Verified Solution • 2025–26 Session moderate

QJEEM19Ch4696: JEE Main 2019 Chemistry - Class 12 JEEM19Ch4696 Solution: NO 2 required for a reaction is produced by the decomposition…

NO2 required for a reaction is produced by the decomposition of N2O5 in CCl4 as per the equation,
2N2O5(g) →\to 4NO2(g) + O2(g).
The initial concentration of N2O5 is 3.00 mol L–1 and it is 2.75 mol L–1 after 30 minutes. The rate of formation of NO2 is :

Multiple Choice Options

  • A 2.083 × 10–3 mol L–1 min–1
  • B 8.333 × 10–3 mol L–1 min–1
  • C 4.167 × 10–3 mol L–1 min–1
  • D 1.667 × 10–2 mol L–1 min–1

Step-by-Step Verified Solution

Rate of disappearance of N2O5 = −Δ[N2O5]Δ[t]- {{\Delta \left[ {{N_2}{O_5}} \right]} \over {\Delta \left[ t \right]}} = −[2.75−3]30- {{\left[ {2.75 - 3} \right]} \over {30}} = +1120+ {1 \over {120}} mol L-1 min-1

Rate of formation of NO2 = +Δ[NO2]Δ[t]+ {{\Delta \left[ {N{O_2}} \right]} \over {\Delta \left[ t \right]}}

As −12Δ[N2O5]Δ[t]=+14Δ[NO2]Δ[t]- {1 \over 2}{{\Delta \left[ {{N_2}{O_5}} \right]} \over {\Delta \left[ t \right]}} = + {1 \over 4}{{\Delta \left[ {N{O_2}} \right]} \over {\Delta \left[ t \right]}}
⇒\Rightarrow Δ[NO2]Δ[t]{{\Delta \left[ {N{O_2}} \right]} \over {\Delta \left[ t \right]}} = +1120×42+ {1 \over {120}} \times {4 \over 2} = 0.01667 = 1.667 × 10–2 mol L–1 min–1

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