 to 0.4 mol L^-1 in 10 minutes.…
In a reaction $\displaystyle 2 \mathrm{~N}_{2} \mathrm{O}_{5}(\mathrm{~g}) \longrightarrow 4 \mathrm{NO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g})$, the concentration of $\displaystyle \mathrm{N}_{2} \mathrm{O}_{5}$ decreases from $\displaystyle 0.5 \mathrm{~mol} \mathrm{~L}^{-1}$ to $\displaystyle 0.4 \mathrm{~mol} \mathrm{~L}^{-1}$ in $\displaystyle 10$ minutes. Calculate the average rate of this reaction and rate of production of $\displaystyle \mathrm{NO}_{2}$ during this period.
Marking-scheme solution
\[\begin{aligned}
& \text { Average rate }=\frac{-1}{2} \frac{\Delta\left(\mathrm{~N}_{2} \mathrm{O}_{5}\right)}{\Delta t} \\
&=\frac{-1}{2} \frac{(0 \cdot 4-0 \cdot 5)}{10} \\
&=\frac{0 \cdot 1}{20}=0.005 \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~min}^{-1} \\
& \frac{\Delta\left[\mathrm{NO}_{2}\right]}{\Delta t}=4 \times 0.005=0.02 \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~min}^{-1}
\end{aligned}
\]
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CBSE Class 12 Chemistry past-paper question from the 2022 board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.