Theromodynamics and Thermochemistry Test

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Theromodynamics and Thermochemistry Test - 7

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Weekly Quiz Competition

Question 1
1 / -0

The gas absorbs 100 J heat and is simultaneously compressed by a constant external pressure of 1.50 atm from 8 litres to 2 litres. Hence ΔE will be

A
–812 J

B
812 J

C
1009 J

D
911 J

Solution

ΔQ_absorb = 100 J
ΔH = +ΔQ_absorb = 100 J
P_ext = 1.50 atm
V₁ = 8
V₂ = 2
ΔE = ΔH + PΔV
= 100J + 1.50 (8 – 2) lit-atm
= 100J+9(101J) = 1009 J nearly
Question 2
1 / -0

A pressure cooker reduces cooking time for food because

A
heat is more evenly distributed in the cooking space

B
boiling point of water involved in cooking is increased

C
the higher pressure inside the cooker crushed the food materia

D
cooking involves chemical changes helped by a rise in temperature

Solution

High pressure inside the cooker increases boiling point, thus heat consistently given is used up by food material instead in boiling and constant boiling evaporation.
Hence, option B is correct.
Question 3
1 / -0

A system is changed from state A to state B by one path and from B to A by another path. If ΔE₁ and ΔE₂ are the corresponding changes in internal energy, then -

A
ΔE₂ + E₂= + ve

B
ΔE₁ + ΔE₂ = – ve

C
ΔE₁ + ΔE₂ = 0

D
None of the above

Solution

ΔE₁ = – ΔE₂
or ΔE₁ + ΔE₂ = 0
Question 4
1 / -0

First four ionization energies of Ni and Pt are given below:
(IE₁ + IE₂) (IE₃ + IE₄)
Ni 2.49 × 10³ kJ mol^–1 8.8 × 10³ kJ mol^–1
Pt 2.66 × 10³ kJ mol^–1 6.7 × 10³ kJ mol^–1
From the data it can be concluded that

A
Ni(II) compounds are thermodynamically more stable than Pt(II) compounds

B
Pt(IV) compounds are thermodynamically more stable than Ni(IV) compounds

C
both correct

D
none of these are correct

Solution

Smaller is the ionization energy of a metal to give a particular oxidation state, greater will be thermodynamic stability of that oxidation state.
Hence, option C is correct.
Question 5
1 / -0

The heat of combustion of sucrose, C₁₂H₂₂O₁₁(s) at constant volume is –1348.9 kcal mol^–1 at 25C, then the heat of reaction at constant pressure, when steam is produced, is

A
– 1342.344 kcal

B
+ 1342.344 kcal

C
+ 1250 kcal

D
None

Solution

The combustion equation of sucrose is
C₁₂H₂₂O₁₁(s) + 12O₂(g) → 12CO₂(g) + 11H₂O (g)
Here,
Δn = sum of gaseous product moles – sum of gaseous reactant moles
Δn = 12 + 11 – 12
Δn = 11
As we Know
ΔH = ΔE + Δn RT, Where ΔH = heat of reaction at constant pressure
ΔE = heat of reaction at constant volume
Here, ΔE = – 1348.9 kcal
R = 2.0 cal, T = 25+273 = 298 K
∴ ΔH = (–1348.9 × 1000) + 11× 2 × 298 = – 1348900 + 6556 = – 1342344 cal = – 1342.344 kcal
Question 6
1 / -0

Given the bond energies N ≡ N, H – H and N – H bonds are 945, 436 and 391 kJ mole^–1 respectively, the enthalpy change of the reaction -
N₂(g) + 3H₂(g) → 2NH₃(g) is -

A
– 93 kJ

B
102 kJ

C
90 kJ

D
105 Kj

Solution

ΔH = (945 + 3 × 436) – (2 × 3 × 391)
= 2253 – 2346
–93 KJ
Question 7
1 / -0

The enthalpies of combustion of carbon and carbon monoxide are -393.5 and -283 kJ mol^-1 respectively. The enthalpy of formation of carbon monoxide per mole is

A
110.5 kJ

B
-110.5 kJ

C
-676.5 kJ

D
676.5 kJ

Solution

C + O₂--→ CO₂ ΔH = -393.5 kJ
2CO + ½ O₂--→ 2CO₂ ΔH = -283 kJ
2C + O₂--→ 2CO ΔH = -110 kJ
Question 8
1 / -0

In the following table, which are correct–
ΔH ΔS Nature of reaction

A
( – ) ( + ) Spontaneous only at high temperature

B
( + ) ( – ) Nonspontaneous regardless of temperature

C
( + ) ( + ) Spontaneous only at low temperature

D
( – ) ( – ) Spontaneous at all Temperatures

Solution

∴ from ΔG = ΔH – TΔS
If ΔG < 0 – spontaneous
If ΔG > 0 – Non spontaneous
If ΔG = 0 – equilibrium.

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Theromodynamics and Thermochemistry Test - 7

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Question 1

–812 J

812 J

1009 J

911 J

Solution

Question 2

heat is more evenly distributed in the cooking space

boiling point of water involved in cooking is increased

the higher pressure inside the cooker crushed the food materia

cooking involves chemical changes helped by a rise in temperature

Solution

Question 3

ΔE2 + E2 = + ve

ΔE1 + ΔE2 = – ve

ΔE1 + ΔE2 = 0

None of the above

Solution

Question 4

Ni(II) compounds are thermodynamically more stable than Pt(II) compounds

Pt(IV) compounds are thermodynamically more stable than Ni(IV) compounds

both correct

none of these are correct

Solution

Question 5

– 1342.344 kcal

+ 1342.344 kcal

+ 1250 kcal

None

Solution

Question 6

– 93 kJ

102 kJ

90 kJ

105 Kj

Solution

Question 7

110.5 kJ

-110.5 kJ

-676.5 kJ

676.5 kJ

Solution

Question 8

( – ) ( + ) Spontaneous only at high temperature

( + ) ( – ) Nonspontaneous regardless of temperature

( + ) ( + ) Spontaneous only at low temperature

( – ) ( – ) Spontaneous at all Temperatures

Solution

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ΔE₂ + E₂= + ve

ΔE₁ + ΔE₂ = – ve

ΔE₁ + ΔE₂ = 0