JEE Advanced (Subjective Type Questions): Chemical Kinetics | Chapter-wise Tests for JEE Main & Advanced PDF Download

subjective solved problems 

Problem : 1 

The rate of change of concentration of C in the reaction 2A+B → 2C+3D was reported as 1.0 mol litre^-1 sec^-1. Calculate the reaction rate as well as rate of change of concentration of A, B and D. 

Sol. We have,

-  =  =  =  = rate of reaction

Therefore,  = 1.0 mol litre^-1 sec^-1

Therefore,  =  = 1.0 mol L^-1sec^-1 

 =  =  = 0.5 mol L^-1 sec^-1 

 =  = 1.5 mol L^-1 sec^-1 

Also,

Q Rate =

Therefore, Rate =  × 1 = 0.5 mol L^-1 sec^-1 

Problem : 2 

For the reaction A B → C, the following data were obtained. In the first experiment, when the initial concentrations of both A and B are 0.1 M, the observed initial rate of formation of C is 1 × 10^-4 mol litre^-1 minute^-1. In the second experiment when the initial concentrations of A and B are 0.1 M and 0.3 M, the initial rate is 9.0 × 10^-4 mol litre^-1 minute^-1. 

(a) Write rate law for this reaction 

(b) Calculate the value of specific rate constant for this reaction. 

Sol. Let Rate = K[A]^m[B]ⁿ

(a) r₁ = 1 × 10^-4 = K[0.1]^m [0.1]^m ...(1)

r₂ = 9 × 10^-4 = K[0.1]^m[0.3]ⁿ ...(2)

r₃ = 2.7 × 10^-3 = K[0.3]^m[0.3]^m ...(3)

By Eqs. (1) and (2),

Therefore, n = 2

By Eqs. (2) and (3),

 m = 1

Therefore, Rate = K[A]¹[B]²

(b) Also by Eq. (1), 1 × 10^-4 = K[0.1]¹ [0.1]²

K = 10^-1 = 0.1 L² mol^-2 min^-1

Problem : 3 

The chemical reaction between K₂C₂O₄ and HgCl₂ is ; 

2HgCl₂ K₂C₂O₄ → 2KCl 2CO₂ Hg₂Cl₂ 

The weights of Hg₂Cl₂ precipitated from different solutions in given time were taken and expressed as following : 

Let the rate law be written as : r = k[HgCl₂]^x [K₂C₂O₄]^y

1. = k[0.0418]^x [0.404]^y

2.  = k[0.0836]^x[0.404]^y

3. = k[0.0836]^x [0.202]^y

Solving the above equations, we get :

x = 1 and y = 2 ⇒ order of reaction w.r.t x = 1 and y = 2 and overall order is 3.

Problem : 4 

The reaction given below, involving the gases is observed to be first order with rate constant
 7.48 × 10^-3 sec^-1. Calculate the time required for the total pressure in a system containing A at an initial pressure of 0.1 atm to rise to 0.145 atm also find the total pressure after 100 sec. 

2A(g) → 4B(g) C(g) 

 Sol.  2A(g) → 4B(g) C(g)

initial P₀ =0 

at time t P₀ - P¢ 2P¢ P¢/2

P_total = P₀ - P¢  2P¢ P¢/2 = P₀

P¢ = (0.145 - 0.1) = 0.03 atm

k =

t =

t = 23.84 sec 

Also, k =

7.48 × 10^-3 =

0.1/0.1 - P¢ = 5

P¢ = 0.08

P_total = 0.1 (0.080) 0.22 atm.
 

Problem : 5  

 The net rate of reaction of the change : 

[Cu(NH₃)₄]²  H₂O  [Cu(NH₃)₃H₂O] ²  NH₃ is, 

 [Cu(NH₃)₄]²  - 3.0 × 10⁵ [Cu(NH₃)₃H₂O] ²  [NH₃] 

calculate : 

(i) rate expression for forward and backward reactions. 

(ii) the ratio of rate constant for forward and backward reactions. 

(iii) the direction of reaction in which the above reaction will be more predominant. 

 Sol. (i) Rate of forward reaction = 2.0 × 10^-4 [Cu(NH₃)₄]²  [H₂O]

Rate of backward reaction = 3.0 × 10⁵ [Cu(NH₃)₃H₂O] ²  [NH₃]

(ii) Also, K_f = 2.0 × 10^-4

K_b = 3.0 × 10⁵

Therefore,  = 6.6 × 10^-10

(iii) More predominant reaction is backward reaction. 

Problem : 6 

The rate law for the decomposition of gaseous N₂O₅, 

N₂O₅(g) → 2NO₂(g) O₂(g) 

is observed to be

r =  = k[N₂O₅]

A reaction mechanism which has been suggested to be consistent with this rate law is

N₂O₅(g)  NO₂(g) NO₃(g) (fast equilibrium)

 NO₂(g) NO(g) O₂(g) (slow)

NO(g) NO₃(g)  2NO₂(g) (fast)

Show that the mechanism is consistent with the observed rate law.

Since the slow step is the rate determining step, hence

r = k₁[NO₂] [NO₃] ...(1)

and from the fast equilibrium step,

Thus, [NO₂] [NO₃] = K[N₂O₅] ...(ii)

Using (ii) in (i), we get :

r = k₁K[N₂O₅] = k[N₂O₅] where k = k₁K

This shows that the mechanism is consistent with the observed rate law.

Problem : 7 

The half life of first order decomposition of nitramide is 2.1 hour at 15°C. 

NH₂NO₂(aq) → N₂O(g) H₂O (l) 

If 6.2 gm of NH₂NO₂ is allowed to decompose, find :

(a) time taken for nitramide to decompose 99%; 

(b) volume of dry N₂O gas produced at this point at STP. 

Sol. (a) Using first order kinetics, we have :

kt = 2.303 log₁₀

⇒  = 2.303 log ⇒ t = 13.96 hours

(b)  6.2 gm of NH₂NO₂≡ 0.1 mol

and 1 mole NH₂NO₂ º 1 mole of N₂O

As 99% of NH₂NO₂ is decomposed

⇒ 0.099 mol of NH₂NO₂ is decomposed

0.099 mol of N₂O are produced º 22.4 × 0.099 = 2.217 L of N₂O at STP.

Problem : 8 

The reaction A OH^- → Products, obeys rate law expression as, 

If initial concentrations of [A] and [OH^-] are 0.002 M and 0.3 M respectively and if it takes 30 sec for 1% A to react at 25°C, calculate the rate constant for the reactions. 

 Sol.  A OH^- → Products

t = 0 0.002 0.3

t = 30

Using

log₁₀

K = 1.12 × 10^-3 L mol^-1 sec^-1 

Problem : 9 

A certain reaction A B → products ; is first order w.r.t. each reactant with k = 5.0 × 10^-3 M^-1s^-1. Calculate the concentration of A remaining after 100s if the initial concentration of A was 0.1 M and that of B was 6.0 M. State any approximation made in obtaining your result. 

Sol. A B → products

Given : Rate = k[A][B] (2^nd Order reaction)

Now, since [B] >> [A], [B] can be assumed to remain constant throughout the reaction. Thus, the rate law for the reaction, becomes :

Rate » k₀[A] where k₀ = k[B] = 5.0 × 10^-3 × 6.0 s^-1 = 3.0 × 10^-2 s^-1

Thus, the reaction is now of first order.

Using, 2.303 log₁₀

⇒ 2.303 log₁₀
 ⇒  [Therefore, log_ex = 2.303 log₁₀x]

⇒

Problem : 10 

Dimethyl ether decomposes according to the following reaction : 

CH₃ - O - CH₃(g) → CH₄(g) CO(g) H₂(g) 

At a certain temperature, when ether was heated in a closed vessel, the increase in pressure with time was noted down. 

(i) Show that the reaction is first order. 

(ii) Compute the pressure of CO(g) after 25 minutes. 

Sol.  CH₃ - O - CH₃ (g) → CH₄(g) CO(g) H₂(g) (all are gases)

 ⇒ P_t = P₀ 2x

⇒ x = (P_t - P₀)

⇒

Now find k₁, k₂ and k₃ using the first order kinetics

k t = 2.303 log₁₀

k₁ = log₁₀ = 0.0129 min^-1

k₂ = log₁₀ = 0.0122 min^-1

k₃ = log₁₀ = 0.0123 min^-1

As k₁ ~ k₂ ~k₃, the reaction is first order.

k_average =  = 0.0127 min^-1

P_CO = x =

Find P after t = 25 min using first order kinetics with k = 0.0127 min^-1

⇒

⇒ P_t = 648.46 mm ⇒ x = 114.23 mm

Problem : 11 

The decomposition of N₂O₅ according to following reaction is first order reaction : 

2N₂O₅(g) → 4NO₂(g) O₂ (g) 

After 30 min. from start of the decomposition in a closed vessel, the total pressure developed is found to be 284.5 mm of Hg and on complete decomposition, the total pressure is 584.5 mm of Hg. Calculate the rate constant of the reaction. 

Sol.  2N₂O₅(g) → 4NO₂(g) O₂(g)

 P₀ : initial pressure ; Let P_t : pressure at 30 min and P_¥ : pressure at the end of decomposition.

⇒ P_t = P₀ 3x ⇒ x = (P_t - P₀)

and P_¥ = 2P₀ =  ⇒ P₀ =

For the first order kinetics

k_eff t = 2.303 log₁₀

A₀ : initial concentration ; A : final concentration

Now  ⇒

⇒ k_eff =  × 2.303 log₁₀ ×  = 5.204 × 10^-3 min^-1

k for the reaction =  = 2.602 × 10^-3 min^-1

Problem : 12 

The gas phase decomposition of N₂O₅ to NO₂ and O₂ is monitored by measurement of total pressure. The following data are obtained. 

Find the average rate of disappearance of N₂O₅ for the time interval between each interval and for the total time interval. [Hint : Integrated rate law is NOT to be used] 

Sol.  2N₂O₅(g) → 4NO₂(g) O₂(g)

Initial Pressure (at t = 0) P₀ 0 0

At equilibrium P₀ - 2x 4x x

Now: P_t = (P₀ - 2x) 4x x ⇒ x =

Thus,  where P_{t 2} and P_{t 1} are the total pressures at time instants t₂ and t₁ (t₂ > t₁) respectively

Problem : 13 

5 ml of ethylacetate was added to a flask containing 100 ml f 0.1 N HCl placed in a thermostat maintained at 30°C. 5 ml of the reaction mixture was withdrawn at different intervals of time and after chilling, titrated against a standard alkali. The following data were obtained : 

Show that hydrolysis of ethyl acetate is a first order reaction. 

Sol. The hydrolysis of ethyl acetate will be a first order reaction if the above data confirm to the equation.

k₁ =

Where V₀, V_t and V_¥ represent the volumes of alkali used at the commencement of the reaction, after time t and at the end of the reaction respectively, Hence

V_∝ - V₀ = 21.05 - 9.62 = 11.43

Time V_∝- V_t k₁ 

75 min 21.05 - 12.10 = 8.95 = 0.003259 min^-1

119 min 21.05 - 13.10 = 7.95  = 0.003051 min^-1

183 min 21.05 - 14.75 = 6.30  = 0.003254 min^-1

A constant value of k shows that hydrolysis of ethyl acetate is a first order reaction.

Problem : 14 

The optical rotations of sucrose in 0.5N HCl at 35°C at various time intervals are given below. Show that the reaction is of first order : 

 Sol. The inversion of sucrose will be first order reaction if the above data confirm to the equation, k₁ = 

log

Where r₀, r_t and r_¥ represent optical rotations initially, at the commencement of the reaction after time t and at the completion of the reaction respectively

In the case a₀ = r₀ - r_¥ = 32.4 - (-11.1) = 43.5

The value of k at different times is calculated as follows :

Time r_t r_t - r_¥  k

10 min 28.8 39.9  = 0.008625 min^-1

20 min 25.5 36.6  = 0.008625 min^-1

30 min 22.4 33.5  = 0.008694 min^-1

40 min 19.6 30.7  = 0.008717 min^-1 

The constancy of k₁ indicates that the inversion of sucrose is a first order reaction.

Problem : 15 

The hydrolysis of ethyl acetate 

CH₃COOC₂H₅ +H₂O  CH₃COOH +C₂H₅OH 

in aqueous solution is first order with respect to ethyl acetate. Upon varying the pH of the solution the first order rate constant varies as follows. 

pH 3 2 1 

k₁ ×10^-4s^-1 1.1 11 110 

what is the order of the reaction with respect of H  and the value of the rate constant? 

Sol. Rate = k[CH₃COOC₂H₅] ^a[H ]^b

[H ] is constant through out the reaction

k₁ = k[H ]^b

Hence,

b = 1

k₁= k[H ]

1.1 × 10^-4 = k(10^-3) ⇒ k = 1.1 × 10^-1 dm³ mol^-1 sec^-1 

Problem : 16 

Two I order reactions having same reactant concentrations proceed at 25°C at the same rate. The temperature coefficient of the rate of the first reaction is 2 and that of second reaction is 3. Find the ratio of the rates of these reactions at 75°C 

 Sol. For I order reaction r₁ = K[C]¹

Therefore,  = temperature coefficient

Let temperature co-efficient be a

Therefore,

Similarly,

For I reaction (R₇₅)_I = 2⁵ × (R₂₅)_I

For II reaction (R₇₅)_II = 3⁵ × (R₂₅)_II
 Therefore,  = 7.9537 [Therefore, (R₂₅)_I = (R₂₅)_II]

Problem : 17 

For the reaction : 

C₂H₅I +OH^- → C₂H₅OH+ I^- 

the rate constant was found to have a value of 5.03 × 10^-2 mol^-1 dm³ s^-1 at 289 K and 6.71 mol^-1 dm³ s^-1 at 333 K. What is the rate constant at 305 K. 

 Sol. k₂ = 5.03 × 10^-2 mol^-1 dm³ s^-1 at T₂ = 289 K

k₁ = 6.71 mol^-1 dm³ s^-1 at T₁ = 333 K

log =

On solving we get, E_a = 88.914 kJ

The rate constant at 305 K may be determined from the relation :

 =

log  =

On solving we get, k₁ = 0.35 mol^-1 dm³ s^-1 

Problem : 18
A secondary alkyl halide (RX) is hydrolysed by alkali simultaneously by S_N1 and S_N2 pathways. A plot of  vs [OH^-] is a straight line with slope equal to 1.0 × 10^-3 mol^-1 L min^-1 and intercept equal to 2.0 × 10^-3 min^-1. Calculate initial rate of consumption of RX when [RX] = 0.5 M and [OH^-] = 1.0 M.  

Sol. 1.5 × 10^-3 mol L^-1 min^-1

For S_N1 pathway:

 K₁ = 1^st order rate constant

For S_N2 pathway:

 = K₂[RX][OH^-] K₂ = 2^nd order rate constant

Thus, the overall rate of consumption of RX is as given below:

= K₁[RX] K₂[RX][OH^-]

or - = K₁ K₂[OH^-]

According to this equation as plot of - vs [OH|-] will be a straight line of the slope equal to K₂ and intercept equal to K₁ . Thus, from question.

K₁ = 2.0 ×10^-3 min^-1

K₂ = 1.0 ×10^-3 min^-1 L min^-1

Thus, - = 2.0 × 10^-3 × 0.5 1.0 × 10^-3 × 0.5 × 1

= 1 × 10^-3 0.5 × 10^-3

= 1.5 × 10^-3 mol L^-1 min^-1 

Problem : 19 

A polymerisation reaction is carried out at 2000 K and the same reaction is carried out at 4000 K with catalyst. The catalyst increases the potential barrier by 20 KJ but the rate of the reaction remains same. Find activation energy of the reaction. 

(Assuming all other parameters to be same.) 

Sol. 0020 

k = Ae^-Ea/R×2000 =

2Ea = Ea 20

Ea = 20 kJ

Problem : 20 

Consider the following first order parallel reaction. 

The concentration of C after time t is :

Sol.

= 3k[A] 5k[A] = 8kA

 = 16 kA

A = A₀ e^{-16 kt} 

,  = 9kA = 9kA₀ e^-16kt

B =

Similarly