JEE Main · 2025 · 23 Jan 2025 · Shift 1 · Chemical Kinetics

For a reaction $a\mathrm{A}\to \text{products}$ that is first order in A, the rate of consumption of A relates to the rate constant of the reaction as $-\frac{d[\mathrm{A}]}{dt}=ak[\mathrm{A}]$. Here $a=2$ for ${\mathrm{N}}_2{\mathrm{O}}_5$, so the decay constant for ${\mathrm{N}}_2{\mathrm{O}}_5$ itself is $k_{{\mathrm{N}}_2{\mathrm{O}}_5}=2\times 4.606\times 10^{-2}=9.212\times 10^{-2}{\mathrm{s}}^{-1}$.

Let $P$ be the partial pressure of ${\mathrm{N}}_2{\mathrm{O}}_5$ that has decomposed by $t=100\mathrm{s}$:

$2{\mathrm{N}}_2{\mathrm{O}}_5(g)\to 2{\mathrm{N}}_2{\mathrm{O}}_4(g)+{\mathrm{O}}_2(g)$, starting at 0.6 atm, 0, 0 and becoming $(0.6-P)$, $P$, $P/2$.

Applying first-order kinetics to ${\mathrm{N}}_2{\mathrm{O}}_5$:

$k_{{\mathrm{N}}_2{\mathrm{O}}_5}t=2.303{\log }_{10}\frac{0.6}{0.6-P}$

$9.212\times 10^{-2}\times 100=9.212=2.303{\log }_{10}\frac{0.6}{0.6-P}\Rightarrow {\log }_{10}\frac{0.6}{0.6-P}=4$

$\frac{0.6}{0.6-P}=10^4\Rightarrow P=0.6(1-10^{-4})=0.59994\mathrm{atm}$

Total pressure at $t=100\mathrm{s}$:

$P_{\text{total}}=(0.6-P)+P+\frac{P}{2}=0.6+\frac{P}{2}=0.6+0.29997=0.89997\mathrm{atm}=899.97\times 10^{-3}\mathrm{atm}$

Rounded to the nearest integer, $x=900$.