Prove that the half-life period of a reaction of n th order in a reaction of the type A → Product, is inversely proportional to the ( n - 1 ) th power of the initial concentration.

We have A → Product Rate of reaction is directly proportional to the concentration of reactants. - dA dt = K A n ∫ A = 0 At d A A n = ∫ t = 0 1 - Kdt ⇒ A 1 - n 1 - n | Ao At = - K ( t - 0 ) A 1 1 - n 1 - n - A 0 1 - n 1 - n = - Kt ............. (1) For t = t 1 / 2 , A 1 = A 0 2 Putting in equation (1) A 0 1 - n 1 - n - A 0 1 - n ( 1 - n ) 2 1 - n = Kt 1 / 2 t 1 / 2 = A 0 1 - n K 1 1 - n - 1 ( 1 - n ) ( 2 ) 1 - n ⇒ t 1 / 2 = 1 K A 0 n - 1 1 ( 1 - n ) - 1 ( 1 - n ) ( 2 ) 1 - n ∴ t 1 / 2 ∝ 1 A 0 n - 1

Prove that the half-life period of a reaction of n th order in a reaction of the type A → Product, is inversely proportional to the ( n - 1 ) th power of the initial concentration.

We have A → Product Rate of reaction is directly proportional to the concentration of reactants. - dA dt = K A n ∫ A = 0 At d A A n = ∫ t = 0 1 - Kdt ⇒ A 1 - n 1 - n | Ao At = - K ( t - 0 ) A 1 1 - n 1 - n - A 0 1 - n 1 - n = - Kt ............. (1) For t = t 1 / 2 , A 1 = A 0 2 Putting in equation (1) A 0 1 - n 1 - n - A 0 1 - n ( 1 - n ) 2 1 - n = Kt 1 / 2 t 1 / 2 = A 0 1 - n K 1 1 - n - 1 ( 1 - n ) ( 2 ) 1 - n ⇒ t 1 / 2 = 1 K A 0 n - 1 1 ( 1 - n ) - 1 ( 1 - n ) ( 2 ) 1 - n ∴ t 1 / 2 ∝ 1 A 0 n - 1

A living plant acquires a definite fraction of C 6 14 nuclei in carbon content. If a freshly cut piece of wood gives 16 · 1 counts per minute gram and an old wooden bowl gives 9 . 6 counts per minute gram of carbon, calculate the age of the wooden bowl. The half-life of C 6 14 is 5770 years .

Initial count rate = A o = 16 . 1 counts / min A t = 9 . 6 count / min t 1 / 2 = 5770 years ∴ K = 0 . 693 t 1 / 2 = 0 . 693 5770 Year - 1 Now we have K = 2 . 303 t log A o A t Putting the values we get 0 . 693 5770 = 2 . 303 t log 16 . 1 9 . 6 t = 2 . 303 0 . 693 × 5770 log 16 . 1 9 . 6 = 4304 year

A substance decomposes according to second order rate law. If the rate constant is 6 · 8 × 10 - 4 L mole s - 1 - 1 , calculate half-life of the substance, if the initial concentration is (i) 0 . 05 mole / L and (ii) 0 . 01 mole / L .

For 2 nd order reactions - d A dt = K A 2 ⇒ ∫ A o A d A A 2 = - ∫ t = 0 t Kdt - 1 A Ao At = - Kt 1 A o - 1 A 1 = - Kt 1 A t = 1 A o + Kt - - - - - ( 1 ) For t = t 1 / 2 , A t = A o 2 ∴ ( 1 ) ⇒ 2 A o = 1 A o + Kt 1 2 ⇒ t 1 2 = 1 K A o (1) concentration is 0 . 05 mole / L ∴ t 1 2 = 1 6 . 8 × 10 - 4 × 0 . 05 = 2 . 94 × 10 4 sec (2) concentration is 0 . 01 mole / L t 1 2 = 1 6 . 8 × 10 - 4 × 0 . 01 = 1 . 47 × 10 4 sec

The following rate data were obtained at 30 ° C for the decomposition of N 2 O 5 in CCl 4 solution: N 2 O 5 (mole/litre) d N 2 O 5 / dt (mole/litre/hour) 0 . 34 0 . 68 1 . 36 0 . 10 0 . 20 0 . 40 Calculate the order of the reaction and the rate constant at 30 ° C .

Since rate is getting doubled on doubling the concentration of dinitrogen pentoxide, we can say that the rate of reaction is directly proportional to the concentration of N 2 O 5 . ∴ - d N 2 O 5 dt = K N 2 O 5 Putting values given in the table 0 . 10 = K ( 0 . 34 ) K = 10 34 hr - 1 = 0 . 29 hr - 1 = 8 . 05 sec - 1 Now if the order of reaction is n then the unit of rate constant is M 1 - n sec - 1 where M is molarity Comparing unit of rate constant we get M 1 - n sec - 1 = M 0 sec - 1 M 1 - n = M 0 1 - n = 0 n = 1 So the order of reaction is 1

HARD

JEE Main

IMPORTANT

Earn 100

Prove that the half-life period of a reaction of $n^{th}$ order in a reaction of the type $A \to$ Product, is inversely proportional to the $(n - 1)^{th}$ power of the initial concentration.

Important Questions on Chemical Kinetics

MEDIUM

JEE Main

IMPORTANT

Modern Approach to Chemical Calculations>Chapter 17 - Chemical Kinetics>PROBLEMS>Q 16

A living plant acquires a definite fraction of

{}_{6}^{14}C

nuclei in carbon content. If a freshly cut piece of wood gives

16 \cdot 1

counts per minute gram and an old wooden bowl gives

9.6

counts per minute gram of carbon, calculate the age of the wooden bowl. The half-life of

{}_{6}^{14}C

5770 years .

HARD

JEE Main

IMPORTANT

Modern Approach to Chemical Calculations>Chapter 17 - Chemical Kinetics>PROBLEMS>Q 17

The following data gives pressure of a gaseous $N_{2} O_{5}$ as a function of time at $45 ° C$ . Determine the rate constant (in SI units).

$\begin{array}{rccc} t (s) & p (mm) & t (s) & p (mm) \\ 0 & 348 & 3600 & 58 \\ 600 & 247 & 4800 & 33 \\ 1200 & 185 & 6000 & 18 \\ 2400 & 105 & 7200 & 10 \end{array}$

If the answer is of type $y \times 10^{- 4}$ , report the value of $y$ correct up to one place of decimal.

HARD

JEE Main

IMPORTANT

Modern Approach to Chemical Calculations>Chapter 17 - Chemical Kinetics>PROBLEMS>Q 18

A substance decomposes according to second order rate law. If the rate constant is

6 \cdot 8 \times 10^{- 4} L

mole

{}^{- 1}s^{- 1},

calculate half-life of the substance, if the initial concentration is (i)

0.05 mole / L

and (ii)

0.01 mole / L

HARD

JEE Main

IMPORTANT

Modern Approach to Chemical Calculations>Chapter 17 - Chemical Kinetics>PROBLEMS>Q 19

For a certain reaction, it takes

5

minutes for the initial concentration of

0.5 mole / L

to become

0.25 mole / L

and another

5

minutes to become

0 \cdot 125

mole / L .

What is the specific rate constant of the reaction?

HARD

JEE Main

IMPORTANT

Modern Approach to Chemical Calculations>Chapter 17 - Chemical Kinetics>PROBLEMS>Q 20

The half-life period of a gaseous substance undergoing thermal decomposition was measured for various initial pressures $(p)$ with the following results: $\begin{array}{l} p (mm) & 250 & 300 & 400 & 450 \\ t_{\frac{1}{2}} (\min) & 136 & 112.5 & 85 & 75.5 \end{array}$

Calculate the order of the reaction.

HARD

JEE Main

IMPORTANT

Modern Approach to Chemical Calculations>Chapter 17 - Chemical Kinetics>PROBLEMS>Q 21

The kinetics of decomposition of

N_{2} O_{5}

{CCl}_{4}

solution is studied by measuring the evolved oxygen. If

24 mL

of the gas was evolved in one hour while

35 mL

of the gas was evolved when no more oxygen was coming out, calculate the fraction of

N_{2} O_{5}

decomposed in one hour.

HARD

JEE Main

IMPORTANT

Modern Approach to Chemical Calculations>Chapter 17 - Chemical Kinetics>PROBLEMS>Q 22

The following rate data were obtained at $30 ° C$ for the decomposition of $N_{2} O_{5}$ in ${CCl}_{4}$ solution:

$[N_{2} O_{5}]$ (mole/litre)	$d [N_{2} O_{5}] / dt$ (mole/litre/hour)
$0.34 0.68 1.36$	$0.10 0.20 0.40$

Calculate the order of the reaction and the rate constant at $30 ° C$ .

HARD

JEE Main

IMPORTANT

Modern Approach to Chemical Calculations>Chapter 17 - Chemical Kinetics>PROBLEMS>Q 23

From the following data calculate the order with respect to each reactant $A, B$ and $C$ . Report the order with respect to $A$ .

$\begin{matrix} \begin{matrix} [A] \\ (mole / L) \end{matrix} & \begin{matrix} [B] \\ (mole / L) \end{matrix} & \begin{matrix} [C] \\ (mole / L) \end{matrix} & \begin{matrix} d [B] / d t \times 10^{- 5} \\ (mole / L / s) \end{matrix} \\ 0.010 & 0.005 & 0.010 & 5.0 \\ 0.015 & 0.005 & 0.010 & 5.0 \\ 0.010 & 0.010 & 0.010 & 2.5 \\ 0.010 & 0.005 & 0.020 & 14.1 \end{matrix}$

Prove that the half-life period of a reaction of nth order in a reaction of the type A→ Product, is inversely proportional to the (n-1)th power of the initial concentration.

Important Questions on Chemical Kinetics

Learn and Prepare for any exam you want !

Prove that the half-life period of a reaction of $n^{th}$ order in a reaction of the type $A \to$ Product, is inversely proportional to the $(n - 1)^{th}$ power of the initial concentration.