Chemical Thermodynamics(12th Grade > Chemistry ) Questions and answers for exam Preparation

12th Grade > Chemistry

CHEMICAL THERMODYNAMICS MCQs

Total Questions : 30 | Page 1 of 3 pages

Question 1. The bond dissociation energy of C - H in

C H_{4}

from the equation
C(g) + 4H(g)

⟶

C H_{4}

(g)

Δ

H = -397.8 Kcal is

+99.45 kcal
-99.45 kcal
+397.8 kcal
-397.8 kcal

Discuss Question

Answer: Option A. -> +99.45 kcal
:
A
The equation for bond dissociation can be given by

C H_{4}

(g)

⟶

C(g) + 4H(g)

Δ

H = 397.8 kcal
There are in total of 4 C - H bonds
Therefore the energy needed to break a C - H bond can be given by

\frac{397.8}{4}

= + 99.45 kcal

Question 2. Calculate entropy change for vaporization of 1 mole of liquid water to steam at

100^{\circ} C

Δ H_{v}

=40.8 kJ

m o l^{- 1}

ΔSv=107.38 JK−1 mol−1
ΔSv=109.48 JK−1 mol−1
ΔSv=109.38 JK−1 mol−1
ΔSv=107.38 JK−1 mol−1

Discuss Question

Answer: Option C. -> ΔSv=109.38 JK−1 mol−1
:
C
For entropy change of vaporization can be given as,

Δ S_{v} = Δ \frac{H_{v}}{T}

Given values are,

Δ H_{v} = 40.8 \times 10^{3} J m o l^{- 1}

; T=373K
Therefore

Δ S_{v} = (40.8 \times \frac{10^{3}}{373})

Δ S_{v} = 109.38 J K^{- 1} m o l^{- 1}

Question 3. Consider an endothermic reaction

X \to Y

with the activation energies

E_{b}

and

E_{f}

for the backward and forward reactions, respectively. IN general

There is no definite relation between EbandEf
Eb=Ef
Eb>Ef
Eb

Discuss Question

Answer: Option D. -> Eb
:
D
Enthalpy of reaction

(Δ H) = E_{a_{(f)}} - E_{a_{(b)}}

for an endothermic reaction

Δ H

= + ve hencefor

Δ H

to be negative

E_{a_{(b)}}

E_{a_{(f)}}

Question 4. A gas mixture

3.67 L

in volume contains

C_{2} H_{4}

and

C H_{4}

is proportion of

2 : 1

by moles and is at

25^{\circ} C

and

1 a t m

. If the

Δ H_{C}

(

C_{2} H_{4})

and

Δ H_{C}

(C H_{4})

are

- 1400

and

- 900 k J / m o l

find heat evolved on burning this mixture

20.91 kJ
50.88 kJ
185 kJ
160 kJ

Discuss Question

Answer: Option C. -> 185 kJ
:
C
Volume of

C_{2} H_{4}

\frac{3.67 \times 2}{3} = 2.45 L i t

Volume of

C H_{4}

1.22 L i t

Mole of

C_{2} H_{5}

P V

n R T n

\frac{P V}{R T}

\frac{1 \times 2.45}{0.082 \times 298} = n = 0.1 m o l e

Mole of

C H_{4}

^{n} C H_{4}

\frac{1 \times 1.22}{0.082 \times 298} = 0.05 m o l e

∵

Energy released due to combusting 1 mole

C_{2} H_{4} = 1400 K J

∴

Energy released due to combusting 1 mole

C_{2} H_{4}

1400 \times 0.1 = 140 k J

Similar energy released in combusting 0.05 mole

C H_{4} = - 900 \times 0.05 = 45

So that heat realized =

140 + 45

185 k J / m o l e

Question 5. The

Δ H

and

Δ S

for a reaction at one atmospheric pressure are +30.558 kJ and 0.066 k J

k^{- 1}

respectively. The temperature at which the free energy change will be zero and below this temperature the nature of reaction would be:

483 K, spontaneous
443 K, non-spontaneous
443 K, spontaneous
463 K, non-spontaneous

Discuss Question

Answer: Option D. -> 463 K, non-spontaneous
:
D
Now, to get T, let's use

△ G

△ H - T △ S

Here,

△ H = + 30.558 K J

△ S = 0.066 k J k^{- 1}

△ G = 0

\Rightarrow 0 = 30.558 - T \times 0.066

\Rightarrow T = \frac{30.558}{0.066} = 463 K

Now, at 463

K \Rightarrow △ G

= 0
If we go below 463 K,

T △ S

value will decrease and

△ G

value will become positive.
Therefore, non-spontaneous

Question 6. Which of the following processes is an iso-entropic process?

Isothermal process
Adiabatic process
Isobaric process
Isochoric process

Discuss Question

Answer: Option B. -> Adiabatic process
:
B
What do you mean by iso-entropic process?

\Rightarrow

The name itself is giving you the answer.
The entropy should be same before and after the process occurs.
Here we know that,

Δ S = \frac{d q}{T}

In case of adiabatic process,
dq = 0
Therefore

Δ S

= 0

Question 7. For the reaction

H_{2} O (s) ⇋ H_{2} O (l)

at 0°C and normal pressure

△H > T△S
△H = T△S
△H = △G
△H < T△S

Discuss Question

Answer: Option B. -> △H = T△S
:
B
Let's look at the reaction.

H_{2} O (s) ⇋ H_{2} O (l)

0^{\circ} C

It will be in equilibrium at

0^{\circ} C

Now, At equilibrium

△

G = 0
&

△

G =

△

H - T

△

\Rightarrow △

H - T

△

S = 0

\Rightarrow △ H = T △ S

Question 8. If the bond energies of H – H, Br – Br and H – Br are 433, 192 and 364 kJ

m o l^{- 1}

respectively, the

Δ H^{\circ}

for the reaction ;

H_{2} (g) + B r_{2} (g) \to 2 H B r (g)

– 261 kJ
+ 103 kJ
+ 261 kJ
– 103 kJ

Discuss Question

Answer: Option D. -> – 103 kJ
:
D

Δ H^{\circ} = \sum B . E (r e a c t a n t s) - \sum B . E (p r o d u c t s) = [B . E (H - H) + B . E (B r - B r)] - [2 B . E (H - B r)] = (433 + 192) - (2 \times 364) = 625 - 728 = - 103 k J

Question 9. Energy required to dissociate 4g of gaseous hydrogen into free gaseous atoms is 208 Kcal at

25^{o} C

. The bond energy of H - H bond will be:

1.04 Kcal
10.4 Kcal
104 Kcal
1040 Kcal

Discuss Question

Answer: Option C. -> 104 Kcal
:
C
Energy needed to dissociate 4g is 208 Kcal

∴

Number of moles of

H_{2}

= 2 mole

∴

Energy needed to dissociate 1 mole =

\frac{208}{2}

= 104 Kcal

∴

The bond energy is 104 Kcal in case of H - H bond.

Question 10. 1 mole of an ideal gas is compressed isothermally and reversibly at 298 K from a pressure of 1 Pa. To do a work of 100J, to what final pressure should we compress it? (R = 8.314 J/K/mol)

0.96 Pa
1.0412 Pa
10 Pa
96 Pa

Discuss Question

Answer: Option B. -> 1.0412 Pa
:
B

W = - 2.303 n R T l o g \frac{P_{1}}{P_{2}} ∴ l o g \frac{P_{1}}{P_{2}} = \frac{W}{- 2.303 n R T} = \frac{100}{- 2.303 \times 1 \times 8.314 \times 298} = - 0.01753 i . e ., l o g P_{1} - l o g P_{2} = - 0.01753 (T h e u n i t s i n w h i c h P_{1} a n d P_{2} a r e e x p r e s s e d m u s t b e i d e n t i c a l) ∴ - l o g P_{2} = - 0.01753 - l o g P_{1} ∴ - l o g P_{2} = - 0.01753 + l o g P_{1} = 0.01753 + l o g 1 = 0.01753 + 0 l o g P_{2} = 0.01753 \Rightarrow P_{2} = A n t i l o g (0.01753) \Rightarrow P_{2} = 1.0412 P a (S i n c e u n i t o f P_{1} i s P a)

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