CBSE 2025 · Region 7 · Set 1 · Q24 · 3 marks
The following data were obtained during the first order thermal decomposition of $\displaystyle \mathrm{N}_{2} \mathrm{O}_{5}(\mathrm{~g})$ at constant volume : \[2 \mathrm{~N}_{2} \mathrm{O}_{5}(\mathrm{~g}) \longrightarrow 2 \mathrm{~N}_{2} \mathrm{O}_{4}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \] S.No. Time/s Total Pressure/atm $\displaystyle 1$ $\displaystyle 0$ $\displaystyle 0.5$ $\displaystyle 2$ $\displaystyle 100$ $\displaystyle 0.625$
Calculate rate constant. [ Given : $\displaystyle \log 2=0 \cdot 3010, \log 10=1$ ]
| S.No. | Time/s | Total Pressure/atm |
| $\displaystyle 1$ | $\displaystyle 0$ | $\displaystyle 0.5$ |
| $\displaystyle 2$ | $\displaystyle 100$ | $\displaystyle 0.625$ |
Marking-scheme solution
Pt= Pi – 2x +2x +x = Pi+x
x= = Pt- Pi
pA= Pi – 2x
= Pi- $\displaystyle 2$(Pt- Pi)
= 3Pi – $\displaystyle 2$ Pt
Where pi=$\displaystyle 0.5$ atm,
pA= 3pi-2pt
= (3X $\displaystyle 0.5$)-(2x0.625)
=0.25atm
= $\displaystyle 6.93$ X $\displaystyle 10$-$\displaystyle 3$ s-$\displaystyle 1$
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CBSE Class 12 Chemistry past-paper question from the 2025board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.