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CHXII04:CHEMICAL KINETICS

320317 A radioactive isotope having a half-life period of 3 days was received after 12 days. If $3 g$ of the isotope is left in the container, what would be the initial mass of the isotope?

1

12 g

2

36 g

3

48 g

4

24 g

Explanation:

Given $t_{1 / 2} = 3$
Total time, $T = 12$
No. of half lives $(n) = \frac{12}{3} = 4$
$\begin{aligned} {(\frac{1}{2})}^{n} = \frac{N}{N_{o}} \\ ∴ {(\frac{1}{2})}^{4} = \frac{3}{N} \end{aligned}$ $\frac{3}{N} = \frac{1}{16}, N = 48 g$

JEE - 2013

CHXII04:CHEMICAL KINETICS

320318 If $60 %$ of a first-order reaction was completed in $60 \min, 50 %$ of the same reaction would be completed in approximately

1

40 \min

2

50 \min

3

45 \min

4

60 \min

Explanation:

$60 = \frac{2.303}{k} \log \frac{100}{40}$
$\begin{aligned} t_{1 / 2} & = \frac{2.303}{k} \log \frac{100}{50} \\ ∴ \frac{60}{t} & = \frac{0.40}{0.310} \\ ∴ t & = 45 minute \end{aligned}$

KCET - 2010

CHXII04:CHEMICAL KINETICS

320319 Which of the following statement is not correct?

1 In zero order reaction, the rate of reaction remains constant throughout.

2 A second order reaction would become a pseudo first order reaction, when one of the reactants is taken in large excess.

3 The units of first order rate constant depends on the units of the concentration terms used.

4 In the first order reaction, the plot of

\log

(a-x) v/s time follows straight ine equation

Explanation:

Units of rate constant of first order is independent of units of concentration. The unit of first order rate constant $K$ is $s^{- 1}$ .

CHXII04:CHEMICAL KINETICS

320320 Naturally occuring potassium consist of $0.01 %$ of $^{40} K$ which has half life of $1.28 \times 10^{9}$ years. Calculate the activity of 1.00 g sample of KCl .

1

4.37 \times 10^{8} y e a r^{- 1}

2

1.28 \times 10^{8} y e a r^{- 1}

3

4.37 \times 10^{9} y e a r^{- 1}

4

1.28 \times 10^{9} y e a r^{- 1}

Explanation:

Rate constant $(k)$
$= \frac{0 .693}{t_{1 / 2}} = \frac{0 .693}{1 .28 \times 1 0^{9}}$
$= 5.414 \times 10^{- 10} y r^{- 1}$
$N u m b e r o f^{40} K a t o m s i n l g K C l$
$= 1 / 74 . 5 m o l$
$= \frac{1}{74.5} \times 6.02 \times 10^{23} m o l e c u l e s$
$= \frac{1}{74.5} \times 6.02 \times 1 0^{23} \times \frac{0.01}{100}^{40} K a t o m s$
$= 8.08 \times 10^{17} a t o m s$
$A c t i v i t y = k λ = 5 .414 \times 1 0^{- 10} = 8 .08 \times 1 0^{17}$
$= 4.37 \times 10^{8} y e a r^{- 1}$

CHXII04:CHEMICAL KINETICS

320321 Which of the following represents the expression for ${\frac{3}{4}}^{th}$ life of a first order reaction?

1

\frac{k}{2 .303} l o g \frac{4}{3}

2

\frac{2 .303}{k} l o g \frac{4}{3}

3

\frac{2 .303}{k} l o g 4

4

\frac{2 .303}{k} l o g 3

Explanation:

$k = \frac{2 .303}{t} l o g (\frac{a}{a - x})$
$t = \frac{2 .303}{k} l o g [\frac{a}{a - \frac{3 a}{4}}] = \frac{2 .303}{k} l o g 4$

CHXII04:CHEMICAL KINETICS

320317 A radioactive isotope having a half-life period of 3 days was received after 12 days. If $3 g$ of the isotope is left in the container, what would be the initial mass of the isotope?

1

12 g

2

36 g

3

48 g

4

24 g

Explanation:

Given $t_{1 / 2} = 3$
Total time, $T = 12$
No. of half lives $(n) = \frac{12}{3} = 4$
$\begin{aligned} {(\frac{1}{2})}^{n} = \frac{N}{N_{o}} \\ ∴ {(\frac{1}{2})}^{4} = \frac{3}{N} \end{aligned}$ $\frac{3}{N} = \frac{1}{16}, N = 48 g$

JEE - 2013

CHXII04:CHEMICAL KINETICS

320318 If $60 %$ of a first-order reaction was completed in $60 \min, 50 %$ of the same reaction would be completed in approximately

1

40 \min

2

50 \min

3

45 \min

4

60 \min

Explanation:

$60 = \frac{2.303}{k} \log \frac{100}{40}$
$\begin{aligned} t_{1 / 2} & = \frac{2.303}{k} \log \frac{100}{50} \\ ∴ \frac{60}{t} & = \frac{0.40}{0.310} \\ ∴ t & = 45 minute \end{aligned}$

KCET - 2010

CHXII04:CHEMICAL KINETICS

320319 Which of the following statement is not correct?

1 In zero order reaction, the rate of reaction remains constant throughout.

2 A second order reaction would become a pseudo first order reaction, when one of the reactants is taken in large excess.

3 The units of first order rate constant depends on the units of the concentration terms used.

4 In the first order reaction, the plot of

\log

(a-x) v/s time follows straight ine equation

Explanation:

Units of rate constant of first order is independent of units of concentration. The unit of first order rate constant $K$ is $s^{- 1}$ .

CHXII04:CHEMICAL KINETICS

320320 Naturally occuring potassium consist of $0.01 %$ of $^{40} K$ which has half life of $1.28 \times 10^{9}$ years. Calculate the activity of 1.00 g sample of KCl .

1

4.37 \times 10^{8} y e a r^{- 1}

2

1.28 \times 10^{8} y e a r^{- 1}

3

4.37 \times 10^{9} y e a r^{- 1}

4

1.28 \times 10^{9} y e a r^{- 1}

Explanation:

Rate constant $(k)$
$= \frac{0 .693}{t_{1 / 2}} = \frac{0 .693}{1 .28 \times 1 0^{9}}$
$= 5.414 \times 10^{- 10} y r^{- 1}$
$N u m b e r o f^{40} K a t o m s i n l g K C l$
$= 1 / 74 . 5 m o l$
$= \frac{1}{74.5} \times 6.02 \times 10^{23} m o l e c u l e s$
$= \frac{1}{74.5} \times 6.02 \times 1 0^{23} \times \frac{0.01}{100}^{40} K a t o m s$
$= 8.08 \times 10^{17} a t o m s$
$A c t i v i t y = k λ = 5 .414 \times 1 0^{- 10} = 8 .08 \times 1 0^{17}$
$= 4.37 \times 10^{8} y e a r^{- 1}$

CHXII04:CHEMICAL KINETICS

320321 Which of the following represents the expression for ${\frac{3}{4}}^{th}$ life of a first order reaction?

1

\frac{k}{2 .303} l o g \frac{4}{3}

2

\frac{2 .303}{k} l o g \frac{4}{3}

3

\frac{2 .303}{k} l o g 4

4

\frac{2 .303}{k} l o g 3

Explanation:

$k = \frac{2 .303}{t} l o g (\frac{a}{a - x})$
$t = \frac{2 .303}{k} l o g [\frac{a}{a - \frac{3 a}{4}}] = \frac{2 .303}{k} l o g 4$

CHXII04:CHEMICAL KINETICS

320317 A radioactive isotope having a half-life period of 3 days was received after 12 days. If $3 g$ of the isotope is left in the container, what would be the initial mass of the isotope?

1

12 g

2

36 g

3

48 g

4

24 g

Explanation:

Given $t_{1 / 2} = 3$
Total time, $T = 12$
No. of half lives $(n) = \frac{12}{3} = 4$
$\begin{aligned} {(\frac{1}{2})}^{n} = \frac{N}{N_{o}} \\ ∴ {(\frac{1}{2})}^{4} = \frac{3}{N} \end{aligned}$ $\frac{3}{N} = \frac{1}{16}, N = 48 g$

JEE - 2013

CHXII04:CHEMICAL KINETICS

320318 If $60 %$ of a first-order reaction was completed in $60 \min, 50 %$ of the same reaction would be completed in approximately

1

40 \min

2

50 \min

3

45 \min

4

60 \min

Explanation:

$60 = \frac{2.303}{k} \log \frac{100}{40}$
$\begin{aligned} t_{1 / 2} & = \frac{2.303}{k} \log \frac{100}{50} \\ ∴ \frac{60}{t} & = \frac{0.40}{0.310} \\ ∴ t & = 45 minute \end{aligned}$

KCET - 2010

CHXII04:CHEMICAL KINETICS

320319 Which of the following statement is not correct?

1 In zero order reaction, the rate of reaction remains constant throughout.

2 A second order reaction would become a pseudo first order reaction, when one of the reactants is taken in large excess.

3 The units of first order rate constant depends on the units of the concentration terms used.

4 In the first order reaction, the plot of

\log

(a-x) v/s time follows straight ine equation

Explanation:

Units of rate constant of first order is independent of units of concentration. The unit of first order rate constant $K$ is $s^{- 1}$ .

CHXII04:CHEMICAL KINETICS

320320 Naturally occuring potassium consist of $0.01 %$ of $^{40} K$ which has half life of $1.28 \times 10^{9}$ years. Calculate the activity of 1.00 g sample of KCl .

1

4.37 \times 10^{8} y e a r^{- 1}

2

1.28 \times 10^{8} y e a r^{- 1}

3

4.37 \times 10^{9} y e a r^{- 1}

4

1.28 \times 10^{9} y e a r^{- 1}

Explanation:

Rate constant $(k)$
$= \frac{0 .693}{t_{1 / 2}} = \frac{0 .693}{1 .28 \times 1 0^{9}}$
$= 5.414 \times 10^{- 10} y r^{- 1}$
$N u m b e r o f^{40} K a t o m s i n l g K C l$
$= 1 / 74 . 5 m o l$
$= \frac{1}{74.5} \times 6.02 \times 10^{23} m o l e c u l e s$
$= \frac{1}{74.5} \times 6.02 \times 1 0^{23} \times \frac{0.01}{100}^{40} K a t o m s$
$= 8.08 \times 10^{17} a t o m s$
$A c t i v i t y = k λ = 5 .414 \times 1 0^{- 10} = 8 .08 \times 1 0^{17}$
$= 4.37 \times 10^{8} y e a r^{- 1}$

CHXII04:CHEMICAL KINETICS

320321 Which of the following represents the expression for ${\frac{3}{4}}^{th}$ life of a first order reaction?

1

\frac{k}{2 .303} l o g \frac{4}{3}

2

\frac{2 .303}{k} l o g \frac{4}{3}

3

\frac{2 .303}{k} l o g 4

4

\frac{2 .303}{k} l o g 3

Explanation:

$k = \frac{2 .303}{t} l o g (\frac{a}{a - x})$
$t = \frac{2 .303}{k} l o g [\frac{a}{a - \frac{3 a}{4}}] = \frac{2 .303}{k} l o g 4$

CHXII04:CHEMICAL KINETICS

320317 A radioactive isotope having a half-life period of 3 days was received after 12 days. If $3 g$ of the isotope is left in the container, what would be the initial mass of the isotope?

1

12 g

2

36 g

3

48 g

4

24 g

Explanation:

Given $t_{1 / 2} = 3$
Total time, $T = 12$
No. of half lives $(n) = \frac{12}{3} = 4$
$\begin{aligned} {(\frac{1}{2})}^{n} = \frac{N}{N_{o}} \\ ∴ {(\frac{1}{2})}^{4} = \frac{3}{N} \end{aligned}$ $\frac{3}{N} = \frac{1}{16}, N = 48 g$

JEE - 2013

CHXII04:CHEMICAL KINETICS

320318 If $60 %$ of a first-order reaction was completed in $60 \min, 50 %$ of the same reaction would be completed in approximately

1

40 \min

2

50 \min

3

45 \min

4

60 \min

Explanation:

$60 = \frac{2.303}{k} \log \frac{100}{40}$
$\begin{aligned} t_{1 / 2} & = \frac{2.303}{k} \log \frac{100}{50} \\ ∴ \frac{60}{t} & = \frac{0.40}{0.310} \\ ∴ t & = 45 minute \end{aligned}$

KCET - 2010

CHXII04:CHEMICAL KINETICS

320319 Which of the following statement is not correct?

1 In zero order reaction, the rate of reaction remains constant throughout.

2 A second order reaction would become a pseudo first order reaction, when one of the reactants is taken in large excess.

3 The units of first order rate constant depends on the units of the concentration terms used.

4 In the first order reaction, the plot of

\log

(a-x) v/s time follows straight ine equation

Explanation:

Units of rate constant of first order is independent of units of concentration. The unit of first order rate constant $K$ is $s^{- 1}$ .

CHXII04:CHEMICAL KINETICS

320320 Naturally occuring potassium consist of $0.01 %$ of $^{40} K$ which has half life of $1.28 \times 10^{9}$ years. Calculate the activity of 1.00 g sample of KCl .

1

4.37 \times 10^{8} y e a r^{- 1}

2

1.28 \times 10^{8} y e a r^{- 1}

3

4.37 \times 10^{9} y e a r^{- 1}

4

1.28 \times 10^{9} y e a r^{- 1}

Explanation:

Rate constant $(k)$
$= \frac{0 .693}{t_{1 / 2}} = \frac{0 .693}{1 .28 \times 1 0^{9}}$
$= 5.414 \times 10^{- 10} y r^{- 1}$
$N u m b e r o f^{40} K a t o m s i n l g K C l$
$= 1 / 74 . 5 m o l$
$= \frac{1}{74.5} \times 6.02 \times 10^{23} m o l e c u l e s$
$= \frac{1}{74.5} \times 6.02 \times 1 0^{23} \times \frac{0.01}{100}^{40} K a t o m s$
$= 8.08 \times 10^{17} a t o m s$
$A c t i v i t y = k λ = 5 .414 \times 1 0^{- 10} = 8 .08 \times 1 0^{17}$
$= 4.37 \times 10^{8} y e a r^{- 1}$

CHXII04:CHEMICAL KINETICS

320321 Which of the following represents the expression for ${\frac{3}{4}}^{th}$ life of a first order reaction?

1

\frac{k}{2 .303} l o g \frac{4}{3}

2

\frac{2 .303}{k} l o g \frac{4}{3}

3

\frac{2 .303}{k} l o g 4

4

\frac{2 .303}{k} l o g 3

Explanation:

$k = \frac{2 .303}{t} l o g (\frac{a}{a - x})$
$t = \frac{2 .303}{k} l o g [\frac{a}{a - \frac{3 a}{4}}] = \frac{2 .303}{k} l o g 4$

CHXII04:CHEMICAL KINETICS

320317 A radioactive isotope having a half-life period of 3 days was received after 12 days. If $3 g$ of the isotope is left in the container, what would be the initial mass of the isotope?

1

12 g

2

36 g

3

48 g

4

24 g

Explanation:

Given $t_{1 / 2} = 3$
Total time, $T = 12$
No. of half lives $(n) = \frac{12}{3} = 4$
$\begin{aligned} {(\frac{1}{2})}^{n} = \frac{N}{N_{o}} \\ ∴ {(\frac{1}{2})}^{4} = \frac{3}{N} \end{aligned}$ $\frac{3}{N} = \frac{1}{16}, N = 48 g$

JEE - 2013

CHXII04:CHEMICAL KINETICS

320318 If $60 %$ of a first-order reaction was completed in $60 \min, 50 %$ of the same reaction would be completed in approximately

1

40 \min

2

50 \min

3

45 \min

4

60 \min

Explanation:

$60 = \frac{2.303}{k} \log \frac{100}{40}$
$\begin{aligned} t_{1 / 2} & = \frac{2.303}{k} \log \frac{100}{50} \\ ∴ \frac{60}{t} & = \frac{0.40}{0.310} \\ ∴ t & = 45 minute \end{aligned}$

KCET - 2010

CHXII04:CHEMICAL KINETICS

320319 Which of the following statement is not correct?

1 In zero order reaction, the rate of reaction remains constant throughout.

2 A second order reaction would become a pseudo first order reaction, when one of the reactants is taken in large excess.

3 The units of first order rate constant depends on the units of the concentration terms used.

4 In the first order reaction, the plot of

\log

(a-x) v/s time follows straight ine equation

Explanation:

Units of rate constant of first order is independent of units of concentration. The unit of first order rate constant $K$ is $s^{- 1}$ .

CHXII04:CHEMICAL KINETICS

320320 Naturally occuring potassium consist of $0.01 %$ of $^{40} K$ which has half life of $1.28 \times 10^{9}$ years. Calculate the activity of 1.00 g sample of KCl .

1

4.37 \times 10^{8} y e a r^{- 1}

2

1.28 \times 10^{8} y e a r^{- 1}

3

4.37 \times 10^{9} y e a r^{- 1}

4

1.28 \times 10^{9} y e a r^{- 1}

Explanation:

Rate constant $(k)$
$= \frac{0 .693}{t_{1 / 2}} = \frac{0 .693}{1 .28 \times 1 0^{9}}$
$= 5.414 \times 10^{- 10} y r^{- 1}$
$N u m b e r o f^{40} K a t o m s i n l g K C l$
$= 1 / 74 . 5 m o l$
$= \frac{1}{74.5} \times 6.02 \times 10^{23} m o l e c u l e s$
$= \frac{1}{74.5} \times 6.02 \times 1 0^{23} \times \frac{0.01}{100}^{40} K a t o m s$
$= 8.08 \times 10^{17} a t o m s$
$A c t i v i t y = k λ = 5 .414 \times 1 0^{- 10} = 8 .08 \times 1 0^{17}$
$= 4.37 \times 10^{8} y e a r^{- 1}$

CHXII04:CHEMICAL KINETICS

320321 Which of the following represents the expression for ${\frac{3}{4}}^{th}$ life of a first order reaction?

1

\frac{k}{2 .303} l o g \frac{4}{3}

2

\frac{2 .303}{k} l o g \frac{4}{3}

3

\frac{2 .303}{k} l o g 4

4

\frac{2 .303}{k} l o g 3

Explanation:

$k = \frac{2 .303}{t} l o g (\frac{a}{a - x})$
$t = \frac{2 .303}{k} l o g [\frac{a}{a - \frac{3 a}{4}}] = \frac{2 .303}{k} l o g 4$

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