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Byju's Answer

Standard XII

Chemistry

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 $0.20 m o l l i t^{- 1}$ and the concentration after $20$ min was $0.18 m o l l i t^{- 1}$ . Then, the rate of formation of $I_{2}$ in $m o l l i t^{- 1} m i n^{- 1}$ would be:

1 \times 10^{- 3}

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5 \times 10^{- 4}

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1 \times 10^{- 4}

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5 \times 10^{- 3}

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Solution

The correct option is A $1 \times 10^{- 3}$
The rate of formation of $I_{2}$ is

$R a t e = \frac{[A_{0}] - [A]}{t}$

$R a t e = \frac{[0.20] - [0.18]}{20}$

$R a t e = 1 \times 10^{- 3} m o l {l i t}^{- 1} {m i n}^{- 1}$

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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 0.2 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

{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 0.20mollit−1 and the concentration after 20 min was 0.18mollit−1. Then, the rate of formation of I2 in mollit−1min−1 would be:

The correct option is A 1×10−3The rate of formation of I2 isRate=[A0]−[A]tRate=[0.20]−[0.18]20Rate=1×10−3 mol lit−1min−1

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 $I_{2}$ in $m o l l i t^{- 1} m i n^{- 1}$ would be:

The correct option is A $1 \times 10^{- 3}$
The rate of formation of $I_{2}$ is

$R a t e = \frac{[A_{0}] - [A]}{t}$

$R a t e = \frac{[0.20] - [0.18]}{20}$

$R a t e = 1 \times 10^{- 3} m o l {l i t}^{- 1} {m i n}^{- 1}$