QBManager

CHXII04:CHEMICAL KINETICS

320313 A first order reaction is $50 %$ completed in $1.26 \times 10^{14} s$ . How much time would it take for $100 %$ completion?

1

1.26 \times 10^{15} s

2

2.52 \times 10^{14} s

3

2.52 \times 10^{28} s

4 Infinite

Explanation:

It is impossible to perform $100 %$ of the reaction. Whole of the substance never reacts because in every half-life, $50 %$ of the substance reacts. Hence, time taken for $100 %$ completion of a reaction is infinite.

NCERT Exemplar

CHXII04:CHEMICAL KINETICS

320314 Decomposition of $H_{2} O_{2}$ follows a first order reaction. In fifty minutes the concentration of $H_{2} O_{2}$ decreases from 0.5 to $0.125 M$ in one such decomposition. When the concentration of $H_{2} O_{2}$ reaches $0.05 M$ , the rate of formation of $O_{2}$ will be :

1

60.5 kJ {mol}^{- 1}

2

2.66 L \min^{- 1}

at

STP

3

1.34 \times 10^{- 2} mol \min^{- 1}

4

6.93 \times 10^{- 4} mol \min^{- 1}

Explanation:

$H_{2} O_{2} \to H_{2} O + \frac{1}{2} O_{2}$
$\begin{aligned} 2 t_{1 / 2} = 50 \min \\ t_{1 / 2} = 25 \min \\ ∴ k = \frac{0.693}{25} \\ - \frac{d [H_{2} O_{2}]}{d t} = \frac{d [O_{2}]}{d t} \times 2 \\ ∴ \frac{d [O_{2}]}{d t} = \frac{1}{2} \times \frac{0.693}{25} \times 0.05 \\ = 6.93 \times 10^{- 4} \end{aligned}$

CHXII04:CHEMICAL KINETICS

320315 The value of rate constant for a first order is $2.303 \times 10^{- 2} \sec^{- 1}$ . What will be the time required to reduce the concentration to $\frac{1}{10}$ th of the initial concentration?

1

100 \sec

2

10 \sec

3

2303 \sec

4

23.03 \sec

Explanation:

$t = \frac{2.303}{k} \log \frac{a}{a - x}$
$t = \frac{2.303}{2.303 \times 10^{- 2}} \log \frac{x}{\frac{x}{10}} = 100 \log 10 = 100 \sec$

CHXII04:CHEMICAL KINETICS

320316 For a reaction,
$X(g) \to Y(g) + Z(g)$
the half-life period is $10 \min$ . In what period of time would the concentration of $X$ be reduced to $10 %$ of original concentration?

1

20 \min

2

33 \min

3

15 \min

4

25 \min

Explanation:

$X(g) \to Y(g) + Z(g)$
The reaction is a first order reaction.
Hence,
$k = \frac{0.693}{t_{1/2}} = \frac{2 .303}{t} \log \frac{a}{(a - x)}$
$\Rightarrow \frac{0.693}{10} = \frac{2 .303}{t} \log \frac{a}{a / 10}$
$or t = \frac{2.303 \times 10}{0.693} \times \log 10 = 33 min$
$[∵ \log 10 = 1]$

CHXII04:CHEMICAL KINETICS

320313 A first order reaction is $50 %$ completed in $1.26 \times 10^{14} s$ . How much time would it take for $100 %$ completion?

1

1.26 \times 10^{15} s

2

2.52 \times 10^{14} s

3

2.52 \times 10^{28} s

4 Infinite

Explanation:

It is impossible to perform $100 %$ of the reaction. Whole of the substance never reacts because in every half-life, $50 %$ of the substance reacts. Hence, time taken for $100 %$ completion of a reaction is infinite.

NCERT Exemplar

CHXII04:CHEMICAL KINETICS

320314 Decomposition of $H_{2} O_{2}$ follows a first order reaction. In fifty minutes the concentration of $H_{2} O_{2}$ decreases from 0.5 to $0.125 M$ in one such decomposition. When the concentration of $H_{2} O_{2}$ reaches $0.05 M$ , the rate of formation of $O_{2}$ will be :

1

60.5 kJ {mol}^{- 1}

2

2.66 L \min^{- 1}

at

STP

3

1.34 \times 10^{- 2} mol \min^{- 1}

4

6.93 \times 10^{- 4} mol \min^{- 1}

Explanation:

$H_{2} O_{2} \to H_{2} O + \frac{1}{2} O_{2}$
$\begin{aligned} 2 t_{1 / 2} = 50 \min \\ t_{1 / 2} = 25 \min \\ ∴ k = \frac{0.693}{25} \\ - \frac{d [H_{2} O_{2}]}{d t} = \frac{d [O_{2}]}{d t} \times 2 \\ ∴ \frac{d [O_{2}]}{d t} = \frac{1}{2} \times \frac{0.693}{25} \times 0.05 \\ = 6.93 \times 10^{- 4} \end{aligned}$

CHXII04:CHEMICAL KINETICS

320315 The value of rate constant for a first order is $2.303 \times 10^{- 2} \sec^{- 1}$ . What will be the time required to reduce the concentration to $\frac{1}{10}$ th of the initial concentration?

1

100 \sec

2

10 \sec

3

2303 \sec

4

23.03 \sec

Explanation:

$t = \frac{2.303}{k} \log \frac{a}{a - x}$
$t = \frac{2.303}{2.303 \times 10^{- 2}} \log \frac{x}{\frac{x}{10}} = 100 \log 10 = 100 \sec$

CHXII04:CHEMICAL KINETICS

320316 For a reaction,
$X(g) \to Y(g) + Z(g)$
the half-life period is $10 \min$ . In what period of time would the concentration of $X$ be reduced to $10 %$ of original concentration?

1

20 \min

2

33 \min

3

15 \min

4

25 \min

Explanation:

$X(g) \to Y(g) + Z(g)$
The reaction is a first order reaction.
Hence,
$k = \frac{0.693}{t_{1/2}} = \frac{2 .303}{t} \log \frac{a}{(a - x)}$
$\Rightarrow \frac{0.693}{10} = \frac{2 .303}{t} \log \frac{a}{a / 10}$
$or t = \frac{2.303 \times 10}{0.693} \times \log 10 = 33 min$
$[∵ \log 10 = 1]$

CHXII04:CHEMICAL KINETICS

320313 A first order reaction is $50 %$ completed in $1.26 \times 10^{14} s$ . How much time would it take for $100 %$ completion?

1

1.26 \times 10^{15} s

2

2.52 \times 10^{14} s

3

2.52 \times 10^{28} s

4 Infinite

Explanation:

It is impossible to perform $100 %$ of the reaction. Whole of the substance never reacts because in every half-life, $50 %$ of the substance reacts. Hence, time taken for $100 %$ completion of a reaction is infinite.

NCERT Exemplar

CHXII04:CHEMICAL KINETICS

320314 Decomposition of $H_{2} O_{2}$ follows a first order reaction. In fifty minutes the concentration of $H_{2} O_{2}$ decreases from 0.5 to $0.125 M$ in one such decomposition. When the concentration of $H_{2} O_{2}$ reaches $0.05 M$ , the rate of formation of $O_{2}$ will be :

1

60.5 kJ {mol}^{- 1}

2

2.66 L \min^{- 1}

at

STP

3

1.34 \times 10^{- 2} mol \min^{- 1}

4

6.93 \times 10^{- 4} mol \min^{- 1}

Explanation:

$H_{2} O_{2} \to H_{2} O + \frac{1}{2} O_{2}$
$\begin{aligned} 2 t_{1 / 2} = 50 \min \\ t_{1 / 2} = 25 \min \\ ∴ k = \frac{0.693}{25} \\ - \frac{d [H_{2} O_{2}]}{d t} = \frac{d [O_{2}]}{d t} \times 2 \\ ∴ \frac{d [O_{2}]}{d t} = \frac{1}{2} \times \frac{0.693}{25} \times 0.05 \\ = 6.93 \times 10^{- 4} \end{aligned}$

CHXII04:CHEMICAL KINETICS

320315 The value of rate constant for a first order is $2.303 \times 10^{- 2} \sec^{- 1}$ . What will be the time required to reduce the concentration to $\frac{1}{10}$ th of the initial concentration?

1

100 \sec

2

10 \sec

3

2303 \sec

4

23.03 \sec

Explanation:

$t = \frac{2.303}{k} \log \frac{a}{a - x}$
$t = \frac{2.303}{2.303 \times 10^{- 2}} \log \frac{x}{\frac{x}{10}} = 100 \log 10 = 100 \sec$

CHXII04:CHEMICAL KINETICS

320316 For a reaction,
$X(g) \to Y(g) + Z(g)$
the half-life period is $10 \min$ . In what period of time would the concentration of $X$ be reduced to $10 %$ of original concentration?

1

20 \min

2

33 \min

3

15 \min

4

25 \min

Explanation:

$X(g) \to Y(g) + Z(g)$
The reaction is a first order reaction.
Hence,
$k = \frac{0.693}{t_{1/2}} = \frac{2 .303}{t} \log \frac{a}{(a - x)}$
$\Rightarrow \frac{0.693}{10} = \frac{2 .303}{t} \log \frac{a}{a / 10}$
$or t = \frac{2.303 \times 10}{0.693} \times \log 10 = 33 min$
$[∵ \log 10 = 1]$

CHXII04:CHEMICAL KINETICS

320313 A first order reaction is $50 %$ completed in $1.26 \times 10^{14} s$ . How much time would it take for $100 %$ completion?

1

1.26 \times 10^{15} s

2

2.52 \times 10^{14} s

3

2.52 \times 10^{28} s

4 Infinite

Explanation:

It is impossible to perform $100 %$ of the reaction. Whole of the substance never reacts because in every half-life, $50 %$ of the substance reacts. Hence, time taken for $100 %$ completion of a reaction is infinite.

NCERT Exemplar

CHXII04:CHEMICAL KINETICS

320314 Decomposition of $H_{2} O_{2}$ follows a first order reaction. In fifty minutes the concentration of $H_{2} O_{2}$ decreases from 0.5 to $0.125 M$ in one such decomposition. When the concentration of $H_{2} O_{2}$ reaches $0.05 M$ , the rate of formation of $O_{2}$ will be :

1

60.5 kJ {mol}^{- 1}

2

2.66 L \min^{- 1}

at

STP

3

1.34 \times 10^{- 2} mol \min^{- 1}

4

6.93 \times 10^{- 4} mol \min^{- 1}

Explanation:

$H_{2} O_{2} \to H_{2} O + \frac{1}{2} O_{2}$
$\begin{aligned} 2 t_{1 / 2} = 50 \min \\ t_{1 / 2} = 25 \min \\ ∴ k = \frac{0.693}{25} \\ - \frac{d [H_{2} O_{2}]}{d t} = \frac{d [O_{2}]}{d t} \times 2 \\ ∴ \frac{d [O_{2}]}{d t} = \frac{1}{2} \times \frac{0.693}{25} \times 0.05 \\ = 6.93 \times 10^{- 4} \end{aligned}$

CHXII04:CHEMICAL KINETICS

320315 The value of rate constant for a first order is $2.303 \times 10^{- 2} \sec^{- 1}$ . What will be the time required to reduce the concentration to $\frac{1}{10}$ th of the initial concentration?

1

100 \sec

2

10 \sec

3

2303 \sec

4

23.03 \sec

Explanation:

$t = \frac{2.303}{k} \log \frac{a}{a - x}$
$t = \frac{2.303}{2.303 \times 10^{- 2}} \log \frac{x}{\frac{x}{10}} = 100 \log 10 = 100 \sec$

CHXII04:CHEMICAL KINETICS

320316 For a reaction,
$X(g) \to Y(g) + Z(g)$
the half-life period is $10 \min$ . In what period of time would the concentration of $X$ be reduced to $10 %$ of original concentration?

1

20 \min

2

33 \min

3

15 \min

4

25 \min

Explanation:

$X(g) \to Y(g) + Z(g)$
The reaction is a first order reaction.
Hence,
$k = \frac{0.693}{t_{1/2}} = \frac{2 .303}{t} \log \frac{a}{(a - x)}$
$\Rightarrow \frac{0.693}{10} = \frac{2 .303}{t} \log \frac{a}{a / 10}$
$or t = \frac{2.303 \times 10}{0.693} \times \log 10 = 33 min$
$[∵ \log 10 = 1]$