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Chemical Kinetics question

2011 · Q64
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Chemical Kinetics question

2011 · Q64

NEETChemistryChemical KineticsMCQ+4 / −1
The rate of the reaction :   2N2O5 →\to→ 4NO2 + O2
can be written in three ways.

−d[N2O5]dt=k[N2O5]{{ - d\left[ {{N_2}{O_5}} \right]} \over {dt}} = k\left[ {{N_2}{O_5}} \right]dt−d[N2​O5​]​=k[N2​O5​]

d[NO2]dt=k′[N2O5];  {{d\left[ {N{O_2}} \right]} \over {dt}} = k'\left[ {{N_2}{O_5}} \right];\,\,dtd[NO2​]​=k′[N2​O5​]; d[O2]dt=k′′[N2O5]{{d\left[ {{O_2}} \right]} \over {dt}} = k''\left[ {{N_2}{O_5}} \right]dtd[O2​]​=k′′[N2​O5​]

The relationship between k and k' and between k and k'' are
  1. A
    k′=2k;k′′=kk' = 2k ; k'' = kk′=2k;k′′=k
  2. B
    k′=2k;k′′=k/2k' = 2k ; k'' = k/2k′=2k;k′′=k/2
  3. C
    k′=2k;k′′=2kk' = 2k ; k'' = 2kk′=2k;k′′=2k
  4. D
    k′=k;k′′=kk' = k ; k'' = kk′=k;k′′=k
View written solutionFree

Correct answer: B

Rate of disappearance of reactants = Rate of appearance of products

- {1 \over 2}$$${{ - d\left[ {{N_2}{O_5}} \right]} \over {dt}}$$ ={1 \over 4}{{d\left[ {N{O_2}} \right]} \over {dt}}$$ = $${{d\left[ {{O_2}} \right]} \over {dt}}$$ <br><br>$ \Rightarrow $ ${1 \over 2}k\left[ {{N_2}{O_5}} \right] = ${1 \over 4}$$$k'\left[ {{N_2}{O_5}} \right] = k′′[N2O5]k''\left[ {{N_2}{O_5}} \right]k′′[N2​O5​]

⇒\Rightarrow⇒ k2=k′4=k′′{k \over 2} = {{k'} \over 4} = k''2k​=4k′​=k′′

⇒\Rightarrow⇒ k′=2k;k′′=k/2k' = 2k ; k'' = k/2k′=2k;k′′=k/2

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