Thermodynamics Test

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Thermodynamics Test - 8

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Weekly Quiz Competition

Question 1
4 / -1

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
4 / -1

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
4 / -1

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
4 / -1

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
4 / -1

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
4 / -1

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
4 / -1

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
4 / -1

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
4 / -1

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
4 / -1

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.
Question 11
4 / -1

Among the following enthalpies, which is always less than zero?

A
Δ_cH°

B
Δ_subH°

C
Δ_mixH°

D
Δ_fH°

Solution

The symbol represents enthalpy of combustion which is always exothermic. For exothermic rxn delta H is always -ve.
Question 12
4 / -1

The solubility of fluorides is much less as compared to corresponding chlorides

A
because of low lattice enthalpy.

B
because of high lattice enthalpy.

C
because of low ionization energy.

D
because of high ionization energy.

Solution

Due to high lattice enthalpy of fluorides, it is difficult to break the lattice of fluoride compounds(due to the small size of fluorine, so the lattice is strong). SO the solubility of fluorides is less than chlorides.
Question 13
4 / -1

Direction (Q. Nos. 13-15) 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 thermodynamic data,

Q. Which oxidising agent will generate the greatest amount of energy per mole of H₂(g)?

A
O₂

B
O₃

C
H₂O₃

D
Equally
Question 14
4 / -1

Based on the following thermodynamic data,

Q. Which oxidising agent will generate the greatest amount of energy per gram of oxidising agent?

A
O₂

B
O₃

C
H₂O₃

D
Equally

Solution

Mass of O₂=32 gm ∆H/gm = -483.636/32 = -15.11
Mass of O₃=48 gm ∆H/gm = -868.2/48 = -18.08
Mass of H₂O₂=34 gm, ∆H/gm = -347.33/34 = -10.21
We can say that II has highest amount of energy per gram of oxidising agent.
Question 15
4 / -1

Based on the following thermodynamic data,

Q. On the total mass basis of reactants, which reaction will generate the greatest amount of heat?

A
I, II

B
II

C
I, III

D
II, III

Solution

From I, total mass of reactant, 36 gm
So, ∆H/mass = -483.636/36 = -13.43 kJ gm^-1
For II, total mass of reactant, 54 gm
So, ∆H/mass = -868.2/54 = -16.07 kJ gm^-1
FOr III, total mass = 36 gm,
So, ∆H/mass = -347.33/36 = -9.64 kJ gm^-1
Therefore, heat released per unit mass is maximum for II.
Question 16
4 / -1

Direction (Q. Nos. 16 - 18) This section contains 3 questions. when worked out will result in an integer from 0 to 9 (both inclusive).

Q. How much heat (in kcal) is required to convert 36 g of diamond into graphite?

A
9

B
8

C
7

D
6

Solution

C(diamond) + O₂(g) → CO₂(g) -----(i) ∆_rH° = -91 kcal mol^-1
C(graphite) + O₂(g) → CO₂(g) -----(ii) ∆_rH° = -94 kcal mol^-1
For 36 gm of C(diamond), we have 3 moles of diamond,
On (i)-(ii), we get → C(graphite) ∆rH° =-91-( -94 ) kcal mol^-1 = 3 kcal mol^-1
So for 3 moles, we have 3×3 = 9 kcal
C(diamond)
Question 17
4 / -1

Based on the following reactions,

Q. Heat of formation of NO₂ (in kcal) is ........

A
1

B
2

C
3

D
4

Solution

N₂(g) + O₂(g) → 2NO ----- (i) ∆_rH° = 43 kcal
2NO(g) + O₂(g) → 2NO₂(g) -----(ii) ∆rH° = -35 kcal
Enthalpy of formation of NO₂ means we need to have 1 mole of NO₂ from its constituting elements.
We will have (i)/2 + (ii)/2
½ N₂(g) + O₂(g) → NO₂(g) ∆_fH° = 4 kcal
Question 18
4 / -1

How many of the following has zero standard molar enthalpy of the formation at 298 K?

[IIT JEE 2010]

A
5

B
4

C
2

D
1

Solution

The standard molar enthalpy of formation of elements in its standard state is taken as zero. Since the standard state of H₂ and Cl₂ are gas, their ΔfHo = 0.