Thermodynamics Test

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Thermodynamics Test - 41

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Weekly Quiz Competition

Question 1
4 / -1

For the spontaniety of a reaction which statement is true?

A
∆G = +ive ; ∆H = +ive

B
∆H = +ive ; ∆S = ‒ive

C
∆G = ‒ive ; ∆S = ‒ive

D
∆H = ‒ive ; ∆S = +ive
Question 2
4 / -1

Which of the following equations correctly represents the standard heat of formation (∆H^o_f ) of methane?

A
C (diamond) + 2H₂(g) = CH₄(g)

B
C (graphite) + 2H₂(g) = CH₄(l)

C
C (graphite) + 2 H₂(g) = CH₄(g)

D
C (graphite) + 4H = CH₄(g)
Question 3
4 / -1

The enthalpy change (- ∆H) for the neutralisation of 1 MHCl by caustic potash in dilute solution at 298 K is

A
68 kJ

B
65 kJ

C
57.3 kJ

D
50 kJ
Question 4
4 / -1

At constant T and P which one of the following statements is correct for the reaction ?
S₈(s) + 8O₂(g) → 8SO₂(g)

A
∆H < ∆E

B
∆H = ∆E

C
∆H > ∆E

D
∆H is independent of the physical state of the reactants.
Question 5
4 / -1

Gibb’s free energy G, enthalpy H and entropy S are related to one another by

A
G = H + TS

B
G = H – TS

C
G – TS = H

D
S = H – G
Question 6
4 / -1

Consider the following reaction occurring in an automobile 2C₈H₁₈(g) + 25O₂(g) → 16CO₂ + 18H₂O(g) the sign of ∆H, ∆S and ∆G would be

A
+, ‒, +

B
‒, +, ‒

C
‒, +, +

D
+, +, ‒

Solution
Question 7
4 / -1

The heat of neutralization of aqueous hydrochloric acid by NaOH is x kcal/mole of HCl. Calculate the heat of neutralisation per mole of aqueous acetic acid.

A
0.5 x kcal

B
x kcal

C
2x kcal

D
cannot be calculated from the given data
Question 8
4 / -1

Heat of neutralisation of strong acid against strong base is constant and is equal to

A
13.7 kcal

B
57 kJ

C
5.7 × 10⁴ J

D
All of the above
Question 9
4 / -1

The heat of formation of the compound in the following reaction is H₂g + Cl₂g → 2HCl(g) + 44 kcal

A
‒ 44 kcal mol^-1

B
‒ 22 kcal mol^-1

C
11 kcal mol^-1

D
‒ 88 kcal mol^-1
Question 10
4 / -1

Heat capacity is

A

B

C

D
None of these

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Thermodynamics Test - 41

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Thermodynamics Test - 41

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Question 1

∆G = +ive ; ∆H = +ive

∆H = +ive ; ∆S = ‒ive

∆G = ‒ive ; ∆S = ‒ive

∆H = ‒ive ; ∆S = +ive

Question 2

C (diamond) + 2H2(g) = CH4(g)

C (graphite) + 2H2(g) = CH4(l)

C (graphite) + 2 H2(g) = CH4(g)

C (graphite) + 4H = CH4(g)

Question 3

68 kJ

65 kJ

57.3 kJ

50 kJ

Question 4

∆H < ∆E

∆H = ∆E

∆H > ∆E

∆H is independent of the physical state of the reactants.

Question 5

G = H + TS

G = H – TS

G – TS = H

S = H – G

Question 6

+, ‒, +

‒, +, ‒

‒, +, +

+, +, ‒

Solution

Question 7

0.5 x kcal

x kcal

2x kcal

cannot be calculated from the given data

Question 8

13.7 kcal

57 kJ

5.7 × 104 J

All of the above

Question 9

‒ 44 kcal mol-1

‒ 22 kcal mol-1

11 kcal mol-1

‒ 88 kcal mol-1

Question 10

None of these

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C (diamond) + 2H₂(g) = CH₄(g)

C (graphite) + 2H₂(g) = CH₄(l)

C (graphite) + 2 H₂(g) = CH₄(g)

C (graphite) + 4H = CH₄(g)

5.7 × 10⁴ J

‒ 44 kcal mol^-1

‒ 22 kcal mol^-1

11 kcal mol^-1

‒ 88 kcal mol^-1