Energy transfer does not take place

Mass transfer does not take place

Both energy $\&$ mass transfer take place

THERMODYNAMICS MCQs for NEET UG CHEMISTRY — Easy, Page 3

61 Easy

61 of 196Question permalink

Understanding +4 | -1

During an isothermal expansion of an ideal gas its:

View answer and explanation

Correct option: D

Internal energy remains constant

Option D is correct because, for an ideal gas undergoing isothermal expansion, the internal energy remains constant as it only depends on temperature, which remains constant in an isothermal process.

Solution approach

Correct Answer + Core Reason: Internal energy remains constant. For an ideal gas, internal energy depends only on temperature, which is constant in an isothermal process.

Step-by-Step Solution:

Step 1: Define an isothermal process. An isothermal process is a thermodynamic process in which the temperature remains constant.

Step 2: Recall the relationship between internal energy and temperature for an ideal gas. For an ideal gas, the internal energy (U) is directly proportional to the temperature (T): $U = f(T)$ .

Step 3: Analyze the situation. Since the temperature remains constant during an isothermal expansion, the internal energy also remains constant.

Why other options are wrong:
- Option A: Enthalpy also depends on temperature for an ideal gas, so it remains constant, not decreases.
- Option B: Internal energy remains constant, not decreases.
- Option C: Internal energy remains constant, not increases.

Key Concepts:
- Isothermal Process
- Internal Energy of an Ideal Gas

Memory Aids:
- Isothermal: Constant temperature
- Ideal gas: Internal energy depends only on temperature

Quick Tips:
- Remember that for an ideal gas, internal energy is a function of temperature only.

62 Easy

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Which among the following is an extensive property?

View answer and explanation

Correct option: A

Volume

Option A (Volume) is correct because volume is an extensive property that depends on the amount of substance.

Solution approach

Correct Answer + Core Reason: Volume. Extensive properties depend on the amount of substance present.

Step-by-Step Solution:

Step 1: Define extensive and intensive properties.\nExtensive properties: Properties that change when the amount of substance changes (e.g., volume, mass).\nIntensive properties: Properties that do not change when the amount of substance changes (e.g., temperature, pressure, density).

Step 2: Analyze each option:\nA. Volume: Depends on the amount of substance. If you have more substance, you have more volume. Therefore, it's extensive.\nB. Viscosity: An intensive property. It doesn't depend on the amount of substance.\nC. Density: An intensive property. It's mass per unit volume and doesn't depend on the amount of substance.\nD. Temperature: An intensive property. It doesn't depend on the amount of substance.

Step 3: Conclusion: Volume is the only extensive property among the given options.

Why other options are wrong:
- Option B: Viscosity is an intensive property, independent of the amount of substance.
- Option C: Density is an intensive property, independent of the amount of substance.
- Option D: Temperature is an intensive property, independent of the amount of substance.

Key Concepts:
- Extensive properties
- Intensive properties

Memory Aids:
- Extensive: "eXtends" with the amount of substance.
- Intensive: "Independent" of the amount of substance.

Quick Tips:
- Remember the definitions of extensive and intensive properties. Consider what happens if you double the amount of substance.

63 Easy

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Thermochemistry is the study of relationship between heat energy and :

View answer and explanation

Correct option: A

Chemical energy

Option A (Chemical energy) is correct because thermochemistry deals with the relationship between heat and chemical changes.

Solution approach

Correct Answer + Core Reason: Chemical energy. Thermochemistry studies the relationship between heat and chemical changes.

Step-by-Step Solution:

Step 1: Define thermochemistry.

Step 2: Thermochemistry is the study of the heat changes that accompany chemical reactions and physical transformations.

Step 3: It explores the relationship between heat energy and chemical energy.

Why other options are wrong:
- Option B: Activation energy is incorrect because while activation energy is related to chemical reactions, thermochemistry focuses on the heat changes during the reaction, not the activation energy itself.
- Option C: Frictional energy is incorrect because frictional energy is related to mechanical processes, not chemical reactions.
- Option D: None of these is incorrect because option A is the correct answer.

Key Concepts:
- Thermochemistry
- Chemical energy
- Heat energy

Memory Aids:
- Thermochemistry: Heat associated with chemical reactions.

Quick Tips:
- Remember the definition of thermochemistry.

64 Easy

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Which of the following are path functions ?
a) Heat capacity b) work c) Entropy

View answer and explanation

Correct option: C

a and b only

Option C (a and b only) is correct because heat capacity and work are path-dependent, while entropy is a state function.

Solution approach

Correct Answer + Core Reason: a and b only. Heat capacity and work are path functions, while entropy is a state function.

Step-by-Step Solution:

Step 1: Define path function. A path function is a property whose value depends on the path taken to reach a particular state.

Step 2: Define state function. A state function is a property whose value depends only on the initial and final states, not on the path taken.

Step 3: Identify heat capacity, work, and entropy. Heat capacity and work are path functions, while entropy is a state function.

Step 4: Conclusion: Therefore, heat capacity and work are path functions.

Why other options are wrong:
- Option A: a and c only: Entropy is a state function, not a path function.
- Option B: b and c only: Entropy is a state function, not a path function.
- Option D: a,b and c: Entropy is a state function, not a path function.

Key Concepts:
- Path function
- State function
- Heat capacity
- Work
- Entropy

Memory Aids:
- Path functions depend on the path; state functions depend only on the initial and final states.

Quick Tips:
- Remember the key examples of path and state functions.

65 Easy

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Entropy of an adiabatic reversible process is:-

View answer and explanation

Correct option: D

Constant

Option D (Constant) is correct because entropy remains constant in a reversible adiabatic process, as there is no heat exchange with the surroundings.

Solution approach

Correct Answer + Core Reason: Constant. In a reversible adiabatic process, there is no heat transfer, hence entropy remains constant.

Step-by-Step Solution:

Step 1: Define adiabatic process: An adiabatic process is one in which there is no heat exchange between the system and the surroundings ( $q = 0$ ).

Step 2: Define reversible process: A reversible process is one that occurs infinitesimally slowly and can be reversed without any net change in the system or surroundings.

Step 3: Entropy change formula: The change in entropy ( $\Delta S$ ) is given by $\Delta S = \frac{q_{rev}}{T}$ , where $q_{rev}$ is the heat exchanged in a reversible process and $T$ is the temperature.

Step 4: Apply to adiabatic reversible process: Since $q = 0$ in an adiabatic process, $\Delta S = \frac{0}{T} = 0$ . Therefore, entropy remains constant.

Why other options are wrong:
- Option A: Positive. Entropy increases when heat is added to the system, which doesn't happen in an adiabatic process.
- Option B: Zero. While the *change* in entropy is zero, this option is less precise than 'Constant'.
- Option C: Negative. Entropy decreases when heat is removed from the system, which doesn't happen in an adiabatic process.

Key Concepts:
- Adiabatic process
- Reversible process
- Entropy

Memory Aids:
- Adiabatic: No heat exchange.
- Reversible: Infinitesimally slow.

Quick Tips:
- Remember that adiabatic processes imply no heat transfer, which directly relates to entropy change.

66 Easy

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Application +4 | -1

For the reaction

aA\left( s \right) + bB\left( g \right)\xrightarrow{\,}dD\left( s \right) + cC\left( g \right),

then

View answer and explanation

Correct option: B

\Delta H - \Delta E = \left( {c - b} \right)RT

Option B (

\Delta H - \Delta E = (c - b)RT

) is correct because it correctly applies the relationship between enthalpy change, internal energy change, and the change in the number of moles of gas.

Solution approach

Correct Answer + Core Reason: $\Delta H - \Delta E = (c - b)RT$ . This equation correctly relates the difference between enthalpy and internal energy changes to the change in moles of gas.

Step-by-Step Solution:
Step 1: Recall the relationship between enthalpy change ( $\Delta H$ ) and internal energy change ( $\Delta E$ ):
$\Delta H = \Delta E + P\Delta V$
Step 2: For a reaction involving gases, we can use the ideal gas law: $PV = nRT$ , so $P\Delta V = \Delta n_g RT$ , where $\Delta n_g$ is the change in the number of moles of gas.
Step 3: Substitute $P\Delta V$ with $\Delta n_g RT$ in the equation for $\Delta H$ :
$\Delta H = \Delta E + \Delta n_g RT$
Step 4: Rearrange the equation to find $\Delta H - \Delta E$ :
$\Delta H - \Delta E = \Delta n_g RT$
Step 5: Determine the change in the number of moles of gas ( $\Delta n_g$ ). In the reaction $aA(s) + bB(g) \rightarrow dD(s) + cC(g)$ , the change in the number of moles of gas is:
$\Delta n_g = \text{moles of gaseous products} - \text{moles of gaseous reactants} = c - b$
Step 6: Substitute $\Delta n_g = c - b$ into the equation:
$\Delta H - \Delta E = (c - b)RT$

Why other options are wrong:
- Option A: This equation is incorrect because it uses (b - d) instead of (c - b) for the change in moles of gas.
- Option C: This equation is incorrect because it includes the moles of solids (a and d) in the calculation of the change in moles.
- Option D: This equation is incorrect because it only considers the moles of solids (a and d) and not the moles of gases.

Key Concepts:
- Enthalpy change
- Internal energy change
- Ideal gas law
- Change in moles of gas

Memory Aids:
- Remember that $\Delta H = \Delta E + P\Delta V$ and $P\Delta V = \Delta n_g RT$ for gases.

Quick Tips:
- Only consider the change in the number of moles of gaseous species when calculating $\Delta n_g$ .

67 Easy

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Entropy of a perfectly crystalline solid at 0 K is

View answer and explanation

Correct option: C

zero

Option C (zero) is correct because the third law of thermodynamics states that the entropy of a perfectly crystalline solid at 0 K is zero.

Solution approach

Correct Answer + Core Reason: zero. According to the Third Law of Thermodynamics, the entropy of a perfectly crystalline solid at absolute zero (0 K) is zero.

Step-by-Step Solution:

Step 1: Recall the Third Law of Thermodynamics: The Third Law of Thermodynamics states that as the temperature approaches absolute zero, the entropy of a system approaches a minimum or zero value. For a perfectly crystalline solid, this minimum value is zero.

Step 2: Explain the concept of entropy: Entropy is a measure of disorder or randomness in a system. In a perfectly crystalline solid at 0 K, all atoms are perfectly ordered, and there is no disorder. Therefore, the entropy is zero.

Why other options are wrong:
- Option A: Entropy cannot be positive at 0K for a perfectly crystalline solid.
- Option B: Entropy cannot be negative. Entropy is zero or positive.
- Option D: Entropy is zero for a perfectly crystalline solid at 0K. It cannot be either positive or negative.

Key Concepts:
- Entropy
- Third Law of Thermodynamics
- Absolute Zero (0 K)
- Perfectly Crystalline Solid

Memory Aids:
- Third Law: Zero Entropy at Zero Kelvin (for perfect crystal)

Quick Tips:
- Remember that the Third Law provides a reference point for entropy calculations.

68 Easy

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Predict the change in internal energy for an isolated system at constant volume

View answer and explanation

Correct option: A

Zero

Option A (Zero) is correct because in an isolated system at constant volume, no heat or work is exchanged with the surroundings, thus the internal energy remains constant.

Solution approach

Correct Answer + Core Reason: Zero. For an isolated system, there is no exchange of energy (heat or work) with the surroundings, and at constant volume, no work is done. Therefore, the change in internal energy is zero.

Step-by-Step Solution:

Step 1: Define an isolated system: An isolated system is a system that cannot exchange energy (heat or work) or matter with its surroundings.

Step 2: Define constant volume: Constant volume means that the volume of the system does not change. Therefore, no work is done by or on the system (W = 0).

Step 3: Recall the First Law of Thermodynamics: The First Law of Thermodynamics states that the change in internal energy (Δ U) of a system is equal to the heat added to the system (q) minus the work done by the system (w): Δ U = q - w.

Step 4: Apply the conditions to the First Law: For an isolated system, q = 0. At constant volume, w = 0. Therefore, Δ U = 0 - 0 = 0.

Why other options are wrong:
- Option B: q would be the change in internal energy if work was zero, but the system is isolated so q=0.
- Option C: w would be the change in internal energy if heat was zero, but the system is isolated and at constant volume so w=0.
- Option D: The change in internal energy is zero.

Key Concepts:
- Isolated System
- Constant Volume
- First Law of Thermodynamics
- Internal Energy

Memory Aids:
- Isolated = No exchange with surroundings
- Constant Volume = No Work

Quick Tips:
- Remember the definitions of isolated system and constant volume conditions.

69 Easy

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A well stoppered thermos flask containing some ice cubes is an example of:

View answer and explanation

Correct option: C

Isolated system

Option C is correct because a thermos flask prevents both heat and mass exchange with the surroundings, making it an isolated system.

Solution approach

Correct Answer + Core Reason: Isolated system. A thermos flask prevents both heat and mass exchange.

Step-by-Step Solution:

Step 1: Define an isolated system. An isolated system is a system that cannot exchange either matter or energy with its surroundings.

Step 2: Define a closed system. A closed system can exchange energy but not matter with its surroundings.

Step 3: Define an open system. An open system can exchange both matter and energy with its surroundings.

Step 4: Analyze the situation. A well-stoppered thermos flask prevents the exchange of both matter (ice cubes cannot escape) and energy (heat transfer is minimized) with the surroundings.

Why other options are wrong:
- Option A: A closed system allows energy exchange, which is minimized in a thermos flask.
- Option B: An open system allows both matter and energy exchange, which is not the case for a thermos flask.
- Option D: A thermos flask is an example of a system, specifically an isolated system.

Key Concepts:
- Isolated System
- Closed System
- Open System

Memory Aids:
- Isolated: No exchange
- Closed: Energy exchange only
- Open: Both matter and energy exchange

Quick Tips:
- Remember the definitions of the three types of systems.

70 Easy

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For the process,

\mathrm{C}{\mathrm{O}_2}\left( \mathrm{s} \right) \to \mathrm{C}{\mathrm{O}_2}\left( \mathrm{g} \right):

View answer and explanation

Correct option: A

Both ∆H and ∆S are +ve

The correct answer is A (Both ∆H and ∆S are +ve) because sublimation requires energy input (endothermic, +∆H) and increases disorder (+∆S).

Solution approach

Correct Answer + Core Reason: Both ∆H and ∆S are +ve. Sublimation is endothermic and increases entropy.

Step-by-Step Solution:

Step 1: Analyze the process: $\mathrm{C}{\mathrm{O}_2}(\mathrm{s}) \to \mathrm{C}{\mathrm{O}_2}(\mathrm{g})$ is sublimation (solid to gas).

Step 2: Consider enthalpy change (∆H): Sublimation requires energy to overcome intermolecular forces, so it's an endothermic process. Therefore, ∆H is positive (+ve).

Step 3: Consider entropy change (∆S): Gases are more disordered than solids. Sublimation increases disorder. Therefore, ∆S is positive (+ve).

Why other options are wrong:
- Option B: ∆H is -ve and ∆S is +ve: ∆H is not negative because sublimation requires energy input.
- Option C: ∆H is +ve and ∆S is -ve: ∆S is not negative because the process increases disorder.
- Option D: Both ∆H and ∆S are -ve: Neither ∆H nor ∆S are negative in this process.

Key Concepts:
- Enthalpy (H)
- Entropy (S)
- Sublimation
- Endothermic process
- Disorder

Memory Aids:
- Solids to gases increase disorder (entropy).
- Sublimation requires energy (endothermic).

Quick Tips:
- Remember that phase transitions from solid to liquid to gas increase entropy.

71 Easy

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Remembering +4 | -1

For a process, entropy change of a system is
expressed a

View answer and explanation

Correct option: B

\frac{q_{\text {rev }}}{T}

Option B (

\frac{{q_{\text{{rev }}}}}{{T}}

) is correct because it represents the definition of entropy change for a reversible process.

Solution approach

Correct Answer + Core Reason: $\frac{q_{\text {rev }}}{T}$ . Entropy change is defined as the heat absorbed in a reversible process divided by the temperature.

Step-by-Step Solution:

Step 1: Recall the definition of entropy change.

Step 2: The entropy change, $\Delta S$ , for a reversible process is given by the formula: $\Delta S = \frac{{q_{\text{{rev}}}}}}{{T}}$ , where $q_{\text{{rev}}}$ is the heat absorbed reversibly and $T$ is the absolute temperature.

Step 3: Therefore, the entropy change of a system is expressed as $\frac{{q_{\text{{rev}}}}}}{{T}}$ .

Why other options are wrong:
- Option A: H - TS represents Gibbs Free Energy, not entropy change.
- Option C: $\frac{T}{q_{\text {rev }}}$ is the inverse of entropy change and is incorrect.
- Option D: $q_{\text {rev }} \times \mathrm{T}$ is not the correct formula for entropy change.

Key Concepts:
- Entropy change
- Reversible process
- Thermodynamics

Memory Aids:
- Remember the formula: $\Delta S = \frac{{q_{\text{{rev}}}}}}{{T}}$

Quick Tips:
- Always remember the definition of entropy change for a reversible process.

72 Easy

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Remembering +4 | -1

In isolated system:

View answer and explanation

Correct option: D

Neither energy nor mass transfer take place

Option D (Neither energy nor mass transfer take place) is correct because an isolated system, by definition, does not exchange energy or mass with its surroundings.

Solution approach

Correct Answer + Core Reason: Neither energy nor mass transfer take place. An isolated system does not exchange energy or mass with its surroundings.

Step-by-Step Solution:

Step 1: Define an isolated system.\nAn isolated system is a thermodynamic system that cannot exchange either energy or matter with its surroundings.

Step 2: Analyze the options based on the definition of an isolated system.\nEnergy transfer does not take place.\nMass transfer does not take place.\nTherefore, neither energy nor mass transfer takes place.

Why other options are wrong:
- Option A: Energy transfer does not take place. This is only partially correct. Mass transfer also does not take place.
- Option B: Mass transfer does not take place. This is only partially correct. Energy transfer also does not take place.
- Option C: Both energy $\& mass transfer take place. This is incorrect because an isolated system does not allow either energy or mass transfer.\n\n**Key Concepts:**\n- Isolated system\n- Thermodynamic systems\n- Energy transfer\n- Mass transfer\n\n**Memory Aids:**\n- Isolated system: no exchange of energy or mass.\n\n**Quick Tips:**\n- Remember the definitions of different types of thermodynamic systems (isolated, closed, open).$

73 Easy

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Application +4 | -1

An ideal gas absorb

50\, \mathrm{cal}

heat and expand in the vaccum from

5 \mathrm{~L}

to

15 \mathrm{~L}

. Change in its internal energy is :-

View answer and explanation

Correct option: B

50\, \mathrm{cal}

Option B (50 cal) is correct because in a vacuum expansion, no external pressure exists, so no work is done, and the heat absorbed equals the change in internal energy.

Solution approach

Correct Answer + Core Reason: $50\, \mathrm{cal}$ . In vacuum expansion, work done is zero, so change in internal energy equals heat absorbed.

Step-by-Step Solution:

Step 1: Identify the given information:
- Heat absorbed, $q = 50\, \mathrm{cal}$
- Expansion in vacuum

Step 2: Recall the definition of work done during expansion: $w = -P_{ext} \Delta V$ , where $P_{ext}$ is the external pressure and $\Delta V$ is the change in volume.

Step 3: Since the expansion occurs in a vacuum, the external pressure $P_{ext} = 0$ .

Step 4: Therefore, the work done $w = -0 \times (15\, \mathrm{L} - 5\, \mathrm{L}) = 0$ .

Step 5: Apply the first law of thermodynamics: $\Delta U = q + w$ , where $\Delta U$ is the change in internal energy.

Step 6: Substitute the values: $\Delta U = 50\, \mathrm{cal} + 0 = 50\, \mathrm{cal}$

Why other options are wrong:
- Option A: $-232\, \mathrm{cal}$ is incorrect. It likely involves an incorrect calculation or misunderstanding of the process.
- Option C: $292\, \mathrm{cal}$ is incorrect. It likely involves an incorrect calculation or misunderstanding of the process.
- Option D: 0 is incorrect because heat is absorbed by the gas.

Key Concepts:
- First law of thermodynamics
- Work done in expansion
- Expansion in vacuum

Memory Aids:
- Vacuum: No pressure, no work.

Quick Tips:
- In vacuum expansion, the work done is always zero.

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Understanding +4 | -1

In a closed container, a liquid is stirred with a paddle to increase the temperature. Which of the following is true?

View answer and explanation

Correct option: A

∆E=W≠0,q=0

Option A, ∆E=W≠0,q=0, is correct because stirring the liquid with a paddle adds work to the system, increasing its internal energy, and since the container is closed, no heat is exchanged.

Solution approach

Correct Answer + Core Reason: ∆E=W≠0,q=0. Work is done on the system, increasing internal energy, and no heat is exchanged.

Step-by-Step Solution:

Step 1: Recall the first law of thermodynamics: $\Delta E = q + W$ , where $\Delta E$ is the change in internal energy, $q$ is the heat added to the system, and $W$ is the work done on the system.

Step 2: The liquid is stirred with a paddle, which means work is done *on* the system. Therefore, $W \neq 0$ .

Step 3: The container is closed, so no heat is exchanged with the surroundings. Therefore, $q = 0$ .

Step 4: Substituting into the first law: $\Delta E = 0 + W = W$

Step 5: Since $W \neq 0$ , $\Delta E \neq 0$ .

Step 6: Final Answer: ∆E=W≠0,q=0

Why other options are wrong:
- Option B: ∆E=W=q≠0 is incorrect because q=0 since no heat is exchanged.
- Option C: ∆E=0,W=q≠0 is incorrect because work is done on the system and q=0, so ∆E cannot be zero.
- Option D: W=0,∆E=q≠0 is incorrect because work is done on the system, so W cannot be zero.

Key Concepts:
- First Law of Thermodynamics
- Internal Energy (E)
- Heat (q)
- Work (W)

Memory Aids:
- Remember that work done *on* the system is positive, and work done *by* the system is negative.

Quick Tips:
- Pay attention to whether work is done on the system or by the system.

75 Easy

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Application +4 | -1

In previous problem if expansion is carried out freely

\left(P_{\mathrm{ext}}=0\right)

, then

\Delta S

is :

View answer and explanation

Correct option: C

R \ln 3

Option C (

R \ln 3

) is correct because for free expansion, the change in entropy is given by

\Delta S = nR \ln(\frac{V_2}{V_1})

. Since the volume increases by a factor of 3 (from the previous problem),

\Delta S = R \ln 3

.

Solution approach

Correct Answer + Core Reason: $R \ln 3$ . The entropy change for free expansion is $\Delta S = nR \ln(\frac{V_2}{V_1})$ .

Step-by-Step Solution:

Step 1: Understand the concept of free expansion. In free expansion, the external pressure is zero ( $P_{\text{ext}} = 0$ ).

Step 2: Recall the formula for entropy change during isothermal expansion: $\Delta S = nR \ln(\frac{V_2}{V_1})$ .

Step 3: From the previous problem (not shown, but implied), assume the volume expands to 3 times its original volume. Therefore, $\frac{V_2}{V_1} = 3$ .

Step 4: Substitute the values into the formula: $\Delta S = (1 \text{ mol}) \times R \times \ln(3)$ .

Step 5: Simplify the expression: $\Delta S = R \ln 3$ .

Final Answer: $R \ln 3$

Why other options are wrong:
- Option A: zero is incorrect. Entropy changes during free expansion even though no work is done.
- Option B: infinity is incorrect. While entropy increases, it doesn't become infinite in this scenario.
- Option D: None is incorrect because $R \ln 3$ is the correct answer.

Key Concepts:
- Entropy change during isothermal expansion: $\Delta S = nR \ln(\frac{V_2}{V_1})$
- Free expansion

Memory Aids:
- Remember that entropy always increases in irreversible processes like free expansion.

Quick Tips:
- For free expansion, $P_{\text{ext}} = 0$ and $W = 0$ , but $\Delta S \neq 0$ .

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Remembering +4 | -1

The correct thermochemical equation is :

View answer and explanation

Correct option: C

\mathrm{C}\left( s \right) + {\mathrm{O}_2}\left( \mathrm{g} \right) \to \mathrm{C}{\mathrm{O}_2}\left( \mathrm{g} \right);\Delta H = - 94\mathrm{kcal}

Option C (

\mathrm{C}( s ) + {\mathrm{O}_2}( \mathrm{g} ) \to \mathrm{C}{\mathrm{O}_2}( \mathrm{g} );\Delta H = - 94\mathrm{kcal}

) is correct because it represents the formation of carbon dioxide from its elements in their standard states (solid carbon and gaseous oxygen) with the correct negative sign for an exothermic reaction.

Solution approach

Correct Answer + Core Reason: $\mathrm{C}( s ) + {\mathrm{O}_2}( \mathrm{g} ) \to \mathrm{C}{\mathrm{O}_2}( \mathrm{g} );\Delta H = - 94\mathrm{kcal}$ . This equation correctly represents the formation of CO2 from its elements in their standard states with a negative $\Delta H$ value.

Step-by-Step Solution:

Step 1: Understand the requirements for a correct thermochemical equation:
- Balanced chemical equation.
- Physical states of reactants and products are specified.
- Enthalpy change ( $\Delta H$ ) is given with the correct sign.

Step 2: Analyze each option:
- Option A: $\mathrm{C} + {\mathrm{O}_2} \to \mathrm{C}{\mathrm{O}_2};\Delta H = - 94\mathrm{kcal}$ - Missing physical states.
- Option B: $\mathrm{C} + {\mathrm{O}_2} \to \mathrm{C}{\mathrm{O}_2};\Delta H = + 94\mathrm{kcal}$ - Missing physical states and incorrect sign for $\Delta H$ (combustion is exothermic).
- Option C: $\mathrm{C}( s ) + {\mathrm{O}_2}( \mathrm{g} ) \to \mathrm{C}{\mathrm{O}_2}( \mathrm{g} );\Delta H = - 94\mathrm{kcal}$ - Correctly represents the reaction with physical states and a negative $\Delta H$ .
- Option D: $\mathrm{C}( s ) + {\mathrm{O}_2}( \mathrm{g} ) \to \mathrm{C}{\mathrm{O}_2}( \mathrm{g} );\Delta H = + 94\mathrm{kcal}$ - Incorrect sign for $\Delta H$ (combustion is exothermic).

Step 3: Select the correct option. Option C is the only one that satisfies all the requirements.

Why other options are wrong:
- Option A: Missing the physical states of the reactants and products. A thermochemical equation must specify the physical states.
- Option B: Missing the physical states and has the wrong sign for $\Delta H$ . The formation of CO2 from C and O2 is an exothermic process, so $\Delta H$ should be negative.
- Option D: Has the correct physical states but the wrong sign for $\Delta H$ . The reaction is exothermic, so $\Delta H$ should be negative.

Key Concepts:
- Thermochemical Equations
- Enthalpy Change ( $\Delta H$ )
- Standard States
- Exothermic Reactions

Memory Aids:
- Exothermic reactions release heat (negative $\Delta H$ ).
- Endothermic reactions absorb heat (positive $\Delta H$ ).

Quick Tips:
- Always include physical states in thermochemical equations.
- Remember that combustion reactions are exothermic.

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Application +4 | -1

The enthalpies of neutralization of a weak base

\mathrm{AOH}

and a strong base

\mathrm{BOH}

by

\mathrm{HCl}

are

-12250 \mathrm{cal} / \mathrm{mol}

and

-13000 \mathrm{cal} / \mathrm{mol}

respectively. When one mole of

\mathrm{HCl}

is added to a solution containing 1 mole of

A O H

and 1 mole of

B O H

, the enthalpy change was

-12500 \mathrm{cal} / \mathrm{mol}

. In what ratio is the acid distribution between

A O H

and

B O H

?

View answer and explanation

Correct option: A

2: 1

Option A (

2: 1

) is correct because the ratio is determined by setting up equations based on the enthalpy changes and solving for the distribution of the acid between the two bases.

Solution approach

Correct Answer + Core Reason: $2: 1$ . The ratio is determined by the enthalpy changes during neutralization.

Step-by-Step Solution:

Step 1: Define variables: Let $x$ be the moles of $\mathrm{HCl}$ reacting with $\mathrm{AOH}$ and $y$ be the moles of $\mathrm{HCl}$ reacting with $\mathrm{BOH}$ .

Step 2: Set up equations based on the given information:
- $x + y = 1$ (total moles of $\mathrm{HCl}$ )
- $-12250x - 13000y = -12500$ (total enthalpy change)

Step 3: Solve the system of equations:
- From the first equation, $y = 1 - x$ .
- Substitute into the second equation: $-12250x - 13000(1 - x) = -12500$ .
- Simplify: $-12250x - 13000 + 13000x = -12500$ .
- Combine like terms: $750x = 500$ .
- Solve for $x$ : $x = \frac{500}{750} = \frac{2}{3}$ .

Step 4: Calculate $y$ : $y = 1 - x = 1 - \frac{2}{3} = \frac{1}{3}$ .

Step 5: Determine the ratio of acid distribution between $\mathrm{AOH}$ and $\mathrm{BOH}$ : $x : y = \frac{2}{3} : \frac{1}{3} = 2 : 1$ .

Final Answer: $2: 1$

Why other options are wrong:
- Option B: $2: 3$ is incorrect. This ratio does not satisfy the enthalpy change equation.
- Option C: $1: 2$ is incorrect. This ratio would result in a different enthalpy change than the given value.
- Option D: None of these is incorrect because $2:1$ is the correct ratio.

Key Concepts:
- Enthalpy of neutralization
- Stoichiometry
- System of equations

Memory Aids:
- Remember that enthalpy is an additive property.

Quick Tips:
- Always set up a system of equations when dealing with multiple reactions and enthalpy changes.
- Pay attention to the signs of enthalpy changes (exothermic vs. endothermic).

78 Easy

78 of 196Question permalink

Understanding +4 | -1

Two atoms of hydrogen combine to form a molecule of hydrogen gas, the energy of the

{H_2}

molecule is :

View answer and explanation

Correct option: C

Lower than that of separate atoms

Option C (Lower than that of separate atoms) is correct because bond formation releases energy, resulting in a lower energy state for the molecule.

Solution approach

Correct Answer + Core Reason: Lower than that of separate atoms. Bond formation releases energy.

Step-by-Step Solution:

Step 1: Describe bond formation.

Step 2: When two atoms combine to form a molecule, a chemical bond is formed. The formation of a chemical bond releases energy, making the molecule more stable than the separate atoms.

Step 3: Therefore, the energy of the $H_2$ molecule is lower than that of the separate hydrogen atoms.

Why other options are wrong:
- Option A: Greater than that of separate atoms is incorrect because bond formation releases energy, lowering the energy of the molecule.
- Option B: Equal to that of separate atoms is incorrect because bond formation involves a change in energy.
- Option D: Sometimes lower and sometimes higher is incorrect because bond formation always releases energy, resulting in a lower energy state.

Key Concepts:
- Chemical bond
- Bond formation
- Energy release

Memory Aids:
- Bond formation = Energy release = Lower energy state

Quick Tips:
- Remember that bond formation is an exothermic process.

79 Easy

79 of 196Question permalink

Remembering +4 | -1

When water is added to quick lime, the reaction is :

View answer and explanation

Correct option: C

Exothermic

Option C (Exothermic) is correct because the reaction of quick lime (calcium oxide) with water releases a significant amount of heat.

Solution approach

Correct Answer + Core Reason: Exothermic. The reaction between quick lime and water releases heat.

Step-by-Step Solution:

Step 1: The reaction between quick lime (CaO) and water (H2O) forms calcium hydroxide (Ca(OH)2).

Step 2: The reaction is: $CaO(s) + H_2O(l) \rightarrow Ca(OH)_2(aq) + heat$

Step 3: This reaction releases a significant amount of heat, making it an exothermic reaction.

Why other options are wrong:
- Option A: Explosive. While the reaction can be vigorous, it is not typically explosive under normal conditions.
- Option B: Endothermic. Endothermic reactions absorb heat, which is the opposite of what happens in this reaction.
- Option D: None of these. The reaction is clearly exothermic.

Key Concepts:
- Exothermic reactions
- Endothermic reactions
- Chemical reactions

Memory Aids:
- Remember that quick lime + water = heat.

Quick Tips:
- Recognize common exothermic and endothermic reactions.

80 Easy

80 of 196Question permalink

Application +4 | -1

What is the enthalpy change for,

2 {H}_2 {O}_2(l) \rightarrow 2 {H}_2 {O}(l)+{O}_2(g)

if heat of formation of

{H}_2 {O}_2(l)

and

{H}_2 {O}(l)

are -188 and

-286 {~kJ} / {mol}

respectively?

View answer and explanation

Correct option: A

-196 {~kJ} / {mol}

Option A (-196 kJ/mol) is correct because it is calculated by subtracting the enthalpy of formation of the reactants from the enthalpy of formation of the products, considering the stoichiometric coefficients.

Solution approach

Correct Answer + Core Reason: $-196 \mathrm{~kJ} / \mathrm{mol}$ . The enthalpy change is calculated using the heats of formation of products and reactants.

Step-by-Step Solution:
Step 1: Write the given reaction:
$2 {H}_2 {O}_2(l) \rightarrow 2 {H}_2 {O}(l)+{O}_2(g)$
Step 2: Recall the formula for calculating the enthalpy change of a reaction using heats of formation:
$\Delta H_{rxn} = \sum \Delta H_f(\text{products}) - \sum \Delta H_f(\text{reactants})$
Step 3: Identify the heats of formation:
$\Delta H_f({H}_2 {O}_2(l)) = -188 \mathrm{~kJ} / \mathrm{mol}$
$\Delta H_f({H}_2 {O}(l)) = -286 \mathrm{~kJ} / \mathrm{mol}$
$\Delta H_f({O}_2(g)) = 0 \mathrm{~kJ} / \mathrm{mol}$ (since it's in its standard state)
Step 4: Substitute the values into the formula:
$\Delta H_{rxn} = [2 \times (-286 \mathrm{~kJ} / \mathrm{mol}) + 1 \times (0 \mathrm{~kJ} / \mathrm{mol})] - [2 \times (-188 \mathrm{~kJ} / \mathrm{mol})]$
$\Delta H_{rxn} = [-572 \mathrm{~kJ} / \mathrm{mol}] - [-376 \mathrm{~kJ} / \mathrm{mol}]$
$\Delta H_{rxn} = -572 + 376 = -196 \mathrm{~kJ} / \mathrm{mol}$

Why other options are wrong:
- Option B: This value has the wrong sign and magnitude. It likely results from adding the heats of formation instead of subtracting or using incorrect stoichiometric coefficients.
- Option C: This value has the wrong sign. It might be obtained by subtracting the products from the reactants incorrectly.
- Option D: This value has the wrong magnitude. It might be obtained by multiplying the heats of formation incorrectly or not considering the stoichiometric coefficients.

Key Concepts:
- Enthalpy change of reaction
- Heat of formation
- Hess's law

Memory Aids:
- Remember that $\Delta H_{rxn} = \sum \Delta H_f(\text{products}) - \sum \Delta H_f(\text{reactants})$ .

Quick Tips:
- Always consider the stoichiometric coefficients when calculating the enthalpy change.
- Remember that the heat of formation of an element in its standard state is zero.

81 Easy

81 of 196Question permalink

Application +4 | -1

A system absorbs 10 kJ of heat and does 4 kJ of work. The internal energy of the system

View answer and explanation

Correct option: B

Increases by 6 kJ

Option B (Increases by 6 kJ) is correct because, according to the first law of thermodynamics, the internal energy increases by the difference between the heat absorbed and the work done by the system.

Solution approach

Correct Answer + Core Reason: Increases by 6 kJ. The internal energy increases by the difference between heat absorbed and work done.

Step-by-Step Solution:

Step 1: Recall the first law of thermodynamics: $\Delta U = q - w$

Step 2: Identify the given values: $q = +10 \text{{ kJ}}$ (heat absorbed), $w = +4 \text{{ kJ}}$ (work done by the system).

Step 3: Substitute the values into the equation: $\Delta U = 10 \text{{ kJ}} - 4 \text{{ kJ}}$

Step 4: Calculate the change in internal energy: $\Delta U = 6 \text{{ kJ}}$

Step 5: Since $\Delta U$ is positive, the internal energy increases.

Why other options are wrong:
- Option A: Decreases by 6 kJ. Incorrect, the internal energy increases because more heat is absorbed than work is done.
- Option C: Decreases by 14 kJ. Incorrect, this would be the case if the system did 24 kJ of work after absorbing 10 kJ of heat.
- Option D: Increases by 14 kJ. Incorrect, this would be the case if the system absorbed 10 kJ of heat and work was done *on* the system (instead of *by* the system) of 4 kJ, or if the work was calculated incorrectly.

Key Concepts:
- First law of thermodynamics, Internal energy, Heat, Work

Memory Aids:
- Remember the sign convention: heat absorbed is positive, heat released is negative; work done by the system is positive, work done on the system is negative.

Quick Tips:
- Always pay attention to the sign conventions for heat and work.

82 Easy

82 of 196Question permalink

Application +4 | -1

The value of

\Delta H

for the reversible isothermal evaporation of

90 \mathrm{~g}

water at

100^{\circ} \mathrm{C}

will be

View answer and explanation

Correct option: A

48600 \mathrm{~cal}

Option A (48600 cal) is correct because it is the heat required to vaporize 90g of water at 100°C, calculated using the latent heat of vaporization.

Solution approach

Correct Answer + Core Reason: 48600 cal. The heat required for vaporization is calculated using the formula $Q = mL$ , where $m$ is the mass and $L$ is the latent heat of vaporization.

Step-by-Step Solution:

Step 1: Identify the given information:\nMass of water, $m = 90 \text{{ g}}$ \nTemperature, $T = 100^{\circ} \text{{C}}$ \nLatent heat of vaporization of water, $L = 540 \text{{ cal/g}}$

Step 2: Apply the formula for heat required for vaporization:
$Q = m \times L$

Step 3: Substitute the values:
$Q = 90 \text{{ g}} \times 540 \text{{ cal/g}}$

Step 4: Calculate the heat required:
$Q = 48600 \text{{ cal}}$

Why other options are wrong:
- Option B (4860 cal): This is incorrect as it divides the correct answer by 10. It might be a calculation error.
- Option C (486 cal): This is incorrect as it divides the correct answer by 100. It might be a calculation error.
- Option D (48.6 cal): This is incorrect as it divides the correct answer by 1000. It might be a calculation error.

Key Concepts:
- Latent heat of vaporization
- Heat transfer during phase change

Memory Aids:
- $Q = mL$ (Heat = mass x latent heat)

Quick Tips:
- Ensure correct units are used in the calculation.
- Remember the latent heat of vaporization of water.

83 Easy

83 of 196Question permalink

Remembering +4 | -1

Which is correct relation between

\mathrm{C}_{\mathrm{p}}

and

\mathrm{C}_{\mathrm{v}}

for one mole of ideal gas

View answer and explanation

Correct option: C

\mathrm{C}_{\mathrm{p}}-\mathrm{C}_{\mathrm{v}}=\mathrm{R}

Option C,

\mathrm{C}_{\mathrm{p}}-\mathrm{C}_{\mathrm{v}}=\mathrm{R}

, is the correct relationship between the molar heat capacity at constant pressure (

C_p

) and constant volume (

C_v

) for one mole of an ideal gas.

Solution approach

Correct Answer + Core Reason: $\mathrm{C}_{\mathrm{p}}-\mathrm{C}_{\mathrm{v}}=\mathrm{R}$ . This equation represents the correct relationship between $C_p$ and $C_v$ for an ideal gas.

Step-by-Step Solution:

Step 1: Recall the relationship between molar heat capacity at constant pressure ( $C_p$ ) and molar heat capacity at constant volume ( $C_v$ ) for an ideal gas.

Step 2: The relationship is given by: $C_p - C_v = R$ , where R is the ideal gas constant.

Step 3: Final Answer: $\mathrm{C}_{\mathrm{p}}-\mathrm{C}_{\mathrm{v}}=\mathrm{R}$

Why other options are wrong:
- Option A: $\mathrm{C}_v=\mathrm{RC}_p$ is incorrect because it implies that $C_v$ is greater than $C_p$ , which is not true for ideal gases.
- Option B: $\mathrm{C}_{\mathrm{p}}+\mathrm{C}_{\mathrm{v}}=\mathrm{R}$ is incorrect because it suggests that the sum of $C_p$ and $C_v$ equals R, which is not the correct relationship.
- Option D: $C_p=R C_v$ is incorrect because it implies that $C_p$ is smaller than $C_v$ , which is not true for ideal gases.

Key Concepts:
- Molar Heat Capacity at Constant Pressure ( $C_p$ )
- Molar Heat Capacity at Constant Volume ( $C_v$ )
- Ideal Gas Constant (R)
- Thermodynamics

Memory Aids:
- Remember that $C_p$ is always greater than $C_v$ for an ideal gas, and the difference is R.

Quick Tips:
- This relationship is fundamental in thermodynamics and is often used in calculations involving ideal gases.

84 Easy

84 of 196Question permalink

Analyzing +4 | -1

Which is not correct?

View answer and explanation

Correct option: D

∆H is less than ∆U for the reaction, C(s)+ (1/2)

{O_2}

(g)→

C{O_2}

(g)

Option D (∆H is less than ∆U for the reaction, C(s)+ (1/2)

{O_2}

(g)→

C{O_2}

(g)) is correct because ∆H is greater than ∆U for this reaction since ∆ng > 0.

Solution approach

Correct Answer + Core Reason: ∆H is less than ∆U for the reaction, C(s)+ (1/2) ${O_2}$ (g)→ $C{O_2}$ (g). This statement is incorrect because ∆H > ∆U for this reaction.

Step-by-Step Solution:

Step 1: Analyze Option A: In an exothermic reaction, the enthalpy of products is less than that of reactants. This is correct because exothermic reactions release heat, so the products have lower energy.

Step 2: Analyze Option B: $\Delta {H_{\mathrm{fusion}}} = \Delta {H_{\mathrm{sublimation}}} - \Delta {H_{\mathrm{vaporisation}}}$ . This is correct based on Hess's Law. Sublimation is the direct conversion from solid to gas, which can also be achieved by melting (fusion) followed by vaporization.

Step 3: Analyze Option C: A reaction for which $\mathrm{\;}\Delta H^\circ < 0$ and $\Delta S^\circ > 0$ is possible at all temperatures. This is correct because $\Delta G = \Delta H - T\Delta S$ . If $\Delta H$ is negative and $\Delta S$ is positive, then $\Delta G$ will always be negative, making the reaction spontaneous at all temperatures.

Step 4: Analyze Option D: ∆H is less than ∆U for the reaction, C(s)+ (1/2) ${O_2}$ (g)→ $C{O_2}$ (g). We know that $\Delta H = \Delta U + \Delta n_g RT$ , where $\Delta n_g$ is the change in the number of moles of gas. In this reaction, $\Delta n_g = 1 - \frac{1}{2} = \frac{1}{2} > 0$ . Therefore, $\Delta H = \Delta U + \frac{1}{2}RT$ , which means $\Delta H > \Delta U$ . So, the statement is incorrect.

Why other options are wrong:
- Option A: This is a correct statement about exothermic reactions.
- Option B: This is a correct application of Hess's Law relating enthalpy changes of fusion, sublimation, and vaporization.
- Option C: This is a correct statement about the spontaneity of reactions based on the signs of ∆H and ∆S.

Key Concepts:
- Enthalpy
- Internal energy
- Hess's Law
- Spontaneity of reactions

Memory Aids:
- $\Delta H = \Delta U + P\Delta V = \Delta U + \Delta n_g RT$
- $\Delta G = \Delta H - T\Delta S$

Quick Tips:
- Remember the relationship between $\Delta H$ and $\Delta U$ and how it depends on $\Delta n_g$ .
- Understand the conditions for spontaneity based on $\Delta H$ and $\Delta S$ .

85 Easy

85 of 196Question permalink

Remembering +4 | -1

The amount of heat measured for a reaction in a bomb calorimeter is

View answer and explanation

Correct option: C

∆E

Option C (∆E) is correct because a bomb calorimeter measures heat at constant volume, which corresponds to the change in internal energy (∆E).

Solution approach

Correct Answer + Core Reason: ∆E. A bomb calorimeter measures heat at constant volume, which is equivalent to the change in internal energy.

Step-by-Step Solution:

Step 1: Bomb Calorimeter: A bomb calorimeter is a device used to measure the heat of reaction at constant volume.

Step 2: Constant Volume Process: In a constant volume process, no work is done ( $\Delta V = 0$ , so $W = -P\Delta V = 0$ ).

Step 3: First Law of Thermodynamics: According to the first law of thermodynamics, $\Delta E = Q + W$ , where $\Delta E$ is the change in internal energy, $Q$ is the heat, and $W$ is the work done.

Step 4: Applying to Bomb Calorimeter: Since $W = 0$ in a bomb calorimeter, $\Delta E = Q_v$ , where $Q_v$ is the heat at constant volume.

Step 5: Conclusion: Therefore, the amount of heat measured in a bomb calorimeter is $\Delta E$ .

Why other options are wrong:
- Option A: ∆G: $\Delta G$ is the Gibbs free energy change, which is relevant at constant pressure and temperature, not constant volume.
- Option B: ∆H: $\Delta H$ is the enthalpy change, which is measured at constant pressure, not constant volume.
- Option D: p.∆V: This represents the work done at constant pressure, which is not measured directly by a bomb calorimeter.

Key Concepts:
- Bomb Calorimetry
- Internal Energy (E)
- First Law of Thermodynamics
- Constant Volume Process

Memory Aids:
- Bomb = Volume = Internal Energy

Quick Tips:
- Remember that a bomb calorimeter operates at constant volume, so the heat measured is $\Delta E$ .

86 Easy

86 of 196Question permalink

Understanding +4 | -1

Entropy decreases during:

View answer and explanation

Correct option: A

Crystallization of sucrose from solution

Option A (Crystallization of sucrose from solution) is correct because crystallization involves the ordering of molecules into a solid lattice, which decreases the disorder and thus the entropy.

Solution approach

Correct Answer + Core Reason: Crystallization of sucrose from solution. Crystallization leads to a more ordered state, thus decreasing entropy.

Step-by-Step Solution:

Step 1: Entropy Definition: Entropy is a measure of the disorder or randomness of a system. Higher disorder means higher entropy.

Step 2: Crystallization: Crystallization is the process by which a solid forms from a solution, melt, or gas. During crystallization, molecules arrange themselves in a highly ordered, repeating pattern (a crystal lattice).

Step 3: Entropy Change in Crystallization: Because crystallization leads to a more ordered state, the entropy decreases during this process.

Step 4: Conclusion: Therefore, entropy decreases during the crystallization of sucrose from solution.

Why other options are wrong:
- Option B: Rusting of iron: Rusting involves the reaction of iron with oxygen and water, leading to an increase in disorder and thus an increase in entropy.
- Option C: Melting of ice: Melting is the transition from a solid to a liquid, which increases disorder and thus increases entropy.
- Option D: Vaporization of camphor: Vaporization is the transition from a solid or liquid to a gas, which significantly increases disorder and thus increases entropy.

Key Concepts:
- Entropy (S)
- Crystallization
- Phase Transitions

Memory Aids:
- Crystallization = Order = Decreased Entropy

Quick Tips:
- Remember that entropy generally increases with phase transitions from solid to liquid to gas.

87 Easy

87 of 196Question permalink

Analyzing +4 | -1

A container has hydrogen and oxygen mixture in ratio of 4∶1 by weight, then:

View answer and explanation

Correct option: A

Entropy of these gases increase

Option A (Entropy of these gases increase) is correct because mixing gases generally leads to an increase in entropy due to the increased disorder.

Solution approach

Correct Answer + Core Reason: Entropy of these gases increase. Mixing of gases always leads to an increase in entropy.

Step-by-Step Solution:

Step 1: Recall that entropy is a measure of disorder or randomness in a system.

Step 2: When two or more gases are mixed, the disorder increases as the different gas molecules intermingle. This leads to an increase in the number of possible microstates, and therefore an increase in entropy.

Step 3: In this case, hydrogen and oxygen are mixed. The mixing process increases the randomness of the system, resulting in an increase in entropy.

Why other options are wrong:
- Option B (Internal energy increases): Mixing ideal gases at constant temperature does not change the internal energy. Internal energy depends on temperature, not on mixing.
- Option C (Internal energy of the gas decreases): Mixing ideal gases at constant temperature does not change the internal energy.
- Option D (Entropy of the gases decrease): Mixing gases always increases entropy due to increased disorder.

Key Concepts:
- Entropy and disorder
- Mixing of gases
- Internal energy of ideal gases

Memory Aids:
- Mixing always increases entropy.

Quick Tips:
- Remember that mixing is generally an irreversible process that increases entropy.

88 Easy

88 of 196Question permalink

Understanding +4 | -1

∆n values in ∆H= ∆U+ ∆nRT may have:

View answer and explanation

Correct option: D

All of these

Option D (All of these) is correct because ∆n, representing the change in the number of moles of gas, can be an integer, a fraction, or positive/negative depending on the reaction stoichiometry.

Solution approach

Correct Answer + Core Reason: All of these. ∆n can be an integer, fractional, positive, or negative depending on the reaction.

Step-by-Step Solution:

Step 1: Define ∆n. ∆n represents the change in the number of moles of gaseous products minus the number of moles of gaseous reactants in a chemical reaction.

Step 2: Analyze the possible values of ∆n.
- Integer nature: If the stoichiometric coefficients are integers, ∆n will be an integer.
- Fractional value: If the stoichiometric coefficients are fractions, ∆n can be a fraction.
- Positive or negative: If the number of moles of gaseous products is greater than the number of moles of gaseous reactants, ∆n is positive. If the opposite is true, ∆n is negative. If they are equal, ∆n is zero.

Step 3: Conclusion: Therefore, ∆n can have integer nature, fractional value, and be positive or negative.

Why other options are wrong:
- Option A: Integer nature: While ∆n can be an integer, it can also be fractional or negative.
- Option B: Fractional value: While ∆n can be a fraction, it can also be an integer or negative.
- Option C: Positive or negative: While ∆n can be positive or negative, it can also be an integer or a fraction.

Key Concepts:
- Change in number of moles (∆n)
- Stoichiometry

Memory Aids:
- ∆n = (moles of gaseous products) - (moles of gaseous reactants).

Quick Tips:
- Calculate ∆n carefully based on the balanced chemical equation.

89 Easy

89 of 196Question permalink

Remembering +4 | -1

Gibbs energy G, enthalpy H and entropy S are related by:

View answer and explanation

Correct option: B

G=H-TS

Option B (G=H-TS) is correct because the Gibbs free energy (G) is defined as G = H - TS, where H is enthalpy, T is temperature, and S is entropy.

Solution approach

Correct Answer + Core Reason: G=H-TS. The Gibbs free energy (G) is defined as G = H - TS.

Step-by-Step Solution:

Step 1: Recall the definition of Gibbs Free Energy: The Gibbs Free Energy (G) is a thermodynamic potential that measures the amount of energy available in a thermodynamic system to do useful work at a constant temperature and pressure.

Step 2: State the formula for Gibbs Free Energy: The formula is given by $G = H - TS$ , where:
- $G$ is the Gibbs Free Energy
- $H$ is the Enthalpy
- $T$ is the absolute temperature
- $S$ is the Entropy

Step 3: Conclusion: The correct relationship between Gibbs energy G, enthalpy H, and entropy S is $G = H - TS$ .

Why other options are wrong:
- Option A: G=H+TS - This is incorrect; the correct formula has a negative sign before the TS term.
- Option C: G-TS=H - This can be rearranged to G = H + TS, which is incorrect.
- Option D: S=H-G - This is incorrect; the correct formula is G = H - TS, which can be rearranged to S = (H - G) / T.

Key Concepts:
- Gibbs Free Energy
- Enthalpy
- Entropy

Memory Aids:
- G = H - TS (Remember the formula for Gibbs Free Energy)

Quick Tips:
- Memorize the formula for Gibbs Free Energy as it is fundamental in thermodynamics.

90 Easy

90 of 196Question permalink

Remembering +4 | -1

The total heat content of a system at constant pressure is

View answer and explanation

Correct option: A

Enthalpy

Option A (Enthalpy) is correct because enthalpy is defined as the total heat content of a system at constant pressure.

Solution approach

Correct Answer + Core Reason: Enthalpy. Enthalpy is the total heat content of a system at constant pressure.

Step-by-Step Solution:

Step 1: Define enthalpy: Enthalpy (H) is a thermodynamic property of a system defined as $H = U + PV$ , where U is internal energy, P is pressure, and V is volume.

Step 2: Relate enthalpy to heat at constant pressure: At constant pressure, the change in enthalpy ( $\Delta H$ ) is equal to the heat absorbed or released by the system ( $q_p$ ): $\Delta H = q_p$ .

Step 3: Interpret the question: The question asks for the total heat content at constant pressure, which is the definition of enthalpy.

Why other options are wrong:
- Option B: Internal energy. Internal energy (U) is the total energy of a system, but it doesn't directly represent the heat content at constant pressure.
- Option C: Entropy. Entropy (S) is a measure of disorder or randomness in a system.
- Option D: Free energy. Free energy (Gibbs free energy, G) is a measure of the amount of energy available to do work.

Key Concepts:
- Enthalpy
- Internal energy
- Entropy
- Free energy
- Constant pressure conditions

Memory Aids:
- Enthalpy: $H = U + PV$
- $\Delta H = q_p$ (at constant pressure)

Quick Tips:
- Remember the definitions of enthalpy, internal energy, entropy, and free energy.

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