Chemical Thermodynamics(11th And 12th > Chemistry ) Questions and answers for exam Preparation

11th And 12th > Chemistry

CHEMICAL THERMODYNAMICS MCQs

Total Questions : 30 | Page 1 of 3 pages

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)

Question 2.

Enthalpy of neutralization of HCl with NaOH is x. The heat evolved when 500 ml of 2N HCl is mixed with 250 ml of 4N NaOH will be

500 x
100 x
x
10 x

Discuss Question

Answer: Option C. -> x
:
C

N o . o f g . e q . o f H C l = \frac{2 \times 500}{1000} = 1 N o . o f g . e q . o f N a O H = \frac{4 \times 500}{1000} = 1 H e a t e v o l v e d = x k J

Question 3.

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

– 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 4.

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 $E_{b} a n d E_{f}$
$E_{b} = E_{f}$
$E_{b} > E_{f}$
$E_{b} < E_{f}$

Discuss Question

Answer: Option D. ->

E_{b} < E_{f}

:
D

Enthalpy of reaction $(Δ H) = E_{a_{(f)}} - E_{a_{(b)}}$ for an endothermic reaction $Δ H$ = + ve hence for $Δ H$ to be negative $E_{a_{(b)}}$ < $E_{a_{(f)}}$

Question 5.

2 mol of zinc is dissolved in $H C l$ at $25^{\circ} C$ . The work done in an open vessel is:

$- 2.447 k J$
$- 4.955 k J$
$\frac{R}{2}$
$3 R$

Discuss Question

Answer: Option B. ->

- 4.955 k J

:
B

The following reaction takes place:

$2 Z n + 4 H C l \to 2 Z n C l_{2} + 2 H_{2}$
It is very impiortant to understand that this is an irreversible process since this reaction happens pretty quickly and hydrogen gas escapes the system.

When $H_{2}$ gas is liberated, it pushes back its surrounding and thus does some work against it.
Since it is an irreversible process, we can use the following equation:

$W = P_{e x t} Δ V = - P_{e x t} (V_{f} - V_{x})$

$V_{i} = 0$ thus $Δ V = V_{f}$

$V_{f} = \frac{n R T}{P_{e x t}}$

$W = - P e x t \times \frac{n R T}{P_{e x t}} = - n R T$

Given that $n = 2, R = 8.314 J / K m o l$

$T = 25 + 273 = 298 K$

$W = - 2 \times 8.314 \times 298 = 4955.144 J = 4.955 k J$
Tip: You can use approximations for $R = 25 / 3$ and $T = 300$ for quicker calculations!

Question 6.

State function among the following is

$q$
$q - w$
$\frac{q}{w}$
$q + w$

Discuss Question

Answer: Option D. ->

q + w

:
D
From the first law of thermodynamics, we know thatL:

q + w = Δ U

Δ U

- internal energy is a state function.

Question 7.

Which of the following acid will release maximum amount of heat when completely neutralized by strong base NaOH?

1M HCl
1M $H N O_{3}$
1M $H C l O_{4}$
1M $H_{2} S O_{4}$

Discuss Question

Answer: Option D. -> 1M

H_{2} S O_{4}

:
D
Ionisation of

H_{2} S O_{4}

gives double amount of

H^{+}

ions as compared to other acids

H_{2} S O_{4} ⟶ 2 H^{+} + S O_{4}^{- 2}

Therefore 1M

H_{2} S O_{4}

release more amount of heat compared to others.

Question 8.

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 9.

A system undergoes a process in which system releases 200 J heat and work done by the surroundings is 300 J. Change in internal energy of the system is

−100 J
100 J
−500 J
500 J

Discuss Question

Answer: Option B. -> 100 J
:
B

$Δ U = q + w$
Since system releases heat, q will be negative. It is also given that work done by the surroundings is 300 J which is the same as the work done on the system. Hence, w has a +ve sign. Keeping this in mind, we get:

= $- 200 + 300 = 100 J$

Question 10.

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.
$H_{2} O (s) ⇋ H_{2} O (l)$ at $0^{\circ} C$
It will be in equilibrium at $0^{\circ} C$
Now, At equilibrium
$△$ G = 0
& $△$ G = $△$ H - T $△$ S
$\Rightarrow △$ H - T $△$ S = 0
$\Rightarrow △ H = T △ S$