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

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

The reaction A + B → C + D ; ΔH = 25 kJ mol - 1 must have an activation energy of (a) - 25 kJ mol - 1 (b) < + 25 kJ mol - 1 (c) > + 25 kJ mol - 1 (d) Either answer (b) or (c), depending upon the experiment.

The enthalpy change value is positive, which indicates that the given reaction is endothermic. Hence, the energy of the products is higher than the energy of the reactants. The energy of the transition state is higher than the energy of reactants and products. Hence, the activation energy will be greater than the enthalpy change, i.e., E a > 25 kJ mol - 1 .

The half-life period for the reaction, N 2 O 5 → 2 NO 2 + 1 2 O 2 is 2 . 4 hours at 30 ° C . What time would be required to reduce 5 × 10 10 molecules of N 2 O 5 to 10 8 molecules?

N 2 O 5 → 2 NO 2 + 1 2 O 2 t 1 / 2 = 2 . 4 hr K = 0 . 693 t 1 / 2 = 0 . 693 2 . 4 Also we have K = 2 . 303 t log C 0 C f Putting the values we get: 0 . 693 2 . 4 = 2 . 303 t log 5 × 10 10 10 8 t = 2 . 303 0 . 693 / 2 . 4 log 500 = 2 . 303 × 2 . 4 0 . 693 log 500 = 21 . 52 hr

HARD

JEE Main

IMPORTANT

Earn 100

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) $0.01 mole / L$ .

Important Questions on Chemical Kinetics

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 $A$ .

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

MEDIUM

JEE Main

IMPORTANT

Modern Approach to Chemical Calculations>Chapter 17 - Chemical Kinetics>PROBLEMS>Q 24

For a given reaction $A + B \to P$ , the orders w.r.t. $A$ and $B$ are $1$ and $2$ respectively. Fill in the blanks from the following data:

$\begin{matrix} Rate ({Ms}^{- 1}) & [A] & [B] \\ 0 \cdot 10 & 1.0 M & 0 \cdot 20 M \\ \dots & 2.0 M & 0 \cdot 20 M \\ \dots & 2 \cdot 0 M & 0 \cdot 40 M \end{matrix}$

MEDIUM

JEE Main

IMPORTANT

Modern Approach to Chemical Calculations>Chapter 17 - Chemical Kinetics>PROBLEMS>Q 25

The reaction $A + B \to C + D; ΔH = 25 kJ {mol}^{- 1}$ must have an activation energy of

(a) $- 25 kJ {mol}^{- 1}$

(b) $< + 25 kJ {mol}^{- 1}$

(d) Either answer (b) or (c), depending upon the experiment.

HARD

JEE Main

IMPORTANT

Modern Approach to Chemical Calculations>Chapter 17 - Chemical Kinetics>PROBLEMS>Q 26

The half-life period for the reaction,

N_{2} O_{5} \to 2 {NO}_{2} + \frac{1}{2} O_{2}

2.4

hours at

30 ° C

. What time would be required to reduce

5 \times 10^{10}

molecules of

N_{2} O_{5}

10^{8}

molecules?

A substance decomposes according to second order rate law. If the rate constant is 6·8×10-4L 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.

Important Questions on Chemical Kinetics

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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) $0.01 mole / L$ .