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← JEE Advanced 2019 · Paper 2
ChemistryNumerical+3Question 29 of 54

JEE Advanced 2019 · Paper 2, Chemistry question 11

The decomposition reaction 2 NX2OX5(g)→Δ2 NX2OX4(g)+OX2(g)\ce{2N2O5(g) ->[\Delta] 2N2O4(g) + O2(g)} is started in a closed cylinder under isothermal isochoric condition at an initial pressure of 1 atm. After Y×103Y\times10^3 s, the pressure inside the cylinder is found to be 1.45 atm. If the rate constant of the reaction is 5×10−4 s−15\times10^{-4}\ \mathrm{s^{-1}}, assuming ideal gas behavior, the value of YY is ______

Answer and worked solution

Official answer: 2.3

Pressures. Let 2x2x atm of NX2OX5\ce{N2O5} decompose:
ptotal=(1−2x)+2x+x=1+x=1.45  ⟹  x=0.45,pNX2OX5=0.10 atm.p_{total}=(1-2x)+2x+x=1+x=1.45\implies x=0.45,\qquad p_{\ce{N2O5}}=0.10\text{ atm}.

Rate law. The rate constant is in s⁻¹, so the reaction is first order, with rate =−12d[NX2OX5]dt=k[NX2OX5]=-\frac12\frac{d[\ce{N2O5}]}{dt}=k[\ce{N2O5}]. Hence
ln⁡10.10=2kt  ⟹  t=2.3032×5×10−4=2.30×103 s.\ln\frac{1}{0.10}=2kt\implies t=\frac{2.303}{2\times5\times10^{-4}}=2.30\times10^3\text{ s}.
Answer: 2.30 (key: 2.30 to 2.31)

Topics:Integrated rate equationsOrder and molecularity
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