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Standard XII

Chemistry

Rate of Reaction

In a reaction...

Question

In a reaction of acidified hydrogen peroxide with potassium iodide, the concentration of iodine formed rises from $0$ to $10^{- 5}$ $m o l$ ${d m}^{- 3}$ in $10$ seconds. What is the rate of reaction?

10^{- 6}

m o l

{d m}^{- 3}

s^{- 1}

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10^{6}

m o l

{d m}^{- 3}

s^{- 1}

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10^{- 5}

m o l

{d m}^{- 3}

s^{- 1}

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10^{4}

m o l

{d m}^{- 3}

s^{- 1}

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Solution

The correct option is A $10^{- 6}$ $m o l$ ${d m}^{- 3}$ $s^{- 1}$
The reaction is
$2 I^{-} + H_{2} O_{2} + 2 H^{+} ⟶ I_{2} + 2 H_{2} O$
Rate $= + \frac{d [I_{2}]}{d t} = \frac{10^{- 5}}{10} = 10^{- 6}$ $m o l$ ${d m}^{- 3}$ $s^{- 1}$

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Similar questions

\begin{matrix} Experiment No. & [A] (m o l d m^{- 3}) & [B] (m o l d m^{- 3}) & [C] (m o l d m^{- 3}) & Rate of the reaction (m o l d m^{- 3} s^{- 1}) 1 & 0.2 & 0.1 & 0.1 & 6.0 \times 10^{- 5} 2 & 0.2 & 0.2 & 0.1 & 6.0 \times 10^{- 5} 3 & 0.2 & 0.1 & 0.2 & 1.2 \times 10^{- 4} 4 & 0.3 & 0.1 & 0.1 & 9.0 \times 10^{- 5} \end{matrix}

The rate of the reaction for

[A] = 0.15 m o l d m^{- 3}, [B] = 0.25 m o l d m^{- 3} a n d [C] = 0.15 m o l d m^{- 3}

is found to be

Y \times 10^{- 5} m o l d m^{- 3} s^{- 1}

.The value of Y is

In a first order reaction the concentration of reactant decreases from $800 m o l / d m^{3}$ to $50 m o l / d m^{3}$ in $2 \times 10^{4} s$ . The rate constant of reaction in $s^{- 1}$ is:

In a first order reaction, the concentration of the reactant decreases from 800 mol $d m^{- 3}$ to 50 mol $d m^{- 3}$ in $2 \times 10^{4} s .$ The rate constant of the reaction in $(s^{- 1})$ is
(IIT-JEE, 2003)

\begin{matrix} Experiment No. & [A] (m o l d m^{- 3}) & [B] (m o l d m^{- 3}) & [C] (m o l d m^{- 3}) & Rate of the reaction (m o l d m^{- 3} s^{- 1}) 1 & 0.2 & 0.1 & 0.1 & 6.0 \times 10^{- 5} 2 & 0.2 & 0.2 & 0.1 & 6.0 \times 10^{- 5} 3 & 0.2 & 0.1 & 0.2 & 1.2 \times 10^{- 4} 4 & 0.3 & 0.1 & 0.1 & 9.0 \times 10^{- 5} \end{matrix}

The rate of the reaction for

[A] = 0.15 m o l d m^{- 3}, [B] = 0.25 m o l d m^{- 3} a n d [C] = 0.15 m o l d m^{- 3}

is found to be

Y \times 10^{- 5} m o l d m^{- 3} s^{- 1}

.The value of Y is

In a first order reaction the concentration of reactant decreases from $800 m o l / d m^{3}$ to $50 m o l / d m^{3}$ in $2 \times 10^{4} s$ . The rate constant of reaction in $s^{- 1}$ is:

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In a reaction of acidified hydrogen peroxide with potassium iodide, the concentration of iodine formed rises from 0 to 10−5 mol dm−3 in 10 seconds. What is the rate of reaction?

The correct option is A 10−6 mol dm−3 s−1The reaction is2I−+H2O2+2H+⟶I2+2H2ORate =+d[I2]dt=10−510=10−6 mol dm−3 s−1

In a reaction of acidified hydrogen peroxide with potassium iodide, the concentration of iodine formed rises from $0$ to $10^{- 5}$ $m o l$ ${d m}^{- 3}$ in $10$ seconds. What is the rate of reaction?

The correct option is A $10^{- 6}$ $m o l$ ${d m}^{- 3}$ $s^{- 1}$
The reaction is
$2 I^{-} + H_{2} O_{2} + 2 H^{+} ⟶ I_{2} + 2 H_{2} O$
Rate $= + \frac{d [I_{2}]}{d t} = \frac{10^{- 5}}{10} = 10^{- 6}$ $m o l$ ${d m}^{- 3}$ $s^{- 1}$