Thermodynamics Test

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Thermodynamics Test - 25

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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. If the enthalpy of HCI (g) and Cl^- (aq) are - 92.30 kJ mol^-1 and - 167.44 kJ mol^-1 respectively, then Δ_rH° for the following reaction is
HCI (g) + aq → H⁺(aq) + Cl^-(aq)

A
-75.14 kJ mol-1

B
+75.14 kJ mol^-1

C
-259.74 kJ mol^-1

D
0.00 kJ mol^-1

Solution

Enthalpy of reaction = ∑enthalpy of products - ∑enthalpy of reactants
Therefore
∆H_reaction = -167.44 - (- 92.30) (∆H_formation of H+ is 0)
= -75.14 kilo J / mole
Question 2
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The enthalpy of neutralisation of HS^- (aq) is - 5.1 kJ mol^-1. Thus, second ionisation energy of H₂S is

A
+52.6 kJ mol^-1

B
-13.7 kJ mol^-1

C
-52.2 kJ mol^-1

D
+52.2 kJ mol^-1

Solution

∆H_{neutralisation} = ∆H_H⁺ _{+ OH}- + ∆_rHionisation
-5.1 = -57.3 + ∆H_ionisation
∆H_ionisation = 52.2 kJ mol^-1
Question 3
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Given, Δ_fH° of HCI (g) is - 22 . 10 kcal mol^-1 and Δ_SolutionH° (heat of solution) of HCI (g) is - 17.9 kcal mol^-1. Thus, Δ_fH ° of Cl^- (aq) is

A
-40.0 kcal mol^-1

B
+40.0 kcal mol^-1

C
+17.9 kcal mol^-1

D
+22.1 kcal mol^-1

Solution

½ H₂(g) + ½ Cl₂(g) → HCl(g) ∆_fH° = -22.10 kcal mol^-1 -----(I)
HCl(g) + H₂O(l) → H⁺(aq) + Cl^-(aq) ∆_solutionH° = -17.9 kcal mol^-1 -----(II)
Let us add (I) and (II), we get
½ H₂(g) + ½ Cl₂(g) → H⁺(aq) + Cl-(aq) ∆_rH° = -40.0 kcal mol^-1
∆_rH° = ∑∆H_product - ∑∆H_reactant = ∆_fH_Cl- (since enthalpy of formation of H, H₂ and Cl₂ are 0)
∆_fH_Cl- = -40.0
Question 4
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Enthalpy of neutralisation of a weak dibasic acid by NaOH is - 24.0 kcal. Thus, enthalpy of ionisation of dibasic acid is

A
10.3 kcal mol^-1

B
3.4 kcal mol^-1

C
13.7 kcal mol^-1

D
-13.7 kcal mol^-1

Solution

Enthalpy of neutralisation = ∆H_{H+ + OH- → H2O +} ∆H_ionisation
-24.0 = -27.4 + ∆H_ionisation (since we have 2 moles of H⁺ and OH^-, so enthalpy will also double)
∆H_ionisation = 3.4 kcal mol-1
Question 5
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Consider the following reaction,
30 mL of 0.10 MNaOH (aq) + 10 mL of 0.10 M HCI (aq), enthalpy change is .
On further adding 20 mL of 0.10 M HCI (aq), enthalpy change is . Thus, these values are

A
a

B
b

C
c

D
d
Question 6
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In the following reactions,
I. 30 mL of 0.1 M Ba(OH)₂(aq) + 30 mL of 0.1 M H₂SO₄ (aq), rise intemperature = ΔT₁
II. 90 mL of 0.1 M Ba(OH)₂(aq) + 90 mL of 0.1 M H₂SO₄ (aq), rise intemperature = ΔT₂. Thus,

A
ΔT₂ = 3ΔT₁

B
ΔT₁ = ΔT₂

C
ΔT₁ = 3ΔT₂

D
ΔT₁ = 2ΔT₂
Question 7
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Given, CH₃COOH(aq) → CH₃COO^-(aq) + H⁺ (aq), Δ_rH° = 0.005 kcal g^-1
Enthalpy change when 1 mole of Ca(OH)₂, a strong base, is completely neutralised by CH₃COOH (aq) in dilute solution is

A
-27.4 kcal mol^-1

B
-13.4 kcal mol^-1

C
-26.8 kcal mol^-1

D
-27.1 kcal mol^-1

Solution

CH₃COOH → CH₃COO^- + H⁺
∆H = 0.3 kcal mol^-1
H⁺ + OH^- → H₂O
∆H = -13.7 kcal mol-1
Equation needed
2CCH₃COH + Ca(OH)₂ → 2CH₃COO + 2H₂O
∆H = 2×EQN(I) + 2×EQN(II)
∆H = (0.3)2 + (-13.7)2 = -26.8 kcal mol^-1
Question 8
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For the reactions,

A
15.08 kcal

B
-15.08 kcal

C
13.36 kcal

D
-13.36 kcal

Solution

For H⁺(aq) + OH^-(aq) → H₂O(l), we need to have -(I)+(II).
On doing the, we have ∆H = -14.22 + 0.66 = -13.56 kcal
Question 9
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Thus, heat of neutralisation of H₂C₂O₄ (aq) is

A
-16.9 kcal mol^-1

B
-10.5 kcal mol^-1

C
-24.2 kcal mol^-1

D
-30.6 kcal mol^-1

Solution

We can say that enthalpy of neutralisation of H₂C₂O4 means that we have 2 moles of H⁺. So, for weak acid, enthalpy of neutralisation = heat released in neutralisation of H⁺ and OH^- + enthalpy of ionisation.
∆H_{neutralisation} = 2 (-13.6) + 3.2 = -24.2 kcal mol^-1
Question 10
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Given,

Thus, for the reaction,
Δ_rH° is

A
-57.3 kJ mol^-1

B
+23.6 kJ mol^-1

C
-3.6 kJ mol^-1

D
+57.3 kJ mol^-1

Solution

H₂S(aq) + OH^-(aq) → HS-(aq) + H₂O(l)
(l) ∆H_r° = -33.7 kJ mol^-1
We also have, H+(aq) + OH-(aq) → H2O(l)
(ll) ∆H_r° = -57.3 kJ mol^-1
For the reaction,
we need (I) -(II)
So the Δ_rH° be = -33.7 - (-57.3) = 23.6 kJ mol^-1
Question 11
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Hydration energies of Li⁺ and Cl^- ions are - 499 and - 382 kJ mol^-1 respectively. If lattice energy of LiCI is - 840 kJ mol^-1 then heat of solution of LiCI is

A
+41 kJ mol ^-1

B
-41 kJ mol ^-1

C
-1621 kJ mol ^-1

D
-1222 kJ mol ^-1

Solution

Li+ + H₂O → Li⁺(aq.)
∆H₁ = -499 kJ mol^-1 ---------- (I)
Cl^- + H₂O →Cl^-(aq)
∆H₂ = -382 kJ mol^-1 ---------- (II)
Li⁺+ Cl^-→ LiCl(s)
∆H₃ = -840 kJ mol^-1---------- (III)
For r the heat of solution, we need the reaction,
LiCl(s) → Li⁺(aq.) + Cl^-(aq)
This can be obtained by (I) + (II) - (III)
So the ∆H value be = -499 - 382 -(-840) = -41 kJ mol^-1.
Question 12
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Enthalpy change on mixing 100 mL of 1 M Ca(OH)₂ with 100 mL of 1 M H₂SO₄ is

A
-13.7 kcal

B
-27.4 kcal

C
-1.37 kcal

D
-2.74 kcal

Solution

For reaction of 1 mole of H⁺ and 1 mole of OH^-, -13.6 kcal energy is released.
100 mL of 1 M Ca(OH)₂ = moles = molarity volume = 0.1× 1 = 0.1
Moles of OH^- = 0.2
100 mL of 1 M H₂SO₄= moles = molarity volume = 0.1× 1 = 0.1
Moles of H⁺ = 0.2
SO 0.2 moles of H⁺ and OH^- reacts.
Energy released, 0.2 ×13.6 =2.74 kcal.
Question 13
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Taking each reactant in 1:1 molar ratio, enthalpy changes are and in the following reactions.
and are (in kcal)

A
a

B
b

C
c

D
d

Solution

Since both reactions are acid base neutralization having strong acid-strong base and in 1:1 ratio of H⁺and OH^-,so will have same enthalpy of reaction equal to-13.6 kcal.
Question 14
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Direction (Q. Nos. 14) 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 thermochemical reactions :

Ionisation of HF(aq)is an exothermic reaction.

Statement II : Enthalpy decrease in hydration of H⁺ (aq) and F^- (aq) exceeds the ionisation energy of HF.

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

On (I) - (II), we have
HF(aq) → H⁺(aq) + F^-(aq)
∆H = -16.4-(-13.7) = -2.7 kcal
So, the ionisation of HF is an exothermic reaction. And so the statement I is true.
The reason behind HF being exothermic as the hydration energy of H⁺and F^- exceeds the lettuce enthalpy of HF and so overall the reaction becomes exothermic.