SATHEE: Chemical Kinetics - Result Question 10

Chemical Kinetics - Result Question 10

####10. For the reaction, $N_{2} + 3 H_{2} ⟶ 2 N H_{3}$ ,

[2009]

$\frac{d [N H_{3}]}{d t} = 2 \times 10^{- 4} m o l L^{- 1} s^{- 1}$ , the value of

$\frac{- d [H_{2}]}{d t}$ would be:

(a) $4 \times 10^{- 4} m o l L^{- 1} s^{- 1}$

(b) $6 \times 10^{- 4} m o l L^{- 1} s^{- 1}$

(d) $3 \times 10^{- 4} m o l L^{- 1} s^{- 1}$

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Solution:

(d) Rate of disappearance of $H_{2} =$ rate of formation of $N H_{3}$ .

$- \frac{1}{3} \frac{d [H_{2}]}{d t} = \frac{1}{2} \frac{d [N H_{3}]}{d t}$

$\Rightarrow \frac{- d [H_{2}]}{d t} = \frac{3}{2} \frac{d [N H_{3}]}{d t} = \frac{3}{2} \times 2 \times 10^{- 4}$

$= 3 \times 10^{- 4} m o l L^{- 1} s^{- 1}$

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Chemical Kinetics - Result Question 10

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