SATHEE: Chemical Kinetics - Result Question 8

Chemical Kinetics - Result Question 8

####8. For the reaction $[N_{2} O_{5} (g) \to 2 N O_{2} (g) + .$ $.1 / 2 O_{2} (g)]$ the value of rate of disappearance of $N_{2} O_{5}$ is given as $6.25 \times 10^{- 3} m o l L^{- 1} S^{- 1}$ . The rate of formation of $N O_{2}$ and $O_{2}$ is given respectively as :

[2010]

(a) $6.25 \times 10^{- 3} m o l L^{- 1} s^{- 1}$ and $6.25 \times 10^{- 3}$ $m o l L^{- 1} s^{- 1}$

(b) $1.25 \times 10^{- 2} m o l L^{- 1} s^{- 1}$ and $3.125 \times 10^{- 3}$ $m o l L L^{- 1} s^{- 1}$

(d) $1.25 \times 10^{- 2} m o l L^{- 1} s^{- 1}$ and $6.25 \times 10^{- 3}$ $m o l L^{- 1} s^{- 1}$

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Solution:

(b) $N_{2} O_{5} (g) ⟶ 2 N O_{2} (g) + 1 / 2 O_{2} (g)$

$- \frac{d}{d t} [N_{2} O_{5}] = + \frac{1}{2} \frac{d}{d t} [N O_{2}] = 2 \frac{d}{d t} [O_{2}]$

$\frac{d}{d t} [N O_{2}] = 1.25 \times 10^{- 2} m o l L^{- 1} s^{- 1}$ and

$\frac{d}{d t} [O_{2}] = 3.125 \times 10^{- 3} m o l L^{- 1} s^{- 1}$

Chemical Kinetics - Result Question 8

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Chemical Kinetics - Result Question 8

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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