Thermodynamics Test

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Thermodynamics Test - 6

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Weekly Quiz Competition

Question 1
1 / -0

The difference between C_p and C_v can be derived using relation H = U + pV. Calculate the difference between C_p and C_v for 10 moles of an ideal gas.

A
8.314 J

B
8.01 J

C
83.14 J

D
0.8314 J

Solution

Relation between C_p and C_vis C_p – C_v = nR.
C_p – C_v = 10 moles x 8. 314 J mol^-1 = 83.14 J
Question 2
1 / -0

∆_fU^o of formation of CH₄ (g) at a certain temperature is – 393 kJ mol^-1. The value of ∆_fH^o is

A
zero

B
< ∆_fU^o

C
> ∆_fU^o

D
equal to ∆_fU^o

Solution

C (s) + 2H₂ (g) → CH₄ (g)
∆n_g = 1 – 2 = -1; ∆_fH^o = ∆_fU^o + ∆n_g RT
∆_fH^o = ∆_fU^o - RT
Therefore, ∆_fH^o < ∆_fU^o
Question 3
1 / -0

The enthalpy of neutralisation of NaOH with HCl is 57.1 kJ, while with CH₃COOH, it is – 55 kJ. This happens because

A
acetic acid is an organic acid

B
acetic acid is less soluble in water

C
acetic acid is a weak acid and requires less NaOH for neutralisation

D
some heat is required to ionise acetic acid completely

Solution

Acetic acid is a weak electrolyte. It does not get completely ionised in water. So, it absorbs some heat to completely ionise. That is why the enthalpy of neutralisation of NaOH with CH₃COOH is less than the enthalpy of neutralisation of NaOH with HCl.
Question 4
1 / -0

The bond energies of C-C, C=C, H-H and C-H linkages are 350, 600, 400, 410 kJ per mol, respectively. The heat of hydrogenation of ethylene is

A
- 170 kJ per mol

B
- 260 kJ per mol

C
- 400 kJ per mol

D
- 40 kJ per mol

Solution

CH₂=CH₂ + H-H → CH₃-CH₃
∆H = 4 B.E (C-H) – B.E (C=C) + B.E (H-H) – 6 B.E (C-C) – B.E (C-C)
On solving, we get
∆H_r = - 170 kJ per mol
Question 5
1 / -0

Energy required to dissociate 4 gm of gaseous hydrogen into free gaseous atoms is 208 kcal at 25^oC. The bond dissociation energy of H–H bond will be

A
104 kcal

B
10.4 kcal

C
1040 kcal

D
104 cal

Solution

4 grams of hydrogen = 2 moles of dihydrogen
2 moles of dihydrogen needs 208 kcal.
1 mole needs 208/2 = 104 kcal.
Thus, the answer is 104 kcal.
Question 6
1 / -0

∆H_f^o for Al₂O₃ is - 1670 kJ. Calculate the enthalpy change for the following reaction:

4Al + 3O₂ → 2Al₂O₃

A
- 2340 kJ

B
- 3340 kJ

C
- 1340 kJ

D
- 4320 kJ

Solution

Enthalpy of formation of 1 mol of Al₂O₃ = - 1670 kJ
According to the given equation, 4 mol of Al gives 2 mol of Al.
Then, the enthalpy of formation of 2 mol of Al₂O₃ is 2 × - 1670 = - 3340 kJ/mol.

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Re-Attempt Weekly Quiz Competition

Thermodynamics Test - 6

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Thermodynamics Test - 6

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SHARING IS CARING

Question 1

8.314 J

8.01 J

83.14 J

0.8314 J

Solution

Question 2

zero

< ∆fUo

> ∆fUo

equal to ∆fUo

Solution

Question 3

acetic acid is an organic acid

acetic acid is less soluble in water

acetic acid is a weak acid and requires less NaOH for neutralisation

some heat is required to ionise acetic acid completely

Solution

Question 4

- 170 kJ per mol

- 260 kJ per mol

- 400 kJ per mol

- 40 kJ per mol

Solution

Question 5

104 kcal

10.4 kcal

1040 kcal

104 cal

Solution

Question 6

- 2340 kJ

- 3340 kJ

- 1340 kJ

- 4320 kJ

Solution

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< ∆_fU^o

> ∆_fU^o

equal to ∆_fU^o