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*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 ^{-1}sec^{-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) r_{1} = 1 × 10^{-4} = K[0.1]^{m} [0.1]^{m} ...(1)

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

r_{3} = 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 L^{2} mol^{-2} min^{-1}**

*Problem : 3 *

*The chemical reaction between K _{2}C_{2}O_{4} and HgCl_{2} is ; *

*2HgCl _{2} K_{2}C_{2}O_{4} → 2KCl 2CO_{2} Hg_{2}Cl_{2} *

*The weights of Hg _{2}Cl_{2} precipitated from different solutions in given time were taken and expressed as following :*

Let the rate law be written as : r = k[HgCl_{2}]^{x} [K_{2}C_{2}O_{4}]^{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 P_{0} =0

at time t P_{0} - P¢ 2P¢ P¢/2

P_{total} = P_{0} - P¢_{ } 2P¢ P¢/2 = P_{0}

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(NH _{3})_{4}]^{2 } H_{2}O [Cu(NH_{3})_{3}H_{2}O]^{ 2 } NH_{3} is, *

* [Cu(NH _{3})_{4}]^{2 } - 3.0 × 10^{5} [Cu(NH_{3})_{3}H_{2}O]^{ 2 } [NH_{3}] *

*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.*

**S****ol. **(i) Rate of forward reaction = 2.0 × 10^{-4} [Cu(NH_{3})_{4}]^{2 } [H_{2}O]

Rate of backward reaction = 3.0 × 10^{5} [Cu(NH_{3})_{3}H_{2}O]^{ 2 } [NH_{3}]

(ii) Also, K_{f} = 2.0 × 10^{-4}

K_{b} = 3.0 × 10^{5}

Therefore, = 6.6 × 10^{-10}

(iii) More predominant reaction is **backward reaction. **

*Problem : 6 *

*The rate law for the decomposition of gaseous N _{2}O_{5}, *

*N _{2}O_{5}(g) → 2NO_{2}(g) O_{2}(g)*

is observed to be

r = = k[N_{2}O_{5}]

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

N_{2}O_{5}(g) NO_{2}(g) NO_{3}(g) (fast equilibrium)

NO_{2}(g) NO(g) O_{2}(g) (slow)

NO(g) NO_{3}(g) 2NO_{2}(g) (fast)

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

Since the slow step is the rate determining step, hence

r = k_{1}[NO_{2}] [NO_{3}] ...(1)

and from the fast equilibrium step,

Thus, [NO_{2}] [NO_{3}] = K[N_{2}O_{5}] ...(ii)

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

r = k_{1}K[N_{2}O_{5}] = k[N_{2}O_{5}] where k = k_{1}K

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. *

*NH _{2}NO_{2}(aq) → N_{2}O(g) H_{2}O (l) *

*If 6.2 gm of NH _{2}NO_{2} is allowed to decompose, find :*

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

*(b) volume of dry N _{2}O gas produced at this point at STP.*

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

kt = 2.303 log_{10}

⇒ = 2.303 log ⇒ t = 13.96 hours

**(b) ** 6.2 gm of NH_{2}NO_{2}≡ 0.1 mol

and 1 mole NH_{2}NO_{2} º 1 mole of N_{2}O

As 99% of NH_{2}NO_{2} is decomposed

⇒ 0.099 mol of NH_{2}NO_{2} is decomposed

0.099 mol of N_{2}O are produced º 22.4 × 0.099 = 2.217 L of N_{2}O 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

log_{10}

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^{-1}s^{-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 » k_{0}[A] where k_{0} = 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 log_{10}

⇒ 2.303 log_{10}

⇒ [Therefore, log_{e}x = 2.303 log_{10}x]

⇒

*Problem : 10*

**Dimethyl ether decomposes according to the following reaction : **

**CH _{3} - O - CH_{3}(g) → CH_{4}(g) CO(g) H_{2}(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. ** CH_{3} - O - CH_{3} (g) → CH_{4}(g) CO(g) H_{2}(g) (all are gases)

⇒ P_{t} = P_{0} 2x

⇒ x = (P_{t} - P_{0})

⇒

Now find k_{1}, k_{2} and k_{3} using the first order kinetics

*k* t = 2.303 log_{10}

k_{1} = log_{10} = 0.0129 min^{-1}

k_{2} = log_{10 } = 0.0122 min^{-1}

k_{3} = log_{10} = 0.0123 min^{-1}

As k_{1} ~ k_{2} ~k_{3}, 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 N _{2}O_{5} according to following reaction is first order reaction : *

*2N _{2}O_{5}(g) → 4NO_{2}(g) O_{2} (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. ** 2N_{2}O_{5}(g) → 4NO_{2}(g) O_{2}(g)

P_{0} : initial pressure ; Let P_{t} : pressure at 30 min and P_{¥ }: pressure at the end of decomposition.

⇒ P_{t} = P_{0} 3x ⇒ x = (P_{t} - P_{0})

and P_{¥ }= 2P_{0} = ⇒ P_{0} =

For the first order kinetics

k_{eff }t = 2.303 log_{10}

A_{0} : initial concentration ; A : final concentration

Now ⇒

⇒ k_{eff} = × 2.303 log_{10} × = 5.204 × 10^{-3} min^{-1}

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

*Problem : 12*

*The gas phase decomposition of N _{2}O_{5} to NO_{2} and O_{2} is monitored by measurement of total pressure. The following data are obtained. *

P total(atm) | 0.154 | 0.215 | 0.260 | 0.315 | 0.346 |

Time (sec) | 1 | 52 | 103 | 205 | 309 |

*Find the average rate of disappearance of N _{2}O_{5} for the time interval between each interval and for the total time interval. [Hint : Integrated rate law is NOT to be used]*

**Sol. ** 2N_{2}O_{5}(g) → 4NO_{2}(g) O_{2}(g)

Initial Pressure (at t = 0) P_{0} 0 0

At equilibrium P_{0} - 2x 4x x

Now: P_{t} = (P_{0} - 2x) 4x x ⇒ x =

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

k_{1} =

Where V_{0}, 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_{∝} - V_{0} = 21.05 - 9.62 = 11.43

**Time V**_{∝}**- V _{t} k_{1} **

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, k

log

Where r_{0}, 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 a_{0} = r_{0} - 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 k_{1} indicates that the inversion of sucrose is a **first order **reaction.

**Problem : 15 **

*The hydrolysis of ethyl acetate *

*CH _{3}COOC_{2}H_{5} +H_{2}O CH_{3}COOH +C_{2}H_{5}OH *

*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 *

*k _{1 }×10^{-4}s^{-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[CH_{3}COOC_{2}H_{5}]^{ a}[H^{ }]^{b}

[H^{ }] is constant through out the reaction

k_{1} = k[H^{ }]^{b}

Hence,

b = 1

k_{1}= k[H^{ }]

1.1 × 10^{-4} = k(10^{-3}) ⇒ **k = 1.1 × 10 ^{-1} dm^{3} 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 r_{1} = K[C]^{1}

Therefore, = temperature coefficient

Let temperature co-efficient be a

Therefore,

Similarly,

For I reaction (R_{75})_{I} = 2^{5} × (R_{25})_{I}

For II reaction (R_{75})_{II} = 3^{5} × (R_{25})_{II}

Therefore, = **7.9537 **[Therefore, (R_{25})_{I} = (R_{25})_{II}]

*Problem : 17*

**For the reaction : **

**C _{2}H_{5}I +OH^{-} → C_{2}H_{5}OH+ I^{-} **

**the rate constant was found to have a value of 5.03 × 10 ^{-2} mol^{-1} dm^{3} s^{-1} at 289 K and 6.71 mol^{-1} dm^{3} s^{-1} at 333 K. What is the rate constant at 305 K. **

**Sol. **k_{2} = 5.03 × 10^{-2} mol^{-1} dm^{3} s^{-1} at T_{2} = 289 K

k_{1} = 6.71 mol^{-1} dm^{3} s^{-1} at T_{1} = 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, k_{1} = 0.35 mol^{-1} dm^{3} s^{-1}** **

*Problem : 18***A secondary alkyl halide (RX) is hydrolysed by alkali simultaneously by S _{N}1^{ }and S_{N}2 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_{N}1 pathway:

K_{1} = 1^{st} order rate constant

For S_{N}2 pathway:

= K_{2}[RX][OH^{-}] K_{2 }= 2^{nd} order rate constant

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

= K_{1}[RX] K_{2}[RX][OH^{-}]

or - = K_{1} K_{2}[OH^{-}]

According to this equation as plot of - vs [OH|-] will be a straight line of the slope equal to K_{2} and intercept equal to K_{1} . Thus, from question.

K_{1} = 2.0 ×10^{-3} min^{-1}

K_{2} = 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 = A_{0} e^{-16 kt }

, = 9kA = 9kA_{0} e^{-16kt}

B =

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