Thermodynamics Test

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Thermodynamics Test - 10

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Weekly Quiz Competition

Question 1
1 / -0

1 kg block of ice at 0^oC is placed into a perfectly insulated, sealed container that has 2 kg of water also at 0^oC. The water and ice completely fill the container, but the container is flexible. After some time, one can expect that

A
the water will freeze so that the mass of the ice will increase

B
both the amount of water and the amount of ice will remain constant

C
the ice will melt so that the mass of the ice will decrease

D
both the amount of water and the amount of ice will decrease

Solution

The correct option is (3). The ice will melt so that the mass of the ice will decrease.
In accordance to second law of thermodynamics, if an irreversible process occurs in a closed system, the entropy of that system always increases; it never decreases. Entropy measures the state of disorder. As the molecules in ice are more ordered than those in water, the ice will melt and as a result, the mass of ice will decrease.
Question 2
1 / -0

For the reaction of one mole of Zn dust with one mole of H₂SO₄ in a bomb calorimeter, ∆U and w correspond to

A
∆U < 0, w = 0

B
∆U < 0, w < 0

C
∆U > 0, w = 0

D
∆U > 0, w > 0

Solution

Bomb calorimeter is commonly used to find the heat of combustion of organic substance which consists of a sealed combustion chamber, called bomb. If a process is run in a sealed container, then no expansion or compression is allowed.

So, w = 0 and ∆U = ∆H

Since ∆H < 0,

∴ ∆U < 0, w = 0
Question 3
1 / -0

The enthalpy of hydrogenation of cyclohexene is -119.5 kJ/mol. If resonance energy of benzene is -150.4 kJ/mol, its enthalpy of hydrogenation would be

A
-208.1 kJ/mol

B
-269.9 kJ/mol

C
-358.5 kJ/mol

D
-508.9 kJ/mol

Solution

∆H_{observed heat of hydration} - ∆H_resonance = ∆H_{Actual heat of hydration}

3 x - 119.5 - (- 150.4) = - 208.1 kJ/mol
Question 4
1 / -0

If the standard state of iodine and chlorine is I₂(s) and Cl₂(g) respectively, then what is the value of standard enthalpy of formation of ICl(g). You are given the following information:

Cl₂(g) → 2Cl(g); ∆H = 242.3 kJ/mol
I₂(g) → 2I(g); ∆H = 151.0 kJ/mol
ICl(g) → I(g) + Cl(g); 211.3 kJ/mol
I₂(s) → I₂(g); 62.76 kJ/mol

A
- 12.6 kJ/mol

B
- 16.73 kJ/mol

C
+ 16.73 kJ/mol

D
+ 222.8 kJ/mol

Solution

Cl₂(g) → 2Cl(g); ∆H = 242.3 kJ/mol ………….. (i)
I₂(g) → 2I(g); ∆H = 151.0 kJ/mol ...……… (ii)
I₂(s) → I₂(g); ∆H = 62.76 kJ/mol ...………(iii)
ICl(g) → I(g) + Cl(g); ∆H = 211.3 kJ/mol …………(iv)
Add (i), (ii) and (iii), we get
Cl₂(g) + I₂(s) → 2 Cl(g) + 2I(g) (∆H = 456.06 kJ/mol) …………(v)
Divide (v) by 2 and subtract equation (iv) from it, we get
½ Cl₂(g) + ½ I₂(s) → ICl (∆H = + 16.73 kJ/mol).
Question 5
1 / -0

If the enthalpies of formation of Al₂O₃ and Cr₂O₃ are -1596 kJ and -1134 kJ, then what is the value of ∆H for the following reaction?

2Al + Cr₂O₃ → 2Cr + Al₂O₃

A
- 462 kJ

B
- 1365 kJ

C
- 2530 kJ

D
+ 2530 kJ

Solution

Given: 2Al + 3/2 O₂ → Al₂O₃, ∆H = - 1596 kJ, ……. (i)
2Cr + 3/2 O₂ → Cr₂O₃, ∆H = - 1134 kJ ……. (ii)
Subtracting (ii) from (i), we get
2Al + Cr₂O₃ → 2Cr + Al₂O_3, ∆H = - 1596 - (- 1134) = - 462 kJ