Go To Your Study Dashboard

Get Answers to all your Questions

header-bg

$NO_{2}$ required for a reaction is produced by the decomposition of $N_{2}O_{5}$ in $CCl_{4}$ as per the equation, $2N_{2}O_{5}(g)\rightarrow 4NO_{2}(g)+O_{2}(g)$ .

The initial concentration of $N_{2}O_{5}$ is $3.00 mol L^{-1}$ and it is $2.75 mol L^{-1}$ after 30 minutes. The rate of formation of $NO_{2}$ is:

Option 1)
$4.167\times 10^{-3}molL^{-1}min^{-1}$

Option 2)
$1.667\times 10^{-2}molL^{-1}min^{-1}$

Option 3)
$8.333\times 10^{-3}molL^{-1}min^{-1}$

Option 4)
$2.083\times 10^{-3}molL^{-1}min^{-1}$

Answers (1)

best_answer

( $1.667\times 10^{-2}molL^{-1}min^{-1}$ )

$2N_{2}O_{5}(g)\rightarrow 4NO_{2}(g)+O_{2}(g)$

$t=0$ $3.0 M$

$t=30$ $2.75 M$

so, $\frac{-\Delta [N_{2}O_{5}]}{\Delta t}=\frac{-[2.75-3.0]}{30-0}=\frac{0.25}{30}$

from the reaction

$-\frac{1}{2}\frac{\Delta [N_{2}O_{5}]}{\Delta t}=\frac{1}{4}\frac{\Delta [NO_{2}]}{\Delta t}=\frac{\Delta [O_{2}]}{\Delta t}$

$\frac{\Delta [NO_{2}]}{\Delta t}=\frac{0.25}{30}\times 2=1.66\times 10^{-2}M/min$

$=1.66\times 10^{-2}molL^{-1}min^{-1}$

Option 1)

$4.167\times 10^{-3}molL^{-1}min^{-1}$

Option 2)

$1.667\times 10^{-2}molL^{-1}min^{-1}$

Option 3)

$8.333\times 10^{-3}molL^{-1}min^{-1}$

Option 4)

$2.083\times 10^{-3}molL^{-1}min^{-1}$

Posted by

solutionqc

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2025 paper 2 final answer key out; NTA drops 4 questions, 2 from BPlanning

anam-khan

JEE Main 2025 session 2 answer key out for BArch, BPlanning; response sheet on jeemain.nta.nic.in

anam-khan

JEE Mains 2025 Paper 2 Answer Key LIVE: BArch final answer key PDF expected soon at jeemain.nta.nic.in

Latest Question