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QUESTION: 1

**Direction (Q. Nos. 1-13) This section contains 13 multiple choice questions. Each question has four choices (a), (b), (c) and (d), out of which ONLY ONE option is correct.**

**Q. Thus, (BE) of bond in O _{2} is**

Solution:

QUESTION: 2

Δ_{f}H° of CS_{2 }= 117.36 kJ mol^{-1}

C(g) = 716.682 kJ mol^{-1}

S(g) = 278.805 kJ mol^{-1}

**Q. Thus, bond enthalpy in CS _{2} is**

Solution:

∆H = (B.E)_{reactant} - (B.E)_{product}

117.36 = 716.682 + 2(278.805) - ∆H_{2C=S}

∆H _{2C=S} = 1156.932

∆H_{C=S} = 1156.932/2 = 578.74 kJ mol^{-1}

QUESTION: 3

The dissociation energy of CH_{4} and C_{2}H_{6} to convert them into gaseous atoms are 360 and 620 kcal mol^{-1} respectively. Thus, bond energy of (C—C) bond is

Solution:

In CH4 there are 4-C-H bonds.

Let C-H bond energy be X. So 4X = 360. or X = 90.

Now, in C2H6 there are 6 C-H bonds and 1C-C bond.

Let C-C B. E be y.

So 6X + Y = 620. or Y = 620-6×90 = 80.

QUESTION: 4

The (C— Cl) bond energy of CCI_{4} (l) can be derived from thermochemical data

Solution:

QUESTION: 5

Bond dissociation energy of = + 590 kJ mol^{-1} and that of (C—C) = + 331 kJ mol^{-1} at 298 K.

Enthalpy change for the polymerisation of ethene to polyethene is (where n is large integral value)

Solution:

The correct answer is option B

During the polymerization of ethylene, one mole of ethylene breaks i.e. one C = C double bond breaks and the two CH_{2}- groups are linked with C-C single bonds thus forming three single bonds.

But in the whole unit of polymer, number of single C-C bonds formed/mole of ethylene is 2.

Energy due to formation of 2 C-C single bonds = 2 X 331 = 662 kJ/mol.

Energy due to dissociation of 1 C = C double bond = 590 kJ/mol.

Thus, the enthalpy of polymerisation per mole of ethylene at 298 K = (590-662) = -72 kJ/mol

QUESTION: 6

Given at 298 K

**Q. The resonance energy of benzene is**

Solution:

C_{6}H_{10}(l) + H_{2}(g) → C_{6}H_{12}(l) , ∆hydrogenationH° = -119 kJ mol-1

6C(s) + 6H_{2}(g) → C_{6}H12(l) , ∆_{f}H° = -156 kJ mol-1

6C(s) + 3H_{2}(g) → C_{6}H_{6}(l) , ∆fH° = +49 kJ mol-1

C6H6(l) → C_{6}H_{12}(l) , ∆hydrogenationH°

(∆_{hydrogenation}H°)_{Actual} = -156-49 = -2055 kJ mol-1

Hypothetical heat of = (heat of hydrogenation = 3 × (Heat of hydrogenation

hydrogenation of benzene of cyclohexatriene) of cyclohexene)

= 3(-119) = -357 kJmol-1

Here, we have assumed that there are only double bonds in benzene(Kekule structure) and in hydrogenating it, energy released will be roughly 3 times of hydrogenating cyclohexene

Clearly, heat of hydrogenation, for theoretical benzene is more negative and thus it is less stable. So, resonance energy should be negative(since actual benzene is stable)

Resonance energy = -357 -(-205) = -152 kJ mol^{-1}

QUESTION: 7

Given the following thermochemical data at 298 K and 1 bar

ΔH°_{vap} (CH_{3}OH) = 38.0 kJ mol^{-1}

Δ_{f}H°: H(g) = 218 kJ mol^{-1}

O(g) = 249 kJ mol^{-1}

C(g) = 715 kJ mol^{-1}

Bond dissociation energy

(C—H) = 415 kJ mol^{-1}

(C—O) = 356 kJ mol^{-1}

(O—H) = 4 6 3 kJ mol^{-1}

**Q. The Δ _{f}H° of liquid methyl alcohol in kJ mol^{-1} is**

Solution:

QUESTION: 8

Heat of hydrogenation of ethene is x_{1} and that of benzene is x_{2}. The resonance energy of benzene is

Solution:

Energy of ethene =x_{1}

Number of ethene in benzene=3.

Therefore, Observed energy of hydrogenation of benzene =3x_{1}

Calculated energy of hydrogenation of benzene =x_{2} (Given)

Therefore,

Resonance energy = observed - calculated =3x_{1}−x_{2}

QUESTION: 9

Given, BE of = 498.8 kJ mol^{-1}

BE of (O—O) in ozone = 302.3 kJ mol^{-1}

**Q. What is enthalpy change of the reaction**

** **

Solution:

3O2(g) → 2O3(g)

∆H = 3(B.E)_{O2} - 2(B.E)_{O3}

= 3(498.8) - 2(302.3)

= 287.2 kJ ml-1

QUESTION: 10

The standard enthalpies of formation of SF_{6} (g), S (g)and F (g) are -1100, + 275 and + 80 kJ mol^{-1}. Thus, average bond energy of (S—F) in SF_{6} is

Solution:

S(g) + 6F(g) → SF_{6}(g)

∆rH = ∑∆Hreactant - ∑∆H_{product}

= (275+6 80) - (-1100)

= 1855

So, bond energy of (S—F) = 1855/6 = 309.16 kJ mol^{-1}

QUESTION: 11

Calculate resonance energy of N_{2}O from the following data

Δ_{f}H° (N_{2}O) = 82 kJ mol^{-1}

Solution:

= 946 + 249 - 607 - 418 = 1195 - 1025 = 170 kJ mo^{l-1}

Resonance energy = Observed heat of formation – calculated heat of formation

= 82 -170 = 88 kJ mol^{-1}

QUESTION: 12

Given, Δ_{f}H° (H_{2}S) = 20.1 kJ mol^{-1}

Δ_{f}H° (H)(g) = +218.0 kJ mol^{-1}

Δ_{f}H° (S) (g) = +277.0 kJ mol^{-1}

and bond dissociation energy of first (H—S) bond in H_{2}S is 376.6 kJ mol^{-1}. Thus, Δ_{f}H° (HS)is

Solution:

QUESTION: 13

Given, BE (H—H) - x_{1}, BE = x_{2}, BE (O—H) = x_{3} and for H_{2}O (l) → H_{2}O (g), ΔH = x_{4} mol^{-1}, then Δ_{f}H° [H_{2}O (l)] is

Solution:

Required reaction= Heat of formation of water – H2(g) +1/2O_{2} (g)= H_{2}O(l)

WRITE THE EQUATION PROPERLY FOR EACH EQUATION

rearranging the reactions we have to get the above equation

B.E. of H--H= H_{2}(g) → 2H(g)..........X1

B.E of O_{2} = O_{2}(g) ---> 2O(g)...........X2

B.E. OF O-H = O-H(aq)--> O(g)+H(g) …......X3

LATENT HEAT OF VAPOURISATION OF WATER = H_{2}O(l)----> H_{2}O(g)..........X4

THUS REVERSING EQTN X_{3} AND X_{4} , HALVING EQUATION X_{2} AND DOUBLING EQUATION X_{3} WE WILL GET THE DESIRED EQUATION.

QUESTION: 14

**Direction (Q. Nos. 14 -15) This section contains a paragraph, wach describing theory, experiments, data etc. three Questions related to paragraph have been given.Each question have only one correct answer among the four given ptions (a),(b),(c),(d).**

**Based on the following thermochemical data of the given process, answer the questions.**

**Q. Bond energy of (C— C) bond is **

Solution:

For C_{3}H_{8} : 3C + 4H_{2} → C_{3}H_{8} ; ∆H_{1}

For C_{3}H_{8} : 2C + 3H_{2} → C_{2}H_{6} ; ∆H_{2}

∆H_{1} = -[2(C-C)+8(C-H)]+[3C_{s→g}+4(H-H)] -(I)

∆H_{2 }= -[1(C-C)+6(C-H)]+[2C_{s→g}+3(H-H)] -(2)

Let bond energy of C-C be x kcal and bond energy of C-H be y kcal.

From eqn I and 2

∆H_{1} = -[2x+8y]+[3×172+4×104] -(3)

∆H_{2} = -[x+6y]+[2×172+3×104] -(4)

Also given, C+O_{2} → CO_{2}; ∆H = -94.0 kcal -(5)

H_{2} + ½ O2 → H_{2}O ; ∆H = -68.0 kcal -(6)

C_{2}H_{6} + 7/2 O_{2} → 2CO_{2} + 3H_{2}O ; ∆H = -372 kcal -(7)

C_{3}H_{8} + 5O_{2} → 3CO_{2} + 4H_{2}O ; ∆H = -530 kcal-(8)

By inspection method,2 × (5) + 3 × (6) - (7) gives

2C + 3H_{2} → C_{2}H_{6} ; ∆H_{2} = -20 kcal

And 3×(5) + 4×(6) -(8)

3C + 4H_{2} → C_{3}H_{8} ; ∆H_{3} = -20 kcal

By eqn. (3), (4), (9) and (10)

x+6y = 676

2x+8y = 956

x = 82 kcal and y = 99 kcal

QUESTION: 15

**Based on the following thermochemical data of the given process, answer the questions.**

**Q. Bond energy of (C—H) bond is **

Solution:

For C_{3}H_{8} : 3C + 4H_{2} → C_{3}H_{8} ; ∆H_{1}

For C_{3}H_{8} : 2C + 3H_{2} → C_{2}H_{6} ; ∆H_{2}

∆H_{1} = -[2(C-C)+8(C-H)]+[3C_{s→g}+4(H-H)] -(I)

∆H_{2 }= -[1(C-C)+6(C-H)]+[2C_{s→g}+3(H-H)] -(2)

Let bond energy of C-C be x kcal and bond energy of C-H be y kcal.

From eqn I and 2

∆H_{1} = -[2x+8y]+[3×172+4×104] -(3)

∆H_{2} = -[x+6y]+[2×172+3×104] -(4)

Also given, C+O_{2} → CO_{2}; ∆H = -94.0 kcal -(5)

H_{2} + ½ O_{2} → H_{2}O ; ∆H = -68.0 kcal -(6)

C_{2}H_{6} + 7/2 O_{2} → 2CO_{2} + 3H_{2}O ; ∆H = -372 kcal -(7)

C_{3}H_{8} + 5O2 → 3CO_{2} + 4H_{2}O ; ∆H = -530 kcal -(8)

By inspection method, 2 × (5) + 3 × (6) - (7) gives

2C + 3H_{2} → C_{2}H_{6} ; ∆H_{2} = -20 kcal

And 3×(5) + 4×(6) -(8)

3C + 4H_{2} → C_{3}H_{8} ; ∆H_{3} = -20 kcal

By eqn. (3), (4), (9) and (10)

x+6y = 676

2x+8y = 956

x = 82 kcal and y = 99 kcal

*Answer can only contain numeric values

QUESTION: 16

**Direction (Q. Nos. 16 and 17) This section contains 2 questions. when worked out will result in an integer from 0 to 9 (both inclusive)**.

Given bond dissociation energy under standard states.

**Q. i. Number of and bonds in (A) is ......**

Solution:

Let us assume the structures as given below and calculate bond energies,

Let us assume the given structures of C_{6}H_{6} are:

A + 4H2 ---> n - hexane

H =5101+4(435) -7451 = -610KJ/mol

B+4H2 ---> n - hexane

H =5138+4(435)-74551 =-573KJ/mol

Hence our assumed structures are correct.

*Answer can only contain numeric values

QUESTION: 17

Given bond dissociation energy under standard states.

**Q. ii. Number of and ) bond in (B) is .......**

Solution:

Number of bonds between C and C are 3

So the correct answer is 3.

QUESTION: 18

**Direction (Q. Nos. 18) Choice the correct combination of elements and column I and coloumn II are given as option (a), (b), (c) and (d), out of which ONE option is correct.**

**Q. Given thermochemical data under standard states at 278 K and 1 bar.**

Match the types of bonds in Column I with their respective value of bond-energy in Column II and select answer from the codes given below the table

Solution:

∆H = ∑(B.E)_{reactant} - ∑(B.E)_{product}

For CH_{4}

C(g) + 4H(g) → CH_{4}

∆H = (718.4) + 4(217.95) - (-74.9)

4 C-H = 1665.1

C-H = 416.3

For C_{2}H_{6}

2C(g) + 6H(g) → C_{2}H_{6}(g)

∆H = 2(718.4) + 6(217.95) - (-84.7)

C-C+ 6C-H = 2828.5

C-C = (2828.5 - 2497.8) = 331.4

From here, we can see that option c is correct.

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