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Standard XII

Chemistry

Introduction to Le Chatelier's Principle

For the react...

Question

For the reaction $: H_{2} + I_{2} ⟶ 2 H I$ , the differential rate law is:

- \frac{d [H_{2}]}{d t} = - \frac{d [I_{2}]}{d t} = 2 \frac{d [H I]}{d t}

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- \frac{d [H_{2}]}{d t} = - \frac{d [I_{2}]}{d t} = \frac{1}{2} \frac{d [H I]}{d t}

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- \frac{d [H_{2}]}{d t} = - \frac{d [I_{2}]}{d t} = \frac{d [H I]}{d t}

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- \frac{d [H_{2}]}{d t} = - 2 \frac{d [I_{2}]}{d t} = \frac{d [H I]}{d t}

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Solution

The correct option is B $- \frac{d [H_{2}]}{d t} = - \frac{d [I_{2}]}{d t} = \frac{1}{2} \frac{d [H I]}{d t}$
$H_{2} + I_{2} ⟶ 2 H I$
Rate of reaction $= \frac{- d [H_{2}]}{d t} = \frac{- d [I_{2}]}{d t} = \frac{1}{2} \frac{d [H I]}{d t}$
or $\frac{- 2 d [H_{2}]}{d t} = - 2 \frac{d [I_{2}]}{d t} = \frac{d [H I]}{d t}$

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Similar questions

Q. For the reaction

I_{2} + 2 S_{2} O_{3}^{2 -} \to S_{4} O_{6}^{2 -} + 2 I^{⊖}

\frac{- d [I_{2}]}{d t} = - \frac{1}{2} \frac{d [S_{2} O_{3}^{2 -}]}{d t}

II.

\frac{- d [I_{2}]}{d t} = - 2 \frac{d [S_{2} O_{3}^{2 -}]}{d t}

III.

\frac{- d [I_{2}]}{d t} = - 2 \frac{d [I^{⊖}]}{d t} \frac{d [S_{2} O_{3}^{2 -}]}{d t}

IV.

\frac{d [S_{4} O_{6}^{2 -}]}{d t} = \frac{1}{2} \frac{d [I^{⊖}]}{d t}

The correct option is:

Q. For the chemical reaction

N_{2}

(g) +

3 H_{2}

(g)

⇌

2 N H_{3}

(g)
The correct option is:

Consider the chemical reaction,
$N_{2} (g) + 3 H_{2} (g) \to 2 N H_{3} (g)$
The rate of this reaction can be expressed in terms of time derivatives of concentration of $N_{2} (g), H_{2} (g) o r N H_{3} (g)$ . Identify the correct relationship amongst the rate expressions.

Consider the chemical reaction

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

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Q. For the reaction,

2 N O (g) + 2 H_{2} (g) \to N_{2} (g) + 2 H_{2} O (g)

the rate expression can be written in the following ways:

{d t [N_{2}] / d t} = k_{1} [N O] [H_{2}]; {d [H_{2} O] / d t) = k [N O] [H_{2}];

{- d [N O] / d t} = k_{1}^{'} [N O] [H_{2}]; {- d [H_{2}] / d t} = k_{1}^{''} [N O] [H_{2}]

The relation ship between

k, k_{1}, k_{1}^{'}

and

k_{1}^{''}

is:

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For the reaction :H2+I2⟶2HI, the differential rate law is:

The correct option is B −d[H2]dt=−d[I2]dt=12d[HI]dtH2+I2⟶2HIRate of reaction =−d[H2]dt=−d[I2]dt=12d[HI]dtor −2d[H2]dt=−2d[I2]dt=d[HI]dt

For the reaction $: H_{2} + I_{2} ⟶ 2 H I$ , the differential rate law is: