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*Problem : 1*

**In gaseous reactions important for understanding of the upper atmosphere H _{2}O and O react bimolecularly to form two OH radicals. DH for this reaction is 72kJ/mol at 500 K and E_{a} = 77 kJ/mol, then E_{a} for two bimolecular recombination of 2OH radicals to form H_{2}O & O is **

**(A) 3 kJ mol ^{-1}_{ }**

(B) 4 kJ mol^{-1}

(C) 5 kJ mol^{-1}

(D) 7 kJ mol^{-1 }

**Sol. **As Î”H is positive, therefore reaction is endothermic

This is the energy profile diagram for an endothermic reaction.

Now when the products is converted back to reactant the energy of activation is x as shown in fig.

Evidently x = E_{a} - Î”H

= (77 - 72) = 5 kJ mol^{-1} Therefore, (C)

*Problem : 2*

**In a certain reaction 10% of the reactant decomposes in the first hour, 20% is second hour, 30% in third hour and so on. What are the dimensions of rate constant. **

**(A) hour ^{-1 } (B) mol lit^{-1} sec^{-1} (C) lit mol^{-1} sec^{-1} (D) mol sec^{-1} **

**Sol. **If the amount of products formed which is 10%, 20% and 30% is plotted against time i.e., 1 hr, 2 hr and 3 hr respectively, it is a straight line passing through the origin.

Therefore, it is a zero order reaction where x = kt â‡’ = k

Therefore, dimensions of k = moles lit^{-1} sec^{-1}

Therefore, (B)

*Problem : 3*

*Two substances A(tÂ½ = 5 mins) and B(tÂ½ = 15 mins) are taken is such a way that initially [A] = 4[B]. The time after which the conentration of both will be equal is *

*(A) 5 min *

*(B) 15 min *

*(C) 20 min*

*(D) concentration can never be equal*

**Sol. **tÂ½ of A is 5 min

Therefore, in 15 mins it will become 1/8 of initial and tÂ½ of B is 15 mins

Therefore, in 15 mins it will become Â½ of initial

Therefore, ratio of [A] : [B] after 15 min is 4 : 1

But given [A] = 4[B]

Therefore, [A] = [B] after 15 min

Therefore, [B]

*Problem : 4 *

*The reaction A(g) 2B(g) â†’ C(g) D(g) is an elementary process. In an experiment, the initial partial pressure of A & B are P _{A} = 0.60 and P_{B} = 0.80 atm. When P_{c} = 0.2 atm the rate of reaction relative to the initial rate is *

*(A) 1/48 *

*(B) 1/24 *

*(C) 9/16 *

*(D) 1/6*

**Sol. **

(Rate)_{i} = k[A] [B]^{2} = k[0.6][0.8]^{2}

(Rate)_{t} = k[0.4] [0.4]^{2}

= Therefore, [D]

*Problem : 5 *

*For a hypothetical reaction A B â†’ C D, the rate = k[A] ^{-1/2}[B]^{3/2}. On doubling the concentration of A and B the rate will be *

*(A) 4 times *

*(B) 2 times *

*(C) 3 times *

*(D) none of these *

** Sol.** k = k[2]

Therefore, [B]

*Problem : 6 *

**An organic compound A decomposes by following two parallel first order mechanisms :**

** ; , k _{1} = 0.693 hr ^{-1} **

**Select the correct statement(s) **

**(A) If three moles of A are completely decomposed then 2 moles of B and 1 mole of C will be formed. **

**(B) If three moles of A are completely decomposed then 1 moles of B and 2 mole of C will be formed. **

**(C) half life for the decomposition of A is 20 min **

**(D) half life for the decomposition of B is 0.33 min **

**Sol. BC **

*Problem : 7*

**For any I ^{st } order gaseous reaction A â†’ 2B Pressure devoloped after 20 min is 16.4 atm and after very long time is 20 atm. What is the total pressure developed after 10 min. **

**(A) 12 **

**(B) 13 **

**(C) 14 **

**(D) 15 **

**Sol. C**

A(g) 2B(g)

t = 0 P_{O} -

t = 10 P_{O} - P 2P

t = 20 P_{O} - P' 2P'

- 2P_{O }

2P_{O} = 20 P_{O} = 10

16.4 = P_{O} - P' 2P'

16.4 = 10 P'

P' = 6.4

for first order reaction at equal time conc/pr is in G. P.

,

3.6 Ã— 10 = (10 - P)^{2}

10 - P = 6

P = 4

P_{10 min} = P_{O} P = 10 4 = 14 atm

*Problem : 8*

**For any reaction, 2A â†’ B, rate constant of reaction is 0.231 min ^{-1}. Time (in sec) when 25% of A will remain unreacted. **

**(A) 150 **

**(B) 180 **

**(C) 200 **

**(D) 140 **

**Sol.** **B **

,

= 3 mm

= 180 sec

*Problem : 9*

**For any reaction A(g) â†’ B(g), rate constant k = 8.21 Ã— 10 ^{-2} atm/min at 300 K. If initial concentration of A is 2M then what is the half life (in hr.)? **

**(A) 5 **

**(B) 6 **

**(C) 7 **

**(D) 8 **

**Sol. A**

Rate = k =

t_{1/2} = = 300 min

= 5 hr.

*Problem : 10 *

**2A B C **

**The mechanism of the above reaction given as **

**2A X 2B (fast) **

**X B C (slow) **

**E _{1} = Activation energy for K_{1} **

**E _{2} = Activation energy for K_{2 }**

**E _{3} = Activation energy for K_{3} **

**Calculate E _{reaction} **

**(given : E _{1} = 10 kJ, E_{3} = 5 kJ, E_{2} = 12 kJ) **

**(A) 5 **

**(B) 6 **

**(C) 7 **

**(D) 3 **

**Sol.** **D**

rate = K_{3} [x][B]

, [X] =

rate = k_{3} . , = k_{3} . .

K_{eff } = , E_{total }= E_{1} E_{3} - E_{2}

= 3kJ

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