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2 nd Law of thermodynamics and Entropy Questions in English

Class 11 Chemistry · Thermodynamics · 2 nd Law of thermodynamics and Entropy

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251
EasyMCQ
For which one of the following reactions,the entropy change is positive?
A
$H_{2(g)} + \frac{1}{2} O_{2(g)} \longrightarrow H_2O_{(l)}$
B
$Na^{+}_{(g)} + Cl^{-}_{(g)} \longrightarrow NaCl_{(s)}$
C
$NaCl_{(l)} \longrightarrow NaCl_{(s)}$
D
$H_2O_{(l)} \longrightarrow H_2O_{(g)}$

Solution

(D) Entropy $(S)$ is a measure of the randomness or disorder of a system. $A$ positive entropy change $(\Delta S > 0)$ occurs when the system becomes more disordered,such as during a phase transition from a more ordered state (liquid) to a less ordered state (gas).
In the reaction $H_2O_{(l)} \longrightarrow H_2O_{(g)}$,liquid water is converted into water vapor. Since gas molecules have much higher randomness compared to liquid molecules,the entropy of the system increases,making $\Delta S$ positive.
In other options,the reactions involve a decrease in the number of moles of gas or a transition from a disordered state to a more ordered state (like liquid to solid or gas to solid),which results in a negative entropy change.
252
DifficultMCQ
What is the entropy change in $J K^{-1}$ during the melting of $27.3 \ g$ of ice at $0^{\circ} C$? (Latent heat of fusion of ice $= 330 \ J g^{-1}$)
A
$330$
B
$12.1$
C
$3.3$
D
$33$

Solution

(D) The entropy change during a phase transition is given by the formula $\Delta S = \frac{q_{rev}}{T} = \frac{m \times L_f}{T}$.
Given:
Mass of ice $(m)$ $= 27.3 \ g$
Latent heat of fusion $(L_f)$ $= 330 \ J g^{-1}$
Temperature $(T)$ $= 0^{\circ} C = 273 \ K$
Substituting the values:
$\Delta S = \frac{27.3 \ g \times 330 \ J g^{-1}}{273 \ K}$
$\Delta S = \frac{9009 \ J}{273 \ K} = 33 \ J K^{-1}$.
253
EasyMCQ
For the following process $H_2O_{(l)} (1 \ bar, 373.15 \ K) \rightleftharpoons H_2O_{(g)} (1 \ bar, 373.15 \ K)$,identify the correct set of thermodynamic parameters.
A
$\Delta G=0, \Delta S=+ve$
B
$\Delta G=0, \Delta S=-ve$
C
$\Delta G=+ve, \Delta S=0$
D
$\Delta G=-ve, \Delta S=+ve$

Solution

(A) The given process represents the phase transition of water from liquid to gas at its boiling point $(373.15 \ K)$ and standard pressure $(1 \ bar)$.
Since the system is in equilibrium at the boiling point,the change in Gibbs free energy is $\Delta G = 0$.
During the phase change from liquid to gas,the disorder of the system increases,which means the entropy change is positive,$\Delta S > 0$ (or $\Delta S = +ve$).
254
MediumMCQ
The enthalpy of vaporisation of a certain liquid at its boiling point of $35^{\circ} C$ is $24.64 \ kJ \ mol^{-1}$. The value of change in entropy for the process is
A
$80 \ J \ K^{-1} \ mol^{-1}$
B
$70 \ J \ K^{-1} \ mol^{-1}$
C
$24.64 \ J \ K^{-1} \ mol^{-1}$
D
$7.04 \ J \ K^{-1} \ mol^{-1}$

Solution

(A) The entropy of vaporisation $(\Delta S_{vap})$ is given by the formula: $\Delta S_{vap} = \frac{\Delta H_{vap}}{T_b}$
Given: $\Delta H_{vap} = 24.64 \ kJ \ mol^{-1} = 24640 \ J \ mol^{-1}$
Boiling point $T_b = 35 + 273 = 308 \ K$
Substituting the values: $\Delta S_{vap} = \frac{24640 \ J \ mol^{-1}}{308 \ K} = 80 \ J \ K^{-1} \ mol^{-1}$
255
EasyMCQ
The change of entropy $(dS)$ is defined as
A
$dS = \frac{\delta q}{T}$
B
$dS = \frac{dH}{T}$
C
$dS = \frac{\delta q_{rev}}{T}$
D
$dS = \frac{dH - dG}{T}$

Solution

(C) Entropy is a state function that measures the degree of randomness or disorder in a system.
For a reversible process,the change in entropy $(dS)$ is defined as the ratio of the heat exchanged reversibly $(\delta q_{rev})$ to the absolute temperature $(T)$ at which the exchange occurs.
Thus,the mathematical expression is $dS = \frac{\delta q_{rev}}{T}$.
256
EasyMCQ
Which of the following statements is correct?
A
Evaporation of water causes an increase in disorder of the system
B
Melting of ice causes a decrease in randomness of the system
C
Condensation of steam causes an increase in disorder of the system
D
There is practically no change in the randomness of the system when water is evaporated

Solution

(A) When water evaporates,it changes from a liquid state to a gaseous state.
In the gaseous state,the molecules have more freedom of movement compared to the liquid state.
Therefore,the disorder or entropy of the system increases during evaporation.
257
EasyMCQ
The second law of thermodynamics states that in a cyclic process:
A
work cannot be converted into heat
B
heat cannot be converted into work
C
work cannot be completely converted into heat
D
heat cannot be completely converted into work

Solution

(D) The second law of thermodynamics,specifically the Kelvin-Planck statement,asserts that it is impossible for any device that operates on a thermodynamic cycle to receive heat from a single thermal reservoir and produce a net amount of work.
In other words,heat cannot be completely converted into work in a cyclic process without leaving some effect on the surroundings.

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