The decomposition of H2O2 can be followed by titration with KMnO4 and is found to be a first order reaction- The rate constant is 4-5 - 10-2- In an experiment- the initial titrate value was 25 mL- The titrate value will be 5 mL after a lapse of -

The correct option is B log_e5/4.5 × 10^ 2 min As we know, t = 2.303/k log V_0/V_1 = 1/k ln V_0)/V_1 = 1/4.5 × 10^ 2 min^ 1 In 25mL/5 ...

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Standard XII

Chemistry

Rate Constant

The decomposi...

Question

The decomposition of $H_{2} O_{2}$ can be followed by titration with $K M n O_{4}$ and is found to be a first order reaction. The rate constant is $4.5 \times 10^{- 2}$ . In an experiment, the initial titrate value was 25 mL. The titrate value will be 5 mL after a lapse of :

4.5 \times 10^{- 2} \times 5 m i n

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\frac{l o g_{e} 5}{4.5 \times 10^{- 2}} m i n

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\frac{l o g_{e} 5 / 4}{4.5 \times 10^{- 2}} m i n

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None of the above

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Solution

The correct option is B $\frac{l o g_{e} 5}{4.5 \times 10^{- 2}} m i n$
As we know,
$t = \frac{2.303}{k} l o g \frac{V_{0}}{V_{1}}$
$= \frac{1}{k} l n \frac{V_{0})}{V_{1}}$
$= \frac{1}{4.5 \times 10^{- 2} m i n^{- 1}} I n \frac{25 m L}{5 m L}$
$= \frac{l o g_{e} 5}{4.5 \times 10^{- 2}} m i n$

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Similar questions

Q. Decomposition of

H_{2} O_{2}

is a first order reaction. A solution of

H_{2} O_{2}

labelled as 20 volumes was left open. Due to this, some

H_{2} O_{2}

decomposed. To determine the new volume strength after 6 hours, 10 mL of this solution was diluted to 100 mL. 10 mL of this diluted solution was titrated against 25 mL of 0.025 N

K M n O_{4}

solution under acidic conditions. The rate constant for decomposition of

H_{2} O_{2}

is:

Q. In a hypothetical reaction,

A (a q) ⇌ 2 B (q) + C (a q) (1^{s t}

order decomposition)

^{'} A^{'}

is optically active (dextro-rototory) while

^{'} B^{'}

and

^{'} C^{'}

are optically inactive but

^{'} B^{'}

takes part in a titration reaction (fast reaction) with

H_{2} O_{2}

. Hence, the progress of reaction can be monitored by measuring rotation of plane polarised light or by measuring volume of

H_{2} O_{2}

consumed in titration.
In an experiment the optical rotation was found to be

θ = 40^{\circ}

t = 20 m i n

and

θ = 10^{\circ}

t = 50 m i n

. from start of the reaction. If the progress would have been monitored by titration method, volume of

H_{2} O_{2}

consumed at

t = 15 m i n

. (from start) is

40 m l

then volume of

H_{2} O_{2}

consumed at

t = 60

min will be:

Q. The rate constant for an isomerisation reaction,

A \to B

4.5 \times 10^{- 3} m i n^{- 1}

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Q. Decomposition of

H_{2} O_{2}

is a first order reaction. A solution of

H_{2} O_{2}

labelled as

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volumes was left open. Due to this some

H_{2} O_{2}

decomposed. To determine the new volume strength after

6

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10 m L

of this solution was diluted to

100 m L . 10 m L

of this diluted solution was titrated against

25 m L

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acidified solution. Calculate the rate constant for decomposition of

H_{2} O_{2}

Q. A student performed a single titration using

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The decomposition of H2O2 can be followed by titration with KMnO4 and is found to be a first order reaction. The rate constant is 4.5×10−2. In an experiment, the initial titrate value was 25 mL. The titrate value will be 5 mL after a lapse of :

The correct option is B loge54.5×10−2minAs we know,t=2.303klogV0V1=1klnV0)V1=14.5×10−2min−1In25mL5mL=loge54.5×10−2min

The decomposition of $H_{2} O_{2}$ can be followed by titration with $K M n O_{4}$ and is found to be a first order reaction. The rate constant is $4.5 \times 10^{- 2}$ . In an experiment, the initial titrate value was 25 mL. The titrate value will be 5 mL after a lapse of :

The correct option is B $\frac{l o g_{e} 5}{4.5 \times 10^{- 2}} m i n$
As we know,
$t = \frac{2.303}{k} l o g \frac{V_{0}}{V_{1}}$
$= \frac{1}{k} l n \frac{V_{0})}{V_{1}}$
$= \frac{1}{4.5 \times 10^{- 2} m i n^{- 1}} I n \frac{25 m L}{5 m L}$
$= \frac{l o g_{e} 5}{4.5 \times 10^{- 2}} m i n$