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Standard XII

Chemistry

Factors Affecting Rate of Reaction

For the react...

Question

For the reaction $C l_{2} + 2 I^{-} \to I_{2} + 2 C l^{-}$ , the initial concentration of $I^{-}$ was 0.2 mol $L^{- 1}$ and the concentration after 20 min was 0.18 mol $L^{- 1}$ . The rate of formation of l2 in mol $L^{- 1} m i n^{- 1}$ would be

1 \times 10^{- 3}

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5 \times 10^{- 3}

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1 \times 10^{- 4}

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5 \times 10^{- 4}

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Solution

The correct option is D $5 \times 10^{- 4}$
$\frac{d [l_{2}]}{d t} = - \frac{1}{2} \frac{d [I^{-}]}{d t}$
$= \frac{- 1}{2} \times \frac{0.18 - 0.2}{20} = 5 \times 10^{- 4}$

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Similar questions

Q. For the reaction,

C l_{2} + 2 I^{-} \to I_{2} + 2 C l^{-}

, the initial concentration of

I^{-}

was

0.20 m o l l i t^{- 1}

and the concentration after

20

min was

0.18 m o l l i t^{- 1}

. Then, the rate of formation of

I_{2}

m o l l i t^{- 1} m i n^{- 1}

would be:

{C l}_{2} + 2 I^{-} \to I_{2} + 2 {C l}^{-}

The initial concentration of

I^{-}

was

0.25 m o l

L^{- 1}

and the concentration after

10

minutes

0.23 m o l

L^{- 1}

. Calculate the rate of disappearance of

I^{-}

and appearance of

I_{2}

Q. The following reaction was carried out in water,

{C l}_{2} + 2 I^{-} ⟶ I_{2} + 2 {C l}^{-}

.
The initial concentration of

I^{-}

was

0.25

m o l

L^{- 1}

and the concentration after

10

minutes was

0.23

m o l

L^{- 1}

. Calculate the rate of disappearance of

I^{-}

and appearance of

I_{2}

Q. Reaction:

C l_{2} (g) + 2 I^{-} (a q) \to I_{2} (g) + 2 C l^{-} (a q)

Initial concentration of

I^{- 1} = 0.30

mole

l i t r e^{- 1}

Final concentration of

I^{- 1} = 0.25

mole

l i t r e^{- 1}

after

1

minutes. Determine Rate of disappearance of

I^{- 1}

. Rate of formation of

I_{2}

Q. For a reaction

A + 2 B \to C + D,

the following data were obtained
Expt. Initial concentration

(m o l e s l i t r e^{- 1})

Initial rate of formation of D

(m o l e s l i t r e^{- 1} m i n^{- 1})

[A] [B]

0.1 0.1 6.0 \times 10^{- 3}

0.3 0.2 7.2 \times 10^{- 2}

0.3 0.4 2.88 \times 10^{- 1}

0.4 0.1 2.4 \times 10^{- 2}

The correct rate law expression will be:

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For the reaction Cl2+2I–→I2+2Cl–, the initial concentration of I– was 0.2 mol L–1 and the concentration after 20 min was 0.18 mol L–1. The rate of formation of l2 in mol L–1 min–1 would be

The correct option is D 5×10−4d[l2]dt=−12d[I−]dt =−12×0.18−0.220=5×10−4

For the reaction $C l_{2} + 2 I^{-} \to I_{2} + 2 C l^{-}$ , the initial concentration of $I^{-}$ was 0.2 mol $L^{- 1}$ and the concentration after 20 min was 0.18 mol $L^{- 1}$ . The rate of formation of l2 in mol $L^{- 1} m i n^{- 1}$ would be

The correct option is D $5 \times 10^{- 4}$
$\frac{d [l_{2}]}{d t} = - \frac{1}{2} \frac{d [I^{-}]}{d t}$
$= \frac{- 1}{2} \times \frac{0.18 - 0.2}{20} = 5 \times 10^{- 4}$