Thermodynamics Test

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Thermodynamics Test - 24

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Weekly Quiz Competition

Question 1
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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 O₂ is

A
249.2 kJ mol^-1

B
498.4 kJ mol^-1

C
747.6 kJ mol^-1

D
-249.2 kJmol^-1
Question 2
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Δ_fH° of CS₂= 117.36 kJ mol^-1
C(g) = 716.682 kJ mol^-1
S(g) = 278.805 kJ mol^-1
Q. Thus, bond enthalpy in CS₂ is

A
1156.93 kJ mol^-1

B
578.47 kJ mol^-1

C
-578.47 kJ mol^-1

D
117.36 kJ mol^-1

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
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The dissociation energy of CH₄ and C₂H₆ to convert them into gaseous atoms are 360 and 620 kcal mol^-1 respectively. Thus, bond energy of (C—C) bond is

A
260 kcal mol^-1

B
180 kcal mol^-1

C
130 kcal mol^-1

D
80 kcal mol^-1

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
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The (C— Cl) bond energy of CCI₄ (l) can be derived from thermochemical data

A

B

C

D
Question 5
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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)

A
72 kJ mol^-1

B
-72 kJ mol^-1

C
279 kJ mol^-1

D
-279 kJ mol^-1

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 CH₂- 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
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Given at 298 K
Q. The resonance energy of benzene is

A
-152.0 kJ mol^-1

B
+152.0 kJ mol^-1

C
-107.0 kJ mol^-1

D
-226.0 kJ mol^-1

Solution

C₆H₁₀(l) + H₂(g) → C₆H₁₂(l) , ∆hydrogenationH° = -119 kJ mol-1
6C(s) + 6H₂(g) → C₆H12(l) , ∆_fH° = -156 kJ mol-1
6C(s) + 3H₂(g) → C₆H₆(l) , ∆fH° = +49 kJ mol-1
C6H6(l) → C₆H₁₂(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
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Given the following thermochemical data at 298 K and 1 bar
ΔH°_vap (CH₃OH) = 38.0 kJ mol^-1
Δ_fH°: 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_fH° of liquid methyl alcohol in kJ mol^-1 is

A
-170 kJ mol^-1

B
+170 kJ mol^-1

C
-266 kJ mol^-1

D
+266 kJ mol^-1
Question 8
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Heat of hydrogenation of ethene is x₁ and that of benzene is x₂. The resonance energy of benzene is

A
x₁ — x₂

B
x₁ + x₂

C
3x₁ — x₂

D
x₁ - 3x₂

Solution

Energy of ethene =x₁
Number of ethene in benzene=3.
Therefore, Observed energy of hydrogenation of benzene =3x₁
Calculated energy of hydrogenation of benzene =x₂ (Given)
Therefore,
Resonance energy = observed - calculated =3x₁−x₂
Question 9
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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

A
286.8 kJ

B
-286.8 kJ

C
891.6 kJ

D
445.8 kJ

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
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The standard enthalpies of formation of SF₆ (g), S (g)and F (g) are -1100, + 275 and + 80 kJ mol^-1. Thus, average bond energy of (S—F) in SF₆ is

A
309.16 kJ mol^-1

B
1855 kJ mol^-1

C
11.130 x 10³ kJ mol^-1

D
-309.16 kJ mol^-1

Solution

S(g) + 6F(g) → SF₆(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
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Calculate resonance energy of N₂O from the following data
Δ_fH° (N₂O) = 82 kJ mol^-1

A
-358 kJ mol^-1

B
170 kJ mol^-1

C
-88 kJ mol^-1

D
-170 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
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Given, Δ_fH° (H₂S) = 20.1 kJ mol^-1
Δ_fH° (H)(g) = +218.0 kJ mol^-1
Δ_fH° (S) (g) = +277.0 kJ mol^-1
and bond dissociation energy of first (H—S) bond in H₂S is 376.6 kJ mol^-1. Thus, Δ_fH° (HS)is

A
138.5 kJ mol^-1

B
-138.5 kJ mol^-1

C
-501.1 kJ mol^-1

D
-515.1 kJ mol^-1
Question 13
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Given, BE (H—H) - x₁, BE = x₂, BE (O—H) = x₃ and for H₂O (l) → H₂O (g), ΔH = x₄ mol^-1, then Δ_fH° [H₂O (l)] is

A

B

C

D
2x₁ + x₂ - 4x₃ - 2x₄

Solution

Required reaction= Heat of formation of water – H2(g) +1/2O₂ (g)= H₂O(l)
WRITE THE EQUATION PROPERLY FOR EACH EQUATION
rearranging the reactions we have to get the above equation
B.E. of H--H= H₂(g) → 2H(g)..........X1
B.E of O₂ = O₂(g) ---> 2O(g)...........X2
B.E. OF O-H = O-H(aq)--> O(g)+H(g) …......X3
LATENT HEAT OF VAPOURISATION OF WATER = H₂O(l)----> H₂O(g)..........X4
THUS REVERSING EQTN X₃ AND X₄ , HALVING EQUATION X₂ AND DOUBLING EQUATION X₃ WE WILL GET THE DESIRED EQUATION.
Question 14
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Direction (Q. Nos. 14) 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

A
a

B
b

C
c

D
d

Solution

∆H = ∑(B.E)_reactant - ∑(B.E)_product
For CH₄
C(g) + 4H(g) → CH₄
∆H = (718.4) + 4(217.95) - (-74.9)
4 C-H = 1665.1
C-H = 416.3
For C₂H₆
2C(g) + 6H(g) → C₂H₆(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.