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Chemical Kinetics question

2010 · Q106
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Chemical Kinetics question

2010 · Q106

NEETChemistryChemical KineticsMCQ+4 / −1
For the reaction N2O5(g) →\to→  2NO2(g) + 1/2O2(g)
the value of rate of disappearance of N2O5 is given as 6.25 ×\times× 10−-−3 mol L−-−1 s−-−1. The rate of formation of NO2 and O2 is given respectively as
  1. A
    6.25 ×\times× 10−-−3 mol L−-−1 s−-−1 and
    6.25 ×\times× 10−-−3 mol L−-−1 s−-−1
  2. B
    1.25 ×\times× 10−-−2 mol L−-−1 s−-−1 and
    3.125 ×\times× 10−-−3 mol L−-−1 s−-−1
  3. C
    6.25 ×\times× 10−-−3 mol L−-−1 s−-−1 and
    3.125 ×\times× 10−-−3 mol L−-−1 s−-−1
  4. D
    1.25 ×\times× 10−-−2 mol L−-−1 s−-−1 and
    6.25 ×\times× 10−-−3 mol L−-−1 s−-−1
View written solutionFree

Correct answer: B

N2O5(g) →\to→  2NO2(g) + 1/2O2(g)

−d[N2O5]dt=12d[NO2]dt=2d[O2]dt - {{d\left[ {{N_2}{O_5}} \right]} \over {dt}} = {1 \over 2}{{d\left[ {N{O_2}} \right]} \over {dt}} = 2{{d\left[ {{O_2}} \right]} \over {dt}}−dtd[N2​O5​]​=21​dtd[NO2​]​=2dtd[O2​]​

⇒\Rightarrow⇒ d[NO2]dt=−2d[N2O5]dt{{d\left[ {N{O_2}} \right]} \over {dt}} = - 2{{d\left[ {{N_2}{O_5}} \right]} \over {dt}}dtd[NO2​]​=−2dtd[N2​O5​]​

= 2 ×\times× 6.25 ×\times× 10 mol l-1 sec-1

= 1.25 ×\times× 10−-−2 mol L−-−1 s−-−1

d[O2]dt=−12d[N2O5]dt{{d\left[ {{O_2}} \right]} \over {dt}} = - {1 \over 2}{{d\left[ {{N_2}{O_5}} \right]} \over {dt}}dtd[O2​]​=−21​dtd[N2​O5​]​

= 6.25×10−32{{6.25 \times {{10}^{ - 3}}} \over 2}26.25×10−3​

= 3.125 ×\times× 10−-−3 mol L−-−1 s−-−1

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