1. Tardigrade
  2. Question
  3. Chemistry
  4. The rate for the reaction 2N2O5 (. g .) arrow 4(NO)2 (. g .)+O2 (. g .) is 2.4× 10- 5mollit- 1sec- 1. If the rate is 3.0× 10- 5 sec- 1 then the concentration of N2O5 in mollit- 1 is: (Fill your anwer by multiply with 10)

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Q. The rate for the reaction $2N_{2}O_{5 \left(\right. g \left.\right)} \rightarrow 4\left(NO\right)_{2 \left(\right. g \left.\right)}+O_{2 \left(\right. g \left.\right)}$ is $2.4\times 10^{- 5}mollit^{- 1}sec^{- 1}.$ If the rate is $3.0\times 10^{- 5}$ $sec^{- 1}$ then the concentration of $N_{2}O_{5}$ in $mollit^{- 1}$ is:
(Fill your anwer by multiply with 10)

NTA AbhyasNTA Abhyas 2022

Solution:

Rate $=k\left[N_{2} O_{5}\right]$
Hence $2.4\times 10^{- 5}=3.0\times 10^{- 5}IN_{2}O_{5}l$
or $\left[N_{2} O_{5}\right]=0.8molL^{- 1}$