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JEE MainChemistryModerateChemical Kinetics

The decomposition of N2O5 in chloroform was followed by
measuring the volume of O2 gas evolved: 2N2O5(CCI4) → 2N2O4(CCI4) +
O2(g). The maximum volume of O2 gas obtained was 100 cm3. In 500
minutes, 90 cm3 of O2 were evolved. The first order rate constant (in
min-1) for the disappearance of N2O5 is

Show answer and solution
kt=ln(C0C1)kt = ln(\frac{C_{0}}{C_{1}})
2N2O5⟶2N2O4+O22N_{2}O_{5} \longrightarrow 2N_{2}O_{4} + O_{2}
t=0200cm3           0            0\begin{matrix} t = 0 & 200{cm}^{3}\ \ \ \ \ \ \ \ \ \ \ 0\ \ \ \ \ \ \ \ \ \ \ \ 0 \end{matrix}
t=t20cm3180cm390cm3\begin{matrix} t = t & 20{cm}^{3} & 180{cm}^{3} & 90{cm}^{3} \end{matrix}
t=∞0        200cm3100cm3\begin{matrix} t = \infty & 0 & \ \ \ \ \ \ \ \ 200{cm}^{3} & 100{cm}^{3} \end{matrix}
Initial volume of N2O5=200cm3N_{2}O_{5} = 200{cm}^{3}
Because Max. volume of=O2=100cm3= O_{2} = 100{cm}^{3}
∴K×500=ln(20020)⇒k=ln10500=2.303500\therefore K \times 500 = ln(\frac{200}{20}) \Rightarrow k = \frac{ln10}{500} = \frac{2.303}{500}

Source: Grass-G / CH-15 Chemistry Paper 11 Nov..docx

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