Following data were obtained for the catalytic decomposition of ammonia: Initial pressure (mm) 50 100 200 Half-life (hrs) 3 . 52 1 . 92 1 . 00 Find the order of reaction.

From the given set of information, we observed that if the value of initial pressure becomes twice, then the half-life of the ammonia decreases to approximately half. So, we can say that half life is decreasing with the increase in pressure and thus half life is inversely proportional to the pressure applied. Also we know, the half-life of the reaction is related to the pressure in the following way: t 1 / 2 α 1 P o n - 1 Assuming if it is a second-order reaction then, n = 2 So, t 1 / 2 α 1 P o Hence, the given reaction is of second order.

The first order rate constant for the decomposition of ethyl iodide by the reaction C 2 H 5 I g ⟶ C 2 H 4 g + HI g at 600 K is 1 . 60 × 10 - 5 s - 1 . Its energy of activation is 209 kJ / mol . Calculate the rate constant of the reaction at 700 K

Given that: T 1 = 600 K , k 1 = 1 . 60 × 0 - 5 s - 1 T 2 = 700 K , k 2 = ? , E a = 209 kJ / mol By the Arrhenius equation, log k 2 k 1 = E a 2 . 303 R T 2 - T 1 T 1 × T 2 Substituting the values, log k 2 k 1 = 209 × 1000 J mol - 1 2 . 303 × 8 . 314 Jk - 1 mol - 1 700 - 600 600 × 700 k = 209000 19 . 147 100 420000 = 2090 19 . 147 × 42 log k 2 k 1 = 2 . 6 k 2 k 1 = antilog ( 2 . 6 ) k 2 = k 1 × 398 . 12 Substitute the value of k 1 k 2 = 1 . 60 × 10 - 5 × 398 . 12 k 2 = 0 . 00636 k 2 = 6 . 36 × 10 - 3 s - Hence, the answer.

The value of rate constant for a second order reaction is 6 . 7 × 10 - 5 mol - L s - at 298 K and 1 . 64 × 10 - 4 mol - L s - at 313 K . Find the Arrhenius frequency factor A and activation energy of the reaction.

Given that: k 1 = 6 . 7 × 10 - 5 mol - 1 L - 1 s - , T 1 = 298 K k 2 = 1 . 64 × 10 - 4 mol - 1 L - 1 s - , T 2 = 313 K By the Arrhenius equation, log k 2 k 1 = E a 2 . 303 R T 2 - T 1 T 1 T 2 Substituting the values, log 6 . 7 × 10 - 5 1 . 64 × 10 - 4 = E a 2 . 303 × 8 . 314 313 - 298 313 × 298 log ( 0 . 4 ) = Ea × 15 19 . 147 × 93274 1 . 8 = E a 8 . 4 × 10 - 6 E a = 46279 J mol - 1 E a = 46 . 279 kJ mol - 1 We know, the rate constant by Arrhenius equation is given by: k = Ae - E a RT Taking log on both sides, we have log k = log A - E a RT log e log k = log A - E a 2 . 303 RT log A = logk + E a 2 . 303 RT log A = log ( 6 . 7 ) + 46279 2 . 303 × 8 . 314 × 298 = 0 . 8260 + 8 . 239 log A = 9 . 065 A = Antilog ( 9 . 065 ) A = 8649 . 48 collisions s - Hence, solved.

HARD

12th West Bengal Board

IMPORTANT

Earn 100

Hydrolysis of methyl acetate in aqueous solution has been studied by titrating the liberated acetic acid against sodium hydroxide. The concentration of the ester at different times is given below:

$t / \min$ $0$ $30$ $60$ $90$

$C / {molL}^{- 1}$ $0.8500$ $0.8004$ $0.7538$ $0.7096$

Show that it follows a pseudo first order reaction, as the concentration of water remains nearly constant $(55 mol L^{- 1})$ during the course of the reaction. What is the value of $k^{'}$ in this equation? Rate $= k^{'} [{CH}_{3} {COOCH}_{3}] [H_{2} O]$

[Hint. $k = k' [H_{2} O] = \frac{2.303}{t} \log \frac{C_{0}}{C_{t}}$ ]

Important Questions on Chemical Kinetics

HARD

12th West Bengal Board

IMPORTANT

NEW SYSTEMATIC MODERN CHEMISTRY>Chapter 4 - Chemical Kinetics>Follow Up Problems>Q 27

The kinetics of hydrolysis of methyl acetate in excess of hydrochloric acid solution at $298 K$ were followed by withdrawing $2 mL$ of the reaction mixture at intervals of time (t), adding $50 mL$ of water and titrating against baryta-water. The following results were obtained:

$t (\min)$	$0$	$10$	$28$	$58$	$115$	$\infty$
Titre (mL)	$18.5$	$19.1$	$20.1$	$21.6$	$24.6$	$34.8$

Determine the velocity constant of the hydrolysis.

[Hint. Substitute $a = 34.8 - 18.5 = 16.3$ , and $a - x = (34.8 - 19.1)$ , $(34.8 - 20.1), (34.8 - 21.6), (34.8 - 24.6)$ at $t = 10, 28, 58$ and $115$ respectively in $k = 2.303 / t \log a / a - x$ and get $k]$

HARD

12th West Bengal Board

IMPORTANT

NEW SYSTEMATIC MODERN CHEMISTRY>Chapter 4 - Chemical Kinetics>Follow Up Problems>Q 28

The half-life period of a substance is

50

minutes at a certain concentration. When the concentration is reduced to one half of the initial concentration, the half-life period is

25

minute. Calculate order of the reaction.

EASY

12th West Bengal Board

IMPORTANT

NEW SYSTEMATIC MODERN CHEMISTRY>Chapter 4 - Chemical Kinetics>Follow Up Problems>Q 29

At a certain temperature, the half-life period for the decomposition for the substance A is as follows:

$P (mm)$ $500$ $700$ $900 mm$

Half-life period $18$ $17.9$ $18$

What is order of reaction ?

EASY

12th West Bengal Board

IMPORTANT

NEW SYSTEMATIC MODERN CHEMISTRY>Chapter 4 - Chemical Kinetics>Follow Up Problems>Q 30

Following data were obtained for the catalytic decomposition of ammonia:

Initial pressure (mm)	$50$	$100$	$200$
Half-life (hrs)	$3.52$	$1.92$	$1.00$

Find the order of reaction.

HARD

12th West Bengal Board

IMPORTANT

NEW SYSTEMATIC MODERN CHEMISTRY>Chapter 4 - Chemical Kinetics>Follow Up Problems>Q 31

The rate constants of a reaction at

500 K

and

700 K

are

0.02 s^{- 1}

and

0.07 s^{- 1}

respectively. Calculate the values of

E_{a}

and A.

HARD

12th West Bengal Board

IMPORTANT

NEW SYSTEMATIC MODERN CHEMISTRY>Chapter 4 - Chemical Kinetics>Follow Up Problems>Q 32

The first order rate constant for the decomposition of ethyl iodide by the reaction

C_{2} H_{5} I_{(g)} ⟶ C_{2} H_{4} (g) + {HI}_{(g)}

600 K

1.60 \times 10^{- 5} s^{- 1}

. Its energy of activation is

209 kJ / mol .

Calculate the rate constant of the reaction at

700 K

HARD

12th West Bengal Board

IMPORTANT

NEW SYSTEMATIC MODERN CHEMISTRY>Chapter 4 - Chemical Kinetics>Follow Up Problems>Q 33

The value of rate constant for a second order reaction is

6.7 \times 10^{- 5} {mol}^{-} L s^{-}

298 K

and

1.64 \times 10^{- 4} {mol}^{-} L s^{-}

313 K .

Find the Arrhenius frequency factor A and activation energy of the reaction.

HARD

12th West Bengal Board

IMPORTANT

NEW SYSTEMATIC MODERN CHEMISTRY>Chapter 4 - Chemical Kinetics>Follow Up Problems>Q 34

Rate constant $k$ of a reaction varies with temperature according to equation:

$\log k =$ constant $- \frac{E_{a}}{2.303 R} \cdot \frac{1}{T}$

What is the activation energy for the reaction. When a graph is plotted for $\log k$ versus $\frac{1}{T},$ a straight line with a slope $- 6670 K$ is obtained. Calculate energy of activation for this reaction $(R = 8.314 {JK}^{- 1} {mol}^{- 1})$

$t / \min$	$0$	$30$	$60$	$90$
$C / {molL}^{- 1}$	$0.8500$	$0.8004$	$0.7538$	$0.7096$

Important Questions on Chemical Kinetics

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