Thermodynamics Test

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Thermodynamics Test - 22

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Weekly Quiz Competition

Question 1
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Direction (Q. Nos. 1-9) This section contains 9 multiple choice questions. Each question has four choices (a), (b), (c) and (d), out of which ONLY ONE option is correct.
Q. The difference between Δ_rH° and Δ_rE° (in kcal) for the reaction
at 298 K in kcal is

A
-1.788

B
+1.788

C
0.596

D
-0.596

Solution

∆rH° = ∆rE° + ∆ngRT
∆rH° - ∆rE° = ∆ngRT
= (12-15)× 2× 298
= -1.788 kcal
Question 2
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The Δ_rH° for CO₂(g), CO(g) , and H₂O(g) are - 393.5°, - 110.5 and - 241.8 kJ mol^-1, respectively. Thus, Δ_rH° for the reaction (in kJ)
[IITJEE 2000]

A
+524.1

B
+41.2

C
-262.5

D
-41.2

Solution

∆H= ∆H_CO+∆H_H2O-∆H_CO2 (as ΔH for H₂ is 0)
= -110.5-241.8+393.5 = +41.2 kJ mol^-1
Question 3
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Enthalpies of formation of CO(g) , CO₂ (g) , N₂O (g) and N₂O₄ (g) are -110, - 393, 81 and 9.7 kJ mol^-1. Thus, Δ_rU for the reaction at 298 K is,

A
+777.7 KJ

B
- 777.7 KJ

C
-211.7 KJ

D
+211.7 KJ

Solution

Δ_rH=Δ_rH_(product)-ΔrH_(reactant)
=3×-393+81+3×110-9.7
=-777.7kJ
ΔH = ΔU+Δn_gRT
AS Δn_g = 0, ΔH = ΔU
So, ΔU = -777.7 kJ
Question 4
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Given

Q. Thus, heat of formation of CH₃OH(/)is

A
- 239.0 kJ mol^-1

B
+ 239.0 kJ mol^-1

C
- 47.0 kJ mol^-1

D
+ 47.0 kJ mol^-1

Solution

Let's number our eqns-
CH₃OH + 3/2 O₂ → CO₂ + 2H₂O ...(1)
ΔH₁ = –726 kJ/mol
C + O₂ → CO₂ ...(2)
ΔH₂ = –393 kJ/mol
H₂ + 1/2 O₂ → H2O ...(3)
ΔH₃ = –286 kJ/mol
Eqn(2) + 2×eqn(3) - eqn(1) :-
C + O₂ + 2H₂ + O₂ + CO₂ + 2H₂0→ CO₂ + 2H₂0 + CH₃OH + 3/2 O₂
C + 1/2 O₂ + 2H₂ → CH₃OH
Thus enthalpy of formation-
∆Hf(CH₃OH) = ∆H₂ + 2∆H₃ - ∆H₁
∆H_f(CH₃OH) = -393 + 2(-286) + 726
∆Hf(CH3OH) = -239 kJ/mol
Question 5
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Hot carbon reacts with steam to produce an equimolar mixture of CO(g)and H₂(g) known as water gas

Q.
Energy released , if water gas is used as fuel is

A
-262.5 KJ

B
-41.0 KJ

C
-242.0 KJ

D
-525.0 KJ

Solution

For this, we will add combustion reactions of both CO and H₂ as both combined will give us energy.
Δ_rH = -283 kJ+ (-242 kJ) = -525 kJ
Question 6
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For a gaseous phase reaction
When equilibrium is set up, K = 0.33
Energy involved (in kJ) is

A
0.595

B
1.785

C
0.7854

D
1.5875
Question 7
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One mole of a non-ideal gas undergoes a change of state (2.0 atm, 3.0 L, 95 K) → (4.0 atm, 5.0 L, 245 K) with a change in internal energy, Δ_rE° = 30.0 L atm. The change in enthalpy (Δ_rH°) of the process in L-atm is
[IIT JEE 2002]

A
44.0

B
42.0

C
40.0

D
system is not defined as pressure is not constant

Solution

ΔH=ΔU+Δ(PV)
ΔH=30+(P₂V₂−P₁V₁)
ΔH=30+(20−6)
ΔH=30+14=44
Question 8
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Which of the following reactions defines Δ_fH° ?

A

B

C

D

Solution

For ∆H formation we have to look for following:
1. formation of "1 mole" of substance.
2. from "stable" constituting elements.
In first option 1 mole of Co₂ should be formed from C(graphite) for enthalpy of formation to be defined.
In third option 2 moles of NH₃ is formed instead of 1 mole. So incorrect.
In fourth option Co₂ is not formed from constituent elements I.e C(gr) and O2(g) so it is not enthalpy of formation.
Question 9
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Ethanol can undergo decomposition to form two sets of products

Q.
if the molar ratio of C₂H₄ to CH₃CHO is 8 : 1 in a set of product gases, then, energy involved in the decomposition process is

A
14.31 KJ

B
57.23 KJ

C
48.14 KJ

D
66.31 KJ

Solution

A/Q molar ratio of C₂H₄ and CH₃CHO is 8:1
So from eqn 1
∆H = 8 45.54
= 364.32 kJ
Similarly from eqn
∆H = 68.91 1
= 68.91 kJ
Total ∆H = 364.32+68.91
Therefore, enthalpy change per mole of C₂H₅OH used 364.32+68.91/9
= 48.14 kJ
Question 10
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Direction (Q. No. 10) This sectionis based on statement I and Statement II. Select the correct answer from the code given below.
Q.
Statement I : Based on the following thermodynamic data

NO₂ is more stable than NO.
Statement II : NO (g) is an endothermic compound while, NO₂(g) is an exothermic compound.

A
Both Statement I and Statement II are correct and Statement II is the correct explanation of Statement I

B
Both Statement I and Statement II are correct and Statement II is not the correct explanation of Statement I

C
Statement I is correct but Statement II is incorrect

D
Statement II is correct but Statement I is incorrect

Solution

Reaction number 1 is less stable because heat is not released thus the product side of reaction gains energy which get stored in the product and decreases its stability
While in reaction number 2, the reaction formed is more stable because energy is released and so the product is in a more stable state.
Now there is no such endothermic or exothermic compound , the reaction is exothermic or endothermic and not the product.