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Introduction to Le Chatelier's Principle

For a reactio...

Question

For a reaction, $2 N H_{3} ⟶ N_{2} + 3 H_{2}$ , it is observed that:

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

What is the relation between $k_{1}, k_{2}$ and $k_{3}$ ?

A

k_{1} = k_{2} = k_{3}

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B

3 k_{1} = 6 k_{2} = 2 k_{3}

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C

2 k_{1} = 3 k_{2} = 6 k_{3}

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D

6 k_{1} = 3 k_{2} = 2 k_{3}

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Solution

The correct option is B $3 k_{1} = 6 k_{2} = 2 k_{3}$

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Similar questions

Q. For the reaction,

N_{2} O_{5} ⟶ 2 N O_{2} + 1 / 2 O_{2}

,

- \frac{d [N_{2} O_{5}]}{d t} = k_{1} [N_{2} O_{5}]

\frac{d [N O_{2}]}{d t} = k_{2} [N_{2} O_{5}]

\frac{d [O_{2}]}{d t} = k_{3} [N_{2} O_{5}]

The relation in between

k_{1}, k_{2}

k_{3}

Q. Consider the reaction,

N_{2} (g) + 3 H_{2} (g) ⟶ 2 N H_{3} (g)

. The equality relationship between

\frac{d [N H_{3}]}{d t}

- \frac{d [H_{2}]}{d t}

Q. For the reaction :

N_{2} O_{5} \to 2 N O_{2} + \frac{1}{2} O_{2} \frac{- d [N_{2} O_{5}]}{d t} = K_{1} [N_{2} O_{5}], \frac{d [N O_{2}]}{d t} = K_{2} [N_{2} O_{5}], a n d \frac{d [O_{2}]}{d t} = K_{3} [N_{2} O_{5}]

K_{1}, K_{2}

K_{3}

are related as :

Q. For a reaction,

N_{2} O_{5} ⟶ 2 N O_{2} + \frac{1}{2} O_{2}

\frac{- d [N_{2} O_{5}]}{d t} = k_{1} [N_{2} O_{5}]

\frac{d [N O_{2}]}{d t} = k_{2} [N_{2} O_{5}]

\frac{d [O_{2}]}{d t} = k_{3} [N_{2} O_{5}]

The relation between

k_{1}, k_{2}

k_{3}

Q. For the reaction;

N_{2} O_{5} ⟶ 2 {N O}_{2} + \frac{1}{2} O_{2}

- \frac{d [N_{2} O_{5}]}{d t} = K_{1} [N_{2} O_{5}]

- \frac{d [{N O}_{2}]}{d t} = K_{2} [{N O}_{2}]

- \frac{d [O_{2}]}{d t} = K_{3} [O_{2}]

The relation between

K_{1}, K_{2}, K_{3}

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