Chemistry · 2019
JEE Main · 12 April 2019, Shift 2 · Q59
NO_2 required for a reaction is produced by the decomposition of N_2 O_5 in CCl_4 as per the equation, 2 N_2 O_5( g ) → 4 NO_2( g )+ O_2( g ). The…
$\displaystyle \mathrm{NO}_2$ required for a reaction is produced by the decomposition of $\displaystyle \mathrm{N}_2 \mathrm{O}_5$ in $\displaystyle \mathrm{CCl}_4$ as per the equation,
$$2 \mathrm{~N}_2 \mathrm{O}_5(\mathrm{~g}) \rightarrow 4 \mathrm{NO}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) .
$$
The initial concentration of $\displaystyle \mathrm{N}_2 \mathrm{O}_5$ is $\displaystyle 3.00 \mathrm{~mol} \mathrm{~L}^{-1}$ and it is $\displaystyle 2.75 \mathrm{~mol} \mathrm{~L}^{-1}$ after $\displaystyle 30$ minutes. The rate of formation of $\displaystyle \mathrm{NO}_2$ is :
Official answer
(4)
$\displaystyle 1.667 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~min}^{-1}$
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JEE Main 2019 Chemistry question, with the answer from NTA’s final answer key. Where our answers come from.