Write the rate law expression and find the rate constant of the reaction A + B → Product . Also, calculate the half-life period. A 0 Mol litre - 1 B 0 Mol litre - 1 Rate R 0 Mol litre - 1 second - 1 0 . 1 0 . 2 0 . 1 0 . 2 0 . 2 0 . 1 0 . 5 0 . 10 0 . 05

Let, according to the rate law, the rate of the reaction be ( r ) . r 0 = k [ A ] 0 x [ B ] 0 y . . . . . ( i ) x and y are order of reaction with respect to A and B , respectively. On the basis of the given data, ( 0 . 05 ) = k ( 0 . 1 ) x ( 0 . 2 ) y . . . . ( ii ) ( 0 . 10 ) = k ( 0 . 2 ) x ( 0 . 2 ) y . . . . . ( iii ) ( 0 . 05 ) = k ( 0 . 1 ) x ( 0 . 1 ) y . . . . ( iv ) On dividing equation ( iii ) by ( ii ) , 0 . 10 0 . 05 = ( 0 . 2 ) x ( 0 . 1 ) 2 2 = ( 2 ) x Thus, x = 1 On dividing equation ( ii ) by equation ( iv ) , ( 0 . 05 ) ( 0 . 05 ) = 0 . 2 0 . 1 y 1 = ( 2 ) y ( 2 ) 0 = ( 2 ) y Thus , y = 0 Thus, rate law expression, r = k [ A ] 1 [ B ] 0 k = r 0 [ A ] 0 = 0 . 05 0 . 10 = 0 . 5 sec - 1 Rate constant, k = 0 . 5 sec - 1 Half-life period, t 1 / 2 = 0 . 693 k = 0 . 693 0 . 5 sec t 1 / 2 = 1 . 386 sec

For a reaction at 800 ° C , 2 NO + 2 H 2 → N 2 + 2 H 2 O , the following data were obtained: [ NO ] / × 10 - 4 Mol litre - 1 H 2 / × 10 - 3 Mol litre - 1 - 1 2 d [ NO ] d t / × 10 - 4 Mol litre - 1 min - 1 ( i ) 1 . 5 4 . 0 4 . 4 ( ii ) 1 . 5 2 . 0 2 . 2 ( iii ) 0 . 5 2 . 0 0 . 24 What is the order of this reaction with respect to NO and H 2 ? Also calculate the rate constant.

Here, - 1 2 d [ NO ] / dt represents the rate of the reaction. Rate law expression for this reaction is: r = k [ NO ] x [ H 2 ] y . . . . . ( i ) Where, x and y are order with respect to NO and H 2 . From the given data: 4 . 4 × 10 - 4 = k ( 1 . 5 × 10 - 4 ) x ( 4 . 0 × 10 - 3 ) y . . . . ( ii ) 2 . 2 × 10 - 4 = k ( 1 . 5 × 10 - 4 ) x ( 2 . 0 × 10 - 3 ) y . . . . ( iii ) 0 . 24 × 10 - 4 = k ( 0 . 5 × 10 - 4 ) x ( 2 . 0 × 10 - 3 ) y . . . . ( iv ) On dividing equation ( ii ) by equation ( iii ) : 2 = ( 2 ) y Thus, y = 1 On dividing equation ( iii ) by equation ( iv ) : 2 . 2 0 . 24 = ( 3 ) x ( 1 ) y = ( 3 ) x ( 1 ) 1 ( 3 ) x = ( 3 ) 2 Thus, x = 2 Therefore, order with respect to NO = 2 and with respect to H 2 = 1 Now, from equation ( i ) r = k [ NO ] 2 [ H 2 ] 1 Putting values from the experiment ( i ) , 4 . 4 × 10 - 4 = k ( 1 . 5 × 10 - 4 ) 2 ( 4 . 0 × 10 - 3 ) k = 4 . 4 × 10 - 4 1 . 5 × 1 . 5 × 4 . 0 × 10 - 11 = 4 . 88 × 10 8 k = 4 . 88 × 10 8 mol - 2 L 2 sec - 1

Rate of a reaction of A + B → Product , is given as a function of different initial concentrations of A and B . [ A ] Mol L - 1 [ B ] Mol L - 1 r 0 Mol L - 1 min - 1 0 . 01 0 . 01 0 . 005 0 . 02 0 . 01 0 . 010 0 . 01 0 . 02 0 . 005 Determine the order of the reaction with respect to A and with respect to B . What is the half-life of A in the reaction?

The rate law expression for the reaction is, r 0 = k [ A ] x [ B ] y . . . . ( i ) Where, x and y are the order with respect to A and B , respectively. From the given data, ( 0 . 005 ) = k ( 0 . 01 ) x ( 0 . 01 ) y . . . . . ( ii ) ( 0 . 010 ) = k ( 0 . 02 ) x ( 0 . 01 ) y . . . . . ( iii ) ( 0 . 005 ) = k ( 0 . 01 ) x ( 0 . 02 ) y . . . . . ( iv ) Equation ( ii ) is divided by equation ( iii ) , then, 1 2 1 = 1 2 x Thus, x = 1 . Equation ( ii ) is divided by equation ( iv ) , then, ( 1 ) = 1 2 y 1 2 0 = 1 2 y Thus, y = 0 . So, the order of reaction with respect to A = 1 , and with respect to B = 0 . Now, from equation ( i ) , r = k A 1 B 0 k = r [ A ] 1 = 0 . 005 0 . 01 = 0 . 5 min - 1 Since the reaction is of the first order , Half-life time t 1 / 2 = 0 . 693 k = 0 . 693 0 . 5 = 1 . 386 min

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In a $1^{st}$ order reaction, amount of the substance left after $n$ half lives and average life of a $1^{st}$ order reaction respectively are:

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Important Questions on Chemical Kinetics

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IMPORTANT

Chapterwise/Topicwise Daily Practice Problems (DPP) Physical Chemistry Part - 2 JEE Main & Advanced>Chapter 4 - Chemical Kinetics>DPP 4.3>Q 16

If the reactant

' A'

is gaseous and pressures are measured instead of molar concentrations, then for reactions of zero order:

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IMPORTANT

Chapterwise/Topicwise Daily Practice Problems (DPP) Physical Chemistry Part - 2 JEE Main & Advanced>Chapter 4 - Chemical Kinetics>DPP 4.4>Q 1

For the chemical reaction between mercuric chloride and potassium dichromate, the amount of ${Hg}_{2} {Cl}_{2}$ precipitated from different situations in a given time at $100 ° C$ is given in the following table. From this data, calculate the order of the reaction.

Experiment Number	${HgCl}_{2} (mol L^{- 1})$	$K_{2} {Cr}_{2} O_{7} (mol L^{- 1})$	Time $(minutes)$	${Hg}_{2} {Cl}_{2}$ $(mol L^{- 1})$
$1 .$	$0.0836$	$0.404$	$60$	$0.0068$
$2 .$	$0.0836$	$0.202$	$120$	$0.0031$
$3 .$	$0.0418$	$0.404$	$60$	$0.0032$

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IMPORTANT

Chapterwise/Topicwise Daily Practice Problems (DPP) Physical Chemistry Part - 2 JEE Main & Advanced>Chapter 4 - Chemical Kinetics>DPP 4.4>Q 2

Write the rate law expression and find the rate constant of the reaction $A + B \to Product$ . Also, calculate the half-life period.

\underset{Mol {litre}^{- 1}}{[A_{0}]}

\underset{Mol {litre}^{- 1}}{[B_{0}]}

\underset{Mol {litre}^{- 1} {second}^{- 1}}{Rate [R_{0}]}

$0.1$

$0.2$

$0.1$

$0.2$

$0.1$

$0.5$

$0.10$

$0.05$

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Chapterwise/Topicwise Daily Practice Problems (DPP) Physical Chemistry Part - 2 JEE Main & Advanced>Chapter 4 - Chemical Kinetics>DPP 4.4>Q 3

For a reaction at $800 ° C, 2 NO + 2 H_{2} \to N_{2} + 2 H_{2} O$ , the following data were obtained:

$\begin{matrix} [NO] / (\times 10^{- 4}) Mol {litre}^{- 1} & [H_{2}] / (\times 10^{- 3}) Mol {litre}^{- 1} & - \frac{1}{2} \frac{d [NO]}{d t} / (\times 10^{- 4}) Mol {litre}^{- 1} \min^{- 1} \\ (i) & 1.5 & 4.0 & 4.4 \\ (ii) & 1.5 & 2.0 & 2.2 \\ (iii) & 0.5 & 2.0 & 0.24 \end{matrix}$

What is the order of this reaction with respect to $NO$ and $H_{2}$ ? Also calculate the rate constant.

HARD

JEE Main

IMPORTANT

Chapterwise/Topicwise Daily Practice Problems (DPP) Physical Chemistry Part - 2 JEE Main & Advanced>Chapter 4 - Chemical Kinetics>DPP 4.4>Q 4

Rate of a reaction of $A + B \to Product$ , is given as a function of different initial concentrations of $A$ and $B$ .

$[A] (Mol L^{- 1})$	$[B] (Mol L^{- 1})$	$r_{0} (Mol L^{- 1} \min^{- 1})$
$0.01$	$0.01$	$0.005$
$0.02$	$0.01$	$0.010$
$0.01$	$0.02$	$0.005$

Determine the order of the reaction with respect to $A$ and with respect to $B$ . What is the half-life of $A$ in the reaction?

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JEE Main

IMPORTANT

Chapterwise/Topicwise Daily Practice Problems (DPP) Physical Chemistry Part - 2 JEE Main & Advanced>Chapter 4 - Chemical Kinetics>DPP 4.4>Q 5

The experimental data for the reaction,

$2 A + B_{2} \to 2 AB,$ is,

Experiment Number	$[A]$	$[B_{2}]$	Rate $(s^{- 1})$
$1 .$	$0.50$	$0.50$	$1.6 \times 10^{- 4}$
$2 .$	$0.50$	$1.00$	$3.2 \times 10^{- 4}$
$3 .$	$1.0$	$1.00$	$3.2 \times 10^{- 4}$

The rate equation for the above data is -

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Chapterwise/Topicwise Daily Practice Problems (DPP) Physical Chemistry Part - 2 JEE Main & Advanced>Chapter 4 - Chemical Kinetics>DPP 4.4>Q 6

For a chemical reaction

A \to B,

the rate of reaction increases by a factor of

1.837

, when the concentration of

A

is increased by

1.5

times. The order of reaction with respect to

A

is:

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JEE Main

IMPORTANT

Chapterwise/Topicwise Daily Practice Problems (DPP) Physical Chemistry Part - 2 JEE Main & Advanced>Chapter 4 - Chemical Kinetics>DPP 4.4>Q 7

For the decomposition of $HI$ at $1000 K (2 HI \to H_{2} + I_{2})$ , the following data were obtained:

$[HI] M$	Rate of decomposition of $HI (mol L^{- 1} s^{- 1})$
$0.1$	$2.75 \times 10^{- 8}$
$0.2$	$11 \times 10^{- 8}$
$0.3$	$24.75 \times 10^{- 8}$

The order of the reaction is.

In a 1st order reaction, amount of the substance left after n half lives and average life of a 1st order reaction respectively are:

Important Questions on Chemical Kinetics

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In a $1^{st}$ order reaction, amount of the substance left after $n$ half lives and average life of a $1^{st}$ order reaction respectively are: