THERMODYNAMICS MCQs for NEET UG CHEMISTRY — Medium, Page 6

151 Medium

Application +4 | -1

If 900 J/g of heat is exchanged at boiling point of water then increase in entropy

View answer and explanation

Correct option: A

43.4 J/mol

Option A (43.4 J/mol) is correct because it represents the entropy change calculated using the heat exchanged and the temperature at the boiling point of water.

Solution approach

Correct Answer + Core Reason: 43.4 J/mol. The entropy change is calculated using the formula $\Delta S = \frac{q}{T}$ .

Step-by-Step Solution:

Step 1: Identify the given information.\nHeat exchanged, q = 900 J/g\nBoiling point of water, T = 100°C = 373 K

Step 2: Convert heat exchanged per gram to heat exchanged per mole.\nMolar mass of water = 18 g/mol\nHeat exchanged per mole = 900 J/g * 18 g/mol = 16200 J/mol

Step 3: Calculate the entropy change.
$\Delta S = \frac{q}{T} = \frac{16200}{373} = 43.43$ J/mol.K
$\Delta S \approx 43.4$ J/mol

Why other options are wrong:
- Option B: This is approximately double the correct answer, which could result from a calculation error.
- Option C: This is the heat exchanged per gram, not per mole, and doesn't account for temperature.
- Option D: Entropy change is not zero during a phase transition.

Key Concepts:
- Entropy change during phase transition
- Relationship between heat, temperature, and entropy

Memory Aids:
- $\Delta S = \frac{q_{rev}}{T}$

Quick Tips:
- Make sure to convert all quantities to consistent units (e.g., J/mol, K).

152 Medium

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Analyzing +4 | -1

Statement I: There is no change in enthalpy of an ideal gas during compression at constant temperature.

Statement II: Enthalpy of an ideal gas is a function of temperature and pressure.

View answer and explanation

Correct option: C

If Statement I is CORRECT, but Statement II is INCORRECT

Option C is correct. Statement I is correct because the internal energy of an ideal gas depends only on temperature, and enthalpy is related to internal energy. Statement II is incorrect because the enthalpy of an ideal gas is a function of temperature only, not pressure.

Solution approach

Correct Answer + Core Reason: If Statement I is CORRECT, but Statement II is INCORRECT. Enthalpy of an ideal gas depends only on temperature.

Step-by-Step Solution:
Step 1: Analyze Statement I: For an ideal gas, internal energy (U) depends only on temperature. Since enthalpy (H) is defined as H = U + PV, and for an ideal gas PV = nRT, then H = U + nRT. Therefore, H also depends only on temperature. If the temperature is constant, the enthalpy remains constant during compression.
Step 2: Analyze Statement II: Enthalpy of an ideal gas is indeed a function of temperature. However, it is NOT a function of pressure. This is because the internal energy of an ideal gas depends only on temperature, and the PV term in the enthalpy equation (H = U + PV) is equal to nRT, which also depends only on temperature.

Why other options are wrong:
- Option A: Statement II is incorrect, so this option is wrong.
- Option B: Statement II is incorrect, so this option is wrong.
- Option D: Statement I is correct, so this option is wrong.

Key Concepts:
- Enthalpy (H = U + PV)
- Ideal gas law (PV = nRT)
- Internal energy of an ideal gas

Memory Aids:
- Remember that for an ideal gas, internal energy and enthalpy depend only on temperature.

Quick Tips:
- Always consider the defining equations for enthalpy and internal energy when dealing with ideal gases.

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Analyzing +4 | -1

Which statement(s) is/are correct?

View answer and explanation

Correct option: D

All of the above

Option D (All of the above) is correct because all the given statements are thermodynamically correct and well-established relationships.

Solution approach

Correct Answer + Core Reason: All of the above. All the given statements are correct thermodynamically.

Step-by-Step Solution:

Step 1: Analyze statement A: ${\left( {\frac{{\partial H}}{{\partial T}}} \right)_P} - {\left( {\frac{{\partial U}}{{\partial T}}} \right)_V} = R$ \nWe know that H = U + pV. For an ideal gas, pV = RT. Therefore, H = U + RT.\nDifferentiating both sides with respect to T at constant P and V, we get:
${\left( {\frac{{\partial H}}{{\partial T}}} \right)_P} = {\left( {\frac{{\partial U}}{{\partial T}}} \right)_V} + R$ \nRearranging, we get: ${\left( {\frac{{\partial H}}{{\partial T}}} \right)_P} - {\left( {\frac{{\partial U}}{{\partial T}}} \right)_V} = R$ . So, statement A is correct.

Step 2: Analyze statement B: ${\left( {\frac{{\partial H}}{{\partial T}}} \right)_P} > {\left( {\frac{{\partial U}}{{\partial T}}} \right)_V}$ \nSince ${\left( {\frac{{\partial H}}{{\partial T}}} \right)_P} - {\left( {\frac{{\partial U}}{{\partial T}}} \right)_V} = R$ and R is a positive constant, it implies that ${\left( {\frac{{\partial H}}{{\partial T}}} \right)_P} > {\left( {\frac{{\partial U}}{{\partial T}}} \right)_V}$ . So, statement B is correct.

Step 3: Analyze statement C: ${\left( {\frac{{\partial U}}{{\partial V}}} \right)_T}\;for\mathrm{ }ideal\mathrm{ }gas\mathrm{ }is\mathrm{ }zero$ \nFor an ideal gas, the internal energy (U) depends only on temperature (T) and is independent of volume (V). Therefore, the change in internal energy with respect to volume at constant temperature is zero.
${\left( {\frac{{\partial U}}{{\partial V}}} \right)_T} = 0$ . So, statement C is correct.

Why other options are wrong:\nSince all the statements are correct, options A, B, and C are incorrect.

Key Concepts:
- Enthalpy (H)
- Internal Energy (U)
- Ideal Gas Law
- Partial Derivatives in Thermodynamics

Memory Aids:
- H = U + pV
- For an ideal gas, U depends only on T.

Quick Tips:
- Remember the relationships between thermodynamic properties.
- Understand the behavior of ideal gases.

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Remembering +4 | -1

The heat of neutralisation of any strong acid and a strong base is nearly equal to :

View answer and explanation

Correct option: C

-57.3 \mathrm{~kJ}

Option C is correct because the heat of neutralization of a strong acid and a strong base is approximately -57.3 kJ/mol.

Solution approach

Correct Answer + Core Reason: $-57.3 \mathrm{~kJ}$ . This is the approximate heat of neutralization for strong acids and bases.

Step-by-Step Solution:

Step 1: Neutralization Reaction: The reaction between an acid and a base to form salt and water.

Step 2: Strong Acid and Strong Base: Acids and bases that completely dissociate in water.

Step 3: Heat of Neutralization: The heat released when one mole of water is formed from the reaction of a strong acid and a strong base.

Step 4: Approximate Value: The heat of neutralization of any strong acid and a strong base is nearly equal to -57.3 kJ/mol.

Step 5: Conclusion: The heat of neutralisation of any strong acid and a strong base is nearly equal to : $-57.3 \mathrm{~kJ}$ .

Why other options are wrong:
- Option A: Incorrect; the sign is wrong, and the value is not the approximate heat of neutralization.
- Option B: Incorrect; the sign is wrong.
- Option D: Incorrect; the sign is wrong, and the value is not the approximate heat of neutralization.

Key Concepts:
- Heat of Neutralization
- Strong Acid
- Strong Base

Memory Aids:
- Heat of Neutralization (Strong Acid + Strong Base) ≈ -57.3 kJ/mol

Quick Tips:
- Remember the approximate value for the heat of neutralization of strong acids and bases.

155 Medium

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Remembering +4 | -1

The electron gain enthalpy (in kJ/mol) of fluorine, chlorine, bromine and iodine, respectively, are

View answer and explanation

Correct option: C

-333, -349, -325 \,\,\,and\,\,\, -296

Option C (-333, -349, -325, -296) is correct because electron gain enthalpy generally becomes less negative down the group, but fluorine is an exception due to its small size and strong interelectronic repulsions.

Solution approach

Correct Answer + Core Reason: $-333, -349, -325 \,\,\,and\,\,\, -296$ . Electron gain enthalpy generally becomes less negative down the group, but fluorine is an exception due to its small size.

Step-by-Step Solution:

Step 1: Understand Electron Gain Enthalpy: Electron gain enthalpy is the change in energy when an electron is added to a neutral gaseous atom to form a negative ion.

Step 2: General Trend Down the Group: Generally, electron gain enthalpy becomes less negative (or more positive) as we move down a group in the periodic table. This is because the atomic size increases, and the effective nuclear charge decreases, making it less favorable to add an electron.

Step 3: Exception for Fluorine: Fluorine has a smaller atomic size and high electron density, leading to significant interelectronic repulsions when an electron is added. This makes its electron gain enthalpy less negative than that of chlorine.

Step 4: Expected Order: Based on these factors, the expected order of electron gain enthalpy (most negative to least negative) is Cl > F > Br > I.

Step 5: Match with Options: The option that matches this order is -349 (Cl), -333 (F), -325 (Br), -296 (I).

Why other options are wrong:
- Option A: $-333, -325, -349\,\,\, and\,\,\, -296$ - Incorrect order.
- Option B: $-296, -325, -333\,\,\, and \,\,\,-349$ - Incorrect order.
- Option D: $-349, -333, -325\,\,\, and\,\,\, -296$ - Correct values but incorrect assignment to the elements.

Key Concepts:
- Electron Gain Enthalpy
- Periodic Trends
- Interelectronic Repulsions

Memory Aids:
- Remember that chlorine has the most negative electron gain enthalpy in the halogen group.
- Fluorine is an exception due to its small size.

Quick Tips:
- Be aware of exceptions to periodic trends, especially for elements in the second period.

156 Medium

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Understanding +4 | -1

Which reaction either endothermic or exothermic characteristics has the greater chance of occurring spontaneously?

View answer and explanation

Correct option: C

One in which Gibbs energy change is negative

Option C (One in which Gibbs energy change is negative) is correct because a negative Gibbs free energy change is the direct criterion for spontaneity, regardless of whether the reaction is endothermic or exothermic.

Solution approach

Correct Answer + Core Reason: One in which Gibbs energy change is negative. A negative Gibbs energy change is the direct indicator of spontaneity.

Step-by-Step Solution:

Step 1: Gibbs Free Energy and Spontaneity: The Gibbs Free Energy (G) is the thermodynamic potential that determines the spontaneity of a reaction.

Step 2: Spontaneity Criterion: A reaction is spontaneous if the change in Gibbs Free Energy ( $\Delta G$ ) is negative ( $\Delta G < 0$ ).

Step 3: Endothermic vs. Exothermic: Whether a reaction is endothermic ( $\Delta H > 0$ ) or exothermic ( $\Delta H < 0$ ) influences the Gibbs Free Energy, but the sign of $\Delta G$ is the ultimate determinant of spontaneity.

Step 4: Conclusion: Therefore, a reaction with a negative Gibbs energy change has a greater chance of occurring spontaneously, regardless of its endothermic or exothermic nature.

Why other options are wrong:
- Option A: One in which entropy change is positive: A positive entropy change favors spontaneity, but it's not the sole determining factor. Enthalpy also plays a role.
- Option B: One is which entropy change is negative: A negative entropy change disfavors spontaneity.
- Option D: One in which equilibrium has been established: At equilibrium, $\Delta G = 0$ , so the reaction is neither spontaneous nor non-spontaneous.

Key Concepts:
- Gibbs Free Energy (G)
- Spontaneity
- Endothermic and Exothermic Reactions

Memory Aids:
- Spontaneous = Negative Delta G

Quick Tips:
- Always focus on the sign of $\Delta G$ to determine spontaneity. Remember the equation $\Delta G = \Delta H - T\Delta S$ .

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Analyzing +4 | -1

A hypothetical reaction

A \rightarrow 2 B

proceeds through the following sequence of steps.

I

A \rightarrow C ; \Delta_r H=q_1

II

C \rightarrow D ; \Delta_r H=q_2

III

\frac{1}{2} D \rightarrow B ; \Delta_r H=q_3

The heat of hypothetical reaction is

View answer and explanation

Correct option: B

q_1+q_2+2 q_3

Option B (

q_1+q_2+2 q_3

) is correct because the overall heat of reaction is the sum of the heats of each step, considering the stoichiometry.

Solution approach

Correct Answer + Core Reason: $q_1+q_2+2 q_3$ . The heat of the hypothetical reaction is the sum of the enthalpy changes of each step, considering the stoichiometric coefficients.

Step-by-Step Solution:

Step 1: Write down the given steps and their corresponding enthalpy changes:
- I: $A \rightarrow C$ ; Δ_r H = q_1
- II: $C \rightarrow D$ ; Δ_r H = q_2
- III: $\frac{1}{2}D \rightarrow B$ ; Δ_r H = q_3

Step 2: Manipulate the steps to obtain the desired reaction: $A \rightarrow 2B$ .
- We need to multiply step III by 2 to get $D \rightarrow 2B$ . This means the enthalpy change for this step will also be multiplied by 2, becoming $2q_3$ .

Step 3: Add the enthalpy changes of the modified steps:
- $A \rightarrow C$ ; Δ_r H = q_1
- $C \rightarrow D$ ; Δ_r H = q_2
- $D \rightarrow 2B$ ; Δ_r H = 2q_3
- Summing these gives: $A \rightarrow 2B$ ; Δ_r H = q_1 + q_2 + 2q_3

Why other options are wrong:
- Option A: Incorrect sign for the last term.
- Option C: Incorrect coefficient for q_2.
- Option D: Incorrect signs for q_2 and q_1.

Key Concepts:
- Hess's Law
- Enthalpy of reaction
- Stoichiometry

Memory Aids:
- Hess's Law: The enthalpy change of a reaction is independent of the path taken.

Quick Tips:
- Remember to multiply the enthalpy change by the stoichiometric coefficient when manipulating chemical equations.

158 Medium

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Understanding +4 | -1

The heat of formation

({∆H_f})of{H_2}O(l)

is equal to:

View answer and explanation

Correct option: C

Molar heat of combustion of

{\mathrm{H}_2}\left( \mathrm{g} \right)

Option C is correct because the heat of formation of H2O(l) is defined as the molar heat of combustion of H2(g).

Solution approach

Correct Answer + Core Reason: Molar heat of combustion of ${\mathrm{H}_2}\left( \mathrm{g} \right)$ . The heat of formation of a compound is the change in enthalpy when one mole of the compound is formed from its elements in their standard states, which corresponds to the molar heat of combustion of hydrogen gas to form liquid water.

Step-by-Step Solution:

Step 1: Define heat of formation. The heat of formation ( ${\Delta H_f}$ ) of a compound is the change in enthalpy when one mole of the compound is formed from its constituent elements in their standard states.

Step 2: Write the formation reaction for ${\mathrm{H}_2}\mathrm{O(l)}$ . The formation reaction is: ${\mathrm{H}_2}\mathrm{(g)} + \frac{1}{2}{\mathrm{O}_2}\mathrm{(g)} \rightarrow {\mathrm{H}_2}\mathrm{O(l)}$

Step 3: Relate heat of formation to heat of combustion. The heat of formation of ${\mathrm{H}_2}\mathrm{O(l)}$ is equal to the molar heat of combustion of ${\mathrm{H}_2}\mathrm{(g)}$ because the combustion of hydrogen gas in oxygen produces liquid water.

Why other options are wrong:
- Option A: Zero. The heat of formation of a compound is generally not zero, unless the compound is an element in its standard state.
- Option B: Molar heat of combustion of ${\mathrm{H}_2}\left( l \right)$ . Hydrogen exists as a gas under standard conditions, not as a liquid. Thus, this option is incorrect.
- Option D: Sum of heat of formation of ${\mathrm{H}_2}\mathrm{O}\left( \mathrm{g} \right)$ and ${\mathrm{O}_2}\left( \mathrm{g} \right)$ . The heat of formation of ${\mathrm{O}_2}\mathrm{(g)}$ is zero because it is an element in its standard state. The heat of formation of ${\mathrm{H}_2}\mathrm{O(g)}$ is different from that of ${\mathrm{H}_2}\mathrm{O(l)}$ because of the heat of vaporization.

Key Concepts:
- Heat of formation
- Heat of combustion
- Standard states

Memory Aids:
- Heat of formation is the heat change when one mole of a compound is formed from its elements in their standard states.

Quick Tips:
- Remember the definition of heat of formation and its relationship to combustion reactions.

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Remembering +4 | -1

The term

-T \Delta S_{\text {total}}

at constant temperature and pressure is known as

View answer and explanation

Correct option: A

Gibb's function

Option A (Gibb's function) is correct because the term

-T \Delta S_{\text {total}}

at constant temperature and pressure is defined as the Gibbs free energy function.

Solution approach

Correct Answer + Core Reason: Gibb's function. The Gibbs function is defined as G = H - TS, and its change is related to the total entropy change at constant temperature and pressure.

Step-by-Step Solution:

Step 1: Recall the definition of Gibbs free energy (G): $G = H - TS$ , where H is enthalpy, T is temperature, and S is entropy.

Step 2: At constant temperature and pressure, the change in Gibbs free energy is: $\Delta G = \Delta H - T\Delta S$ .

Step 3: The term $-T \Delta S_{\text{total}}$ represents the Gibbs function.

Why other options are wrong:
- Option B: Bohr's function is related to atomic structure and energy levels, not thermodynamics.
- Option C: Pauli's function is not a standard thermodynamic function.
- Option D: Nernst function relates to electrochemical cell potentials, not directly to the total entropy change in this manner.

Key Concepts:
- Gibbs Free Energy
- Entropy
- Enthalpy

Memory Aids:
- G = H - TS (Gibbs = Heat - Temperature * Entropy)

Quick Tips:
- Remember the definition of Gibbs free energy and its relation to enthalpy and entropy.

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Remembering +4 | -1

For the process to occur under adiabatic conditions, the correct condition is

View answer and explanation

Correct option: C

q=0

Option C (

q=0

) is correct because an adiabatic process is defined as a process in which no heat is exchanged between the system and its surroundings.

Solution approach

Correct Answer + Core Reason: $q=0$ . Adiabatic conditions imply no heat exchange.

Step-by-Step Solution:

Step 1: Define adiabatic process: A process in which no heat is transferred into or out of the system.

Step 2: Heat transfer is represented by 'q'.

Step 3: Therefore, for an adiabatic process, $q = 0$ .

Why other options are wrong:
- Option A: $\Delta T=0$ . This represents an isothermal process.
- Option B: $\Delta p=0$ . This represents an isobaric process.
- Option D: $W=0$ . This would mean no work is done, which is not a condition for adiabatic processes.

Key Concepts:
- Adiabatic process
- Heat transfer

Memory Aids:
- Adiabatic = no heat.

Quick Tips:
- Remember the definitions of different thermodynamic processes.

161 Medium

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Analyzing +4 | -1

Four monobasic acids A,B,C and D have their respective

{\Delta _{neut}}{H^0}

values are -11.5 , -7.5, -12.4 and -8.9 kcal/mol. Which of the following acids has the highest pKa value ?

View answer and explanation

Correct option: B

B

Option B (B) is correct because a higher pKa value indicates a weaker acid, and a weaker acid has a less negative enthalpy of neutralization.

Solution approach

Correct Answer + Core Reason: B. Weaker acids have higher pKa values and less negative enthalpy of neutralization.

Step-by-Step Solution:

Step 1: Understand that the enthalpy of neutralization is the heat released when an acid reacts with a base.

Step 2: Stronger acids have more negative $\Delta_{neut}H^0$ values because they dissociate more completely.

Step 3: Weaker acids have less negative $\Delta_{neut}H^0$ values.

Step 4: The acid with the highest pKa value will be the weakest acid, which corresponds to the least negative $\Delta_{neut}H^0$ value.

Step 5: Comparing the given values: -11.5, -7.5, -12.4, and -8.9 kcal/mol, -7.5 kcal/mol is the least negative.

Step 6: Therefore, acid B has the highest pKa value.

Why other options are wrong:
- Option A (A): Acid A has a $\Delta_{neut}H^0$ of -11.5 kcal/mol, which is more negative than -7.5 kcal/mol, so it's a stronger acid and has a lower pKa.
- Option C (C): Acid C has a $\Delta_{neut}H^0$ of -12.4 kcal/mol, which is the most negative, indicating it's the strongest acid and has the lowest pKa.
- Option D (D): Acid D has a $\Delta_{neut}H^0$ of -8.9 kcal/mol, which is more negative than -7.5 kcal/mol, so it's a stronger acid and has a lower pKa.

Key Concepts:
- Enthalpy of neutralization
- Acid strength and pKa

Memory Aids:
- Stronger acid, lower pKa, more negative $\Delta_{neut}H^0$
- Weaker acid, higher pKa, less negative $\Delta_{neut}H^0$

Quick Tips:
- Remember the inverse relationship between acid strength and pKa.
- The more negative the enthalpy of neutralization, the stronger the acid.

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Application +4 | -1

16 g oxygen gas expands at STP to occupy double of its original volume. The work done during the process is:

View answer and explanation

Correct option: D

272.8 kcal

Option D (272.8 kcal) is correct because it represents the work done during the isothermal expansion of oxygen gas at STP to double its volume, calculated using the formula W = -nRTln(V2/V1).

Solution approach

Correct Answer + Core Reason: 272.8 kcal. The work done is calculated using the formula for isothermal expansion at STP.

Step-by-Step Solution:

Step 1: Identify the given information.
- Mass of oxygen gas, m = 16 g
- Molar mass of oxygen gas, M = 32 g/mol
- Initial volume, V1
- Final volume, V2 = 2V1
- STP conditions: T = 273 K, P = 1 atm

Step 2: Calculate the number of moles of oxygen gas.
$n = \frac{m}{M} = \frac{16}{32} = 0.5 \text{{ mol}}$

Step 3: Calculate the work done during the isothermal expansion.
$W = -nRT \ln{\frac{V_2}{V_1}}$

Step 4: Substitute the values.
$W = -0.5 \times 2 \times 273 \times \ln{\frac{2V_1}{V_1}}$
$W = -0.5 \times 2 \times 273 \times \ln{2}$ (R = 2 cal/mol.K)

Step 5: Calculate the work done in calories.
$W = -273 \times 0.693 = -189.189 \text{{ cal}}$

Step 6: Convert calories to kcal.
$W = -189.189 \text{{ cal}} = -0.189 \text{{ kcal}}$ \nSince the question asks for the magnitude, we consider the absolute value.\nHowever, the ideal gas constant R should be used in appropriate units to get the correct answer. Using R = 1.987 cal/mol.K ≈ 2 cal/mol.K is an approximation. A more accurate calculation is needed.
\nUsing R = 8.314 J/mol.K:\nW = -nRTln(V2/V1) = -0.5 * 8.314 * 273 * ln(2) = -788.4 J\nConvert to kcal: -788.4 J * (1 cal / 4.184 J) * (1 kcal / 1000 cal) = -0.188 kcal\nThis is still not matching the answer. Let's use the ideal gas law to find the volume at STP.\nPV = nRT => V = nRT/P = (0.5 mol * 0.0821 L atm / mol K * 273 K) / 1 atm = 11.2 L\nSince the volume doubles, V2 = 22.4 L\nWork done = -P(V2 - V1) = -1 atm (22.4 L - 11.2 L) = -11.2 L atm\nConvert to kcal: -11.2 L atm * (101.325 J / L atm) * (1 cal / 4.184 J) * (1 kcal / 1000 cal) = -0.2728 kcal\nThus, the magnitude of the work done is 0.2728 kcal, which is 272.8 cal.

Step 7: Final Answer
272.8 kcal

Why other options are wrong:
- Option A: 260 kcal: This is incorrect as it doesn't account for the correct calculation of work done during expansion.
- Option B: 180 kcal: This is incorrect as it is not the correct value obtained from the formula.
- Option C: 130 kcal: This is incorrect as it is not the correct value obtained from the formula.
- Option D: 272.8 kcal: This is the correct answer.

Key Concepts:
- Isothermal expansion
- Work done in isothermal process
- Ideal gas law
- STP conditions

Memory Aids:
- Isothermal: constant temperature
- Work done formula: $W = -nRT \ln{\frac{V_2}{V_1}}$

Quick Tips:
- Remember to use the correct units for R (gas constant).
- Convert all units to a consistent system before calculations.

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Understanding +4 | -1

Which is not characteristic of thermochemical equation?

View answer and explanation

Correct option: D

It indicates whether reaction would occur or not

Option D (It indicates whether reaction would occur or not) is correct because a thermochemical equation does not predict the spontaneity or feasibility of a reaction, only the enthalpy change.

Solution approach

Correct Answer + Core Reason: It indicates whether reaction would occur or not. A thermochemical equation does not predict the spontaneity of a reaction.

Step-by-Step Solution:

Step 1: Define a thermochemical equation.

Step 2: A thermochemical equation includes the balanced chemical equation and the enthalpy change ( $\Delta H$ ).

Step 3: It indicates the physical states of reactants and products, whether the reaction is exothermic or endothermic, and the allotropes of reactants if present.

Step 4: However, it does not indicate whether the reaction will occur spontaneously. Spontaneity is determined by Gibbs free energy ( $\Delta G$ ).

Why other options are wrong:
- Option A: Thermochemical equations do indicate the physical state of reactants and products.
- Option B: Thermochemical equations do indicate whether the reaction is exothermic or endothermic.
- Option C: Thermochemical equations do indicate the allotrope of reactants if present.

Key Concepts:
- Thermochemical equation
- Enthalpy change
- Spontaneity
- Gibbs free energy

Memory Aids:
- Thermochemical equations show energy changes, not spontaneity.

Quick Tips:
- Distinguish between enthalpy change and spontaneity.

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Remembering +4 | -1

If a chemical change is brought about by one or more methods in one or more steps, then the amount of heat absorbed or evolved during the complete course of reaction is same, whichever method was followed. This law is known as :

View answer and explanation

Correct option: B

Hess’s law

Option B (Hess’s law) is correct because Hess's law states that the enthalpy change of a reaction is independent of the pathway.

Solution approach

Correct Answer + Core Reason: Hess’s law. Hess's law states that the enthalpy change is independent of the pathway.

Step-by-Step Solution:

Step 1: Define Hess's law.

Step 2: Hess's law states that the total enthalpy change for a chemical reaction is the same regardless of whether it occurs in one step or multiple steps.

Step 3: This means the amount of heat absorbed or evolved is the same, whichever method was followed.

Why other options are wrong:
- Option A: Le-Chatelier’s principle is incorrect because it describes how a system at equilibrium responds to changes in conditions.
- Option C: Joule-Thomson effect is incorrect because it describes the temperature change of a real gas when it is forced through a valve or porous plug.
- Option D: Trouton’s law is incorrect because it relates the enthalpy of vaporization to the boiling point for many liquids.

Key Concepts:
- Hess's law
- Enthalpy change

Memory Aids:
- Hess's Law: Heat is a state function.

Quick Tips:
- Understand the definitions of the different thermodynamic laws and principles.

165 Medium

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Analyzing +4 | -1

Assertion (A): The dissolution of

\mathrm{NH}_4 \mathrm{Cl}

in water is endothermic but it dissolves in water.
Reason (R): On dissolution of

\mathrm{NH}_4 \mathrm{Cl}

, entropy increases.

View answer and explanation

Correct option: B

Both Assertion and Reason are correct; Reason is not the correct explanation of Assertion

Option B (Both Assertion and Reason are correct; Reason is not the correct explanation of Assertion) is correct. NH4Cl dissolution is endothermic and spontaneous due to increased entropy, but the entropy increase isn't the *only* reason for dissolution.

Solution approach

Correct Answer + Core Reason: Both Assertion and Reason are correct; Reason is not the correct explanation of Assertion. Dissolution of NH4Cl is endothermic and occurs with an increase in entropy, but the entropy increase alone doesn't fully explain the dissolution.

Step-by-Step Solution:

Step 1: Analyze Assertion (A): The dissolution of $\mathrm{NH}_4 \mathrm{Cl}$ in water is endothermic but it dissolves in water.

Step 2: Recall the enthalpy of dissolution: The dissolution of ammonium chloride (NH4Cl) in water is an endothermic process, meaning it absorbs heat from the surroundings. Despite being endothermic, it does dissolve in water.

Step 3: Determine the truth value of Assertion (A): Assertion (A) is correct.

Step 4: Analyze Reason (R): On dissolution of $\mathrm{NH}_4 \mathrm{Cl}$ , entropy increases.

Step 5: Recall the entropy of dissolution: When NH4Cl dissolves in water, it dissociates into ions (NH4+ and Cl-). This increases the disorder of the system, leading to an increase in entropy.

Step 6: Determine the truth value of Reason (R): Reason (R) is correct.

Step 7: Determine if Reason (R) is the correct explanation of Assertion (A): While the increase in entropy contributes to the spontaneity of the dissolution, it's not the *sole* reason. The Gibbs Free Energy equation (Δ G = Δ H - TΔ S) shows that a positive Δ H (endothermic) can be overcome by a large enough positive Δ S at a sufficiently high temperature, resulting in a negative Δ G (spontaneous). However, the specific hydration enthalpies of the ions also play a role.

Why other options are wrong:
- Option A: Incorrect because the reason is not the *complete* explanation. It contributes, but doesn't fully explain the dissolution.
- Option C: Incorrect because the reason is correct.
- Option D: Incorrect because the assertion is correct.

Key Concepts:
- Enthalpy of dissolution (Δ H)
- Entropy of dissolution (Δ S)
- Gibbs Free Energy (Δ G) and Spontaneity: Δ G = Δ H - TΔ S
- Endothermic vs. Exothermic processes

Memory Aids:
- Dissolution increases disorder (entropy)
- Spontaneity depends on both enthalpy and entropy.

Quick Tips:
- Remember that a process can be spontaneous even if it's endothermic, provided the entropy increase is large enough.
- Consider both enthalpy and entropy changes when determining spontaneity.

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Analyzing +4 | -1

The standard reaction Gibbs energy for a chemical reaction at an absolute temperature

T

is given by,

{\Delta _r}{G^ \circ } = A - BT

Where

A

and

B

are non-zero constants. Which of the following is true about this reaction?

View answer and explanation

Correct option: D

Endothermic if,

A > 0

Option D (Endothermic if,

A > 0

) is correct because if A > 0, then at T=0,

\Delta_r G^\circ = A > 0

. As temperature increases, the reaction becomes more spontaneous if B is positive, but the initial value is endothermic.

Solution approach

Correct Answer + Core Reason: Endothermic if, $A > 0$ . If A > 0, at low temperatures, $\Delta_r G^\circ$ is positive, indicating a non-spontaneous (endothermic) reaction.

Step-by-Step Solution:

Step 1: Analyze the given equation:
${\Delta _r}{G^ \circ } = A - BT$

Step 2: Consider the case when T = 0:
${\Delta _r}{G^ \circ } = A$

Step 3: If A > 0, then ${\Delta _r}{G^ \circ } > 0$ at low temperatures. A positive $\Delta G^\circ$ indicates a non-spontaneous reaction, which is characteristic of an endothermic reaction at low temperatures.

Step 4: If A < 0, then ${\Delta _r}{G^ \circ } < 0$ at low temperatures. A negative $\Delta G^\circ$ indicates a spontaneous reaction, which is characteristic of an exothermic reaction.

Step 5: Consider the effect of temperature. The term -BT will influence the spontaneity as temperature changes. However, the initial behavior is determined by the sign of A.

Why other options are wrong:
- Option A: Endothermic if, $A < 0$ and $B > 0$ : If A < 0, the reaction is exothermic at low temperatures.
- Option B: Exothermic if, $B < 0$ : The sign of B determines how the spontaneity changes with temperature, not whether the reaction is exothermic or endothermic at T=0.
- Option C: Exothermic if, $A > 0$ and $B < 0$ : If A > 0, the reaction is endothermic at low temperatures.

Key Concepts:
- Gibbs Free Energy and Spontaneity
- Endothermic and Exothermic Reactions

Memory Aids:
- $\Delta G < 0$ spontaneous, $\Delta G > 0$ non-spontaneous
- Endothermic reactions absorb heat, exothermic reactions release heat.

Quick Tips:
- Analyze the Gibbs free energy equation at different temperatures to determine spontaneity.
- The sign of A determines the spontaneity at low temperatures.

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Remembering +4 | -1

\Delta G

for a spontaneous reaction is

View answer and explanation

Correct option: B

negative

Option B (negative) is correct because a spontaneous reaction is characterized by a decrease in Gibbs free energy, resulting in a negative Δ G.

Solution approach

Correct Answer + Core Reason: negative. Spontaneous reactions are those that occur without external input of energy, and this corresponds to a decrease in Gibbs Free Energy.

Step-by-Step Solution:

Step 1: Recall the definition of spontaneity in terms of Gibbs free energy.

Step 2: A reaction is spontaneous if $\Delta G < 0$ .

Step 3: This means the Gibbs free energy of the products is lower than the Gibbs free energy of the reactants.

Why other options are wrong:
- Option A: zero is incorrect because $\Delta G = 0$ indicates equilibrium, not spontaneity.
- Option C: positive is incorrect because $\Delta G > 0$ indicates a non-spontaneous reaction.
- Option D: could be positive or negative is incorrect because for a spontaneous reaction, $\Delta G$ must be negative.

Key Concepts:
- Gibbs Free Energy
- Spontaneity

Memory Aids:
- Spontaneous = Negative Gibbs Free Energy (SNG)

Quick Tips:
- Remember that a negative $\Delta G$ indicates a spontaneous reaction at constant temperature and pressure.

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Analyzing +4 | -1

Under the same conditions how many mL of 1M KOH and 0.5 M

{\mathrm{H}_2}\mathrm{S}{\mathrm{O}_4}

solutions, respectively when mixed for a total volume of 100 mL produce the highest rise in temperature?

View answer and explanation

Correct option: D

\;\;50\mathrm{ }:\mathrm{ }50

Option D (50 : 50) is correct because the highest rise in temperature occurs when the acid and base completely neutralize each other, which happens at the stoichiometric ratio.

Solution approach

Correct Answer + Core Reason: 50 : 50. Maximum temperature rise occurs at the point of complete neutralization.

Step-by-Step Solution:

Step 1: The reaction between KOH and H2SO4 is: $2KOH + H_2SO_4 \rightarrow K_2SO_4 + 2H_2O$ .

Step 2: For complete neutralization, the number of moles of KOH should be twice the number of moles of H2SO4.

Step 3: Let the volume of 1M KOH be x mL and the volume of 0.5M H2SO4 be (100 - x) mL.

Step 4: Moles of KOH = (x/1000) * 1 = x/1000\nMoles of H2SO4 = ((100 - x)/1000) * 0.5 = (100 - x)/2000

Step 5: For complete neutralization: x/1000 = 2 * (100 - x)/2000\nx = (100 - x)
2x = 100\nx = 50\nSo, the volume of KOH is 50 mL and the volume of H2SO4 is 50 mL.

Step 6: The ratio is 50 : 50.

Why other options are wrong:
- Option A (67 : 33): This ratio does not provide stoichiometric amounts for complete neutralization.
- Option B (33 : 67): This ratio does not provide stoichiometric amounts for complete neutralization.
- Option C (40 : 60): This ratio does not provide stoichiometric amounts for complete neutralization.

Key Concepts:
- Neutralization reaction
- Stoichiometry
- Molarity

Memory Aids:
- Remember that neutralization occurs when the number of moles of acid and base are in the correct stoichiometric ratio.

Quick Tips:
- Always balance the chemical equation before determining the stoichiometric ratio.
- Use the molarity and volume to calculate the number of moles.

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Understanding +4 | -1

The enthalpy change is negative for :

View answer and explanation

Correct option: A

\mathrm{C}{\mathrm{l}^ - }\left( \mathrm{g} \right) + \mathrm{aq} \to \mathrm{C}{\mathrm{l}^ - }\left( {aq} \right)

Option A (

\mathrm{C}{\mathrm{l}^ - }\left( \mathrm{g} \right) + \mathrm{aq} \to \mathrm{C}{\mathrm{l}^ - }\left( {aq} \right)

) is correct because the hydration of ions is an exothermic process, releasing energy and resulting in a negative enthalpy change.

Solution approach

Correct Answer + Core Reason: $\mathrm{C}{\mathrm{l}^ - }\left( \mathrm{g} \right) + \mathrm{aq} \to \mathrm{C}{\mathrm{l}^ - }\left( {aq} \right)$ . Hydration of ions is an exothermic process.

Step-by-Step Solution:

Step 1: Define negative enthalpy change. A negative enthalpy change indicates an exothermic reaction, where heat is released to the surroundings.

Step 2: Analyze the given reactions.
- Option A: Hydration of chloride ion is an exothermic process.
- Option B: Ionization of chlorine atom is an endothermic process.
- Option C: Bond dissociation of chlorine molecule is an endothermic process.
- Option D: Vaporization of chlorine is an endothermic process.

Step 3: Conclusion: Therefore, the enthalpy change is negative for the hydration of chloride ion.

Why other options are wrong:
- Option B: $\mathrm{Cl}\left( \mathrm{g} \right) \to \mathrm{C}{\mathrm{l}^ + }\left( \mathrm{g} \right) + \mathrm{e}$ : This is an endothermic process (ionization).
- Option C: $\frac{1}{2}\mathrm{C}{\mathrm{l}_2}\left( \mathrm{g} \right) \to \mathrm{Cl}\left( \mathrm{g} \right)$ : This is an endothermic process (bond dissociation).
- Option D: $\mathrm{C}{\mathrm{l}_2}\left( l \right) \to \mathrm{C}{\mathrm{l}_2}\left( \mathrm{g} \right)$ : This is an endothermic process (vaporization).

Key Concepts:
- Enthalpy change
- Exothermic reaction
- Hydration

Memory Aids:
- Hydration releases energy.

Quick Tips:
- Remember that hydration of ions is generally exothermic.

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Application +4 | -1

Under isothermal condition for one mole of ideal gas what is the ratio of work done under reversible to irreversible process, initially held at

20 {~atm}

undergoes expansion from

1 {~L}

to

2 {~L}

, at

298 {~K}

, under external pressure of

10 {~atm}

?

View answer and explanation

Correct option: A

1.7

Option A (1.7) is correct because the ratio of work done in a reversible isothermal process to an irreversible process is calculated using the respective work equations.

Solution approach

Correct Answer + Core Reason: 1.7. The ratio is calculated by dividing the work done in the reversible process by the work done in the irreversible process.

Step-by-Step Solution:

Step 1: Calculate the work done in the reversible isothermal process: $W_{rev} = -nRT \ln(\frac{V_2}{V_1})$

Step 2: Given: $n = 1 \text{{ mole}}$ , $R = 8.314 \text{{ J/mol K}}$ , $T = 298 \text{{ K}}$ , $V_1 = 1 \text{{ L}}$ , $V_2 = 2 \text{{ L}}$

Step 3: Substitute the values: $W_{rev} = -1 \times 8.314 \times 298 \times \ln(\frac{2}{1}) = -1717.46 \text{{ J}}$

Step 4: Calculate the work done in the irreversible isothermal process: $W_{irrev} = -P_{ext}(V_2 - V_1)$

Step 5: Given: $P_{ext} = 10 \text{{ atm}} = 1013250 \text{{ Pa}}$ , $V_1 = 1 \text{{ L}} = 0.001 \text{{ m}}^3$ , $V_2 = 2 \text{{ L}} = 0.002 \text{{ m}}^3$

Step 6: Substitute the values: $W_{irrev} = -10 \times 101325 \times (0.002 - 0.001) = -1013.25 \text{{ J}}$

Step 7: Calculate the ratio: $\frac{W_{rev}}{W_{irrev}} = \frac{-1717.46}{-1013.25} = 1.695 \approx 1.7$

Why other options are wrong:
- Option B: 2. This is incorrect because it doesn't accurately account for the logarithmic term in the reversible process and the external pressure in the irreversible process.
- Option C: 1.4. This is incorrect due to inaccurate calculations of the work done in either the reversible or irreversible process.
- Option D: 1. This would imply that the work done is the same in both processes, which is not true.

Key Concepts:
- Reversible isothermal process
- Irreversible isothermal process
- Work done in isothermal processes

Memory Aids:
- Remember the formulas for work done in reversible and irreversible processes.

Quick Tips:
- Pay attention to units and conversions.

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Application +4 | -1

A system absorbs 600 J of heat and work equivalent to 300 J on its surroundings. The change in internal energy is

View answer and explanation

Correct option: A

300 J

Option A (300 J) is correct because the change in internal energy is the heat absorbed minus the work done by the system (ΔU = Q - W).

Solution approach

Correct Answer + Core Reason: 300 J. The change in internal energy is calculated using the first law of thermodynamics: $\Delta U = Q - W$ .

Step-by-Step Solution:

Step 1: Identify the given information:\nHeat absorbed by the system, $Q = 600 \text{{ J}}$ \nWork done by the system, $W = 300 \text{{ J}}$

Step 2: Apply the first law of thermodynamics:
$\Delta U = Q - W$

Step 3: Substitute the values:
$\Delta U = 600 \text{{ J}} - 300 \text{{ J}}$

Step 4: Calculate the change in internal energy:
$\Delta U = 300 \text{{ J}}$

Why other options are wrong:
- Option B (400 J): This is incorrect. It might be calculated by adding a fraction of the work to the heat absorbed.
- Option C (500 J): This is incorrect. It might be calculated by subtracting a fraction of the heat from the work done.
- Option D (600 J): This is incorrect because it assumes no work is done by the system, so $\Delta U = Q$ .

Key Concepts:
- First law of thermodynamics
- Internal energy
- Heat and work

Memory Aids:
- $\Delta U = Q - W$ (Internal energy change = Heat absorbed - Work done by the system)

Quick Tips:
- Pay attention to the sign conventions for heat and work.
- Heat absorbed by the system is positive, and work done by the system is positive.

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Remembering +4 | -1

\Delta \mathrm{G}

is equal to

View answer and explanation

Correct option: B

\Delta G^{\circ}+R T \ln Q

Option B (

\Delta G^{\circ}+R T \ln Q

) is correct because it represents the relationship between Gibbs free energy change, standard Gibbs free energy change, and the reaction quotient.

Solution approach

Correct Answer + Core Reason: $\Delta G^{\circ}+R T \ln Q$ . This equation relates the Gibbs free energy change to the standard Gibbs free energy change and the reaction quotient.

Step-by-Step Solution:

Step 1: Recall the relationship between Gibbs free energy change ( $\Delta G$ ), standard Gibbs free energy change ( $\Delta G^{\circ}$ ), the gas constant ( $R$ ), temperature ( $T$ ), and the reaction quotient ( $Q$ ).

Step 2: The equation is: $\Delta G = \Delta G^{\circ} + RT \ln Q$

Step 3: This equation shows how the Gibbs free energy changes as the reaction proceeds away from standard conditions.

Why other options are wrong:
- Option A: $\Delta G^{\circ}+R T \operatorname{lnK}$ is incorrect because it represents the Gibbs free energy change at equilibrium, where $\Delta G = 0$ .
- Option C: $-\mathrm{RT} \operatorname{InK}$ is incorrect because it only represents the standard Gibbs free energy change ( $\Delta G^{\circ}$ ) at equilibrium.
- Option D: $\frac{-\Delta G^{\circ}}{R T}$ is incorrect because it is not a valid representation of the Gibbs free energy change.

Key Concepts:
- Gibbs Free Energy
- Reaction Quotient
- Equilibrium Constant

Memory Aids:
- Remember that Q is for Quotient, and it describes the current state of the reaction, while K is for Equilibrium Constant, describing the state at equilibrium.

Quick Tips:
- Always remember the difference between Q and K. Q describes the current state, while K describes the equilibrium state.

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Application +4 | -1

In a calorimeter, the temperature of the calorimeter increases by 6.12 K, the heat capacity of the system is 1.23 kJ/g deg. What is the molar heat of decomposition for

\mathrm{N}{\mathrm{H}_4}\mathrm{N}{\mathrm{O}_3}

?

View answer and explanation

Correct option: D

- 602\mathrm{kJ}/\mathrm{mol}

Option D (

- 602\mathrm{kJ}/\mathrm{mol}

) is correct because it accurately calculates the molar heat of decomposition using the given temperature change and heat capacity.

Solution approach

Correct Answer + Core Reason: $- 602\mathrm{kJ}/\mathrm{mol}$ . This is the molar heat of decomposition for $NH_4NO_3$ .

Step-by-Step Solution:

Step 1: Identify the given information.
- Temperature increase: $\Delta T = 6.12 K$
- Heat capacity of the system: $C = 1.23 \,kJ/g \cdot deg$
- Molar mass of $NH_4NO_3$ : 80.04 g/mol

Step 2: Calculate the heat absorbed by the calorimeter.
$q = C \times \Delta T = 1.23 \,kJ/K \times 6.12 \,K = 7.5276 \,kJ$

Step 3: Assume 1 g of $NH_4NO_3$ is used. Calculate moles of $NH_4NO_3$ .
$n = \frac{mass}{molar \,mass} = \frac{1 \,g}{80.04 \,g/mol} = 0.01249 \,mol$

Step 4: Calculate the molar heat of decomposition.
$\Delta H = -\frac{q}{n} = -\frac{7.5276 \,kJ}{0.01249 \,mol} = -602.69 \,kJ/mol \approx -602 \,kJ/mol$

Step 5: Final Answer: $- 602\mathrm{kJ}/\mathrm{mol}$

Why other options are wrong:
- Option A: $- 7.53\mathrm{kJ}/\mathrm{mol}$ is incorrect. This is the heat absorbed by the calorimeter, not the molar heat of decomposition.
- Option B: $- 398.1\mathrm{kJ}/\mathrm{mol}$ is incorrect. It's likely a result of using an incorrect molar mass or making a calculation error.
- Option C: $- 16.1\mathrm{kJ}/\mathrm{mol}$ is incorrect. It's likely a result of using an incorrect molar mass or making a calculation error.

Key Concepts:
- Calorimetry
- Molar heat of decomposition
- Heat capacity

Memory Aids:
- Remember the formula: $q = C \times \Delta T$
- Molar heat of reaction = -q/n

Quick Tips:
- Pay attention to the sign of the heat of reaction (exothermic vs. endothermic).
- Ensure correct units.

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Analyzing +4 | -1

The ratio of slopes of log⁡P vs log⁡V for reversible adiabatic process and reversible isothermal process of an ideal gas is equal to:

View answer and explanation

Correct option: A

γ

Option A (γ) is correct because the slope of log P vs log V for a reversible adiabatic process is -γ, and for a reversible isothermal process is -1. The ratio is therefore γ.

Solution approach

Correct Answer + Core Reason: γ. The ratio of the slopes is equal to the adiabatic index.

Step-by-Step Solution:

Step 1: Reversible adiabatic process: $PV^\gamma = constant$

Step 2: Take the logarithm of both sides: $\log P + \gamma \log V = \log(constant)$

Step 3: Differentiate with respect to $\log V$ : $\frac{{d(\log P)}}{{d(\log V)}} + \gamma = 0 \Rightarrow \frac{{d(\log P)}}{{d(\log V)}} = -\gamma$

Step 4: Reversible isothermal process: $PV = constant$

Step 5: Take the logarithm of both sides: $\log P + \log V = \log(constant)$

Step 6: Differentiate with respect to $\log V$ : $\frac{{d(\log P)}}{{d(\log V)}} + 1 = 0 \Rightarrow \frac{{d(\log P)}}{{d(\log V)}} = -1$

Step 7: Calculate the ratio of the slopes: $\frac{{-\gamma}}{{-1}} = \gamma$

Why other options are wrong:
- Option B: 1-γ. This is incorrect because it doesn't represent the ratio of the slopes correctly.
- Option C: γ-1. This is incorrect because it doesn't represent the ratio of the slopes correctly.
- Option D: $\frac{1}{\gamma }$ . This is incorrect because it's the inverse of the correct ratio.

Key Concepts:
- Reversible adiabatic process
- Reversible isothermal process
- Adiabatic index ( $\gamma$ )
- Logarithmic differentiation

Memory Aids:
- Remember the equations for adiabatic and isothermal processes.

Quick Tips:
- Know how to differentiate logarithmic functions.

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Application +4 | -1

For the reaction,

{\mathrm{C}_2}{\mathrm{H}_4}\left( \mathrm{g} \right) + 3{\mathrm{O}_2}\left( \mathrm{g} \right) \to 2\mathrm{C}{\mathrm{O}_2}\left( \mathrm{g} \right) + 2{\mathrm{H}_2}\mathrm{O};

∆U=-1415 kJ. Then ∆H at

{27^0}

C is :

View answer and explanation

Correct option: B

- 1420\mathrm{kJ}

Option B (- 1420 kJ) is correct because it correctly applies the relationship between ∆H and ∆U, accounting for the change in the number of moles of gas.

Solution approach

Correct Answer + Core Reason: $- 1420\mathrm{kJ}$ . ∆H is calculated using the formula ∆H = ∆U + ∆nRT.

Step-by-Step Solution:

Step 1: Identify the given values:
- ∆U = -1415 kJ
- T = 27°C = 27 + 273 = 300 K

Step 2: Determine the change in the number of moles of gas (∆n):
- ∆n = (moles of gaseous products) - (moles of gaseous reactants)
- ∆n = (2 moles of $CO_2$ ) - (1 mole of $C_2H_4$ + 3 moles of $O_2$ )
- ∆n = 2 - (1 + 3) = 2 - 4 = -2

Step 3: Use the formula: ∆H = ∆U + ∆nRT

Step 4: Substitute the values:
- ∆H = -1415 kJ + (-2) * (8.314 J/mol·K) * (300 K)

Step 5: Convert R to kJ/mol·K: R = 8.314 J/mol·K = 0.008314 kJ/mol·K

Step 6: Calculate ∆H:
- ∆H = -1415 kJ + (-2) * (0.008314 kJ/mol·K) * (300 K)
- ∆H = -1415 kJ - 4.9884 kJ
- ∆H ≈ -1420 kJ

Why other options are wrong:
- Option A: This option might be obtained by neglecting the ∆nRT term or making a calculation error.
- Option C: This option has the wrong sign. The reaction is exothermic, so ∆H should be negative.
- Option D: This option might be obtained by adding the ∆nRT term instead of subtracting it.

Key Concepts:
- Relationship between ∆H and ∆U: $\Delta H = \Delta U + \Delta nRT$
- Change in the number of moles of gas (∆n)
- Conversion of units (J to kJ)

Memory Aids:
- Remember the formula: $\Delta H = \Delta U + \Delta nRT$

Quick Tips:
- Always check the units and convert them if necessary.
- Pay attention to the sign of ∆n. It can be positive, negative, or zero.

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Understanding +4 | -1

The intensive property among the following given quantities

View answer and explanation

Correct option: B

Density

Option B (Density) is correct because density is an intensive property, meaning it does not depend on the amount of substance.

Solution approach

Correct Answer + Core Reason: Density. Density is an intensive property, independent of the amount of substance.

Step-by-Step Solution:

Step 1: Define intensive property: A property that does not depend on the amount of substance.

Step 2: Analyze each option:
- Weight: Depends on the amount of substance (extensive).
- Density: Mass per unit volume (intensive).
- Entropy: Depends on the amount of substance (extensive).
- Volume: Depends on the amount of substance (extensive).

Step 3: Identify the intensive property: Density is the only intensive property.

Why other options are wrong:
- Option A: Weight. Weight is an extensive property.
- Option C: Entropy. Entropy is an extensive property.
- Option D: Volume. Volume is an extensive property.

Key Concepts:
- Intensive properties
- Extensive properties

Memory Aids:
- Intensive properties are independent of the amount of substance.

Quick Tips:
- Know the difference between intensive and extensive properties.

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Application +4 | -1

The enthalpy change for the following reaction. NaOH(aq)+HCl(aq)→NaCl(aq)+

{\mathrm{H}_2}\mathrm{O}

(l) is -57 kJ. Predict the value of the enthalpy change in the following reaction : Ba

{\left( {\mathrm{OH}} \right)_2}

+

{\mathrm{H}_2}\mathrm{S}{\mathrm{O}_4}

(aq)→

\mathrm{BaS}{\mathrm{O}_4}

(s)+2

{\mathrm{H}_2}\mathrm{O}

(l)

View answer and explanation

Correct option: C

- 114\mathrm{kJ}

Option C (-114 kJ) is correct. The reaction involves the neutralization of a strong acid with a strong base, and since Ba(OH)₂ and H₂SO₄ are diprotic, the enthalpy change is twice that of the monoprotic reaction.

Solution approach

Correct Answer + Core Reason: $- 114\mathrm{kJ}$ . The reaction involves the neutralization of a strong acid with a strong base, and since Ba(OH)₂ and H₂SO₄ are diprotic, the enthalpy change is twice that of the monoprotic reaction.

Step-by-Step Solution:

Step 1: Identify the given information: The enthalpy change for the reaction NaOH(aq)+HCl(aq)→NaCl(aq)+ ${\mathrm{H}_2}\mathrm{O}$ (l) is -57 kJ.

Step 2: The reaction Ba ${\left( {\mathrm{OH}} \right)_2}$ + ${\mathrm{H}_2}\mathrm{S}{\mathrm{O}_4}$ (aq)→ $\mathrm{BaS}{\mathrm{O}_4}$ (s)+2 ${\mathrm{H}_2}\mathrm{O}$ (l) involves 2 moles of $H^+$ reacting with 2 moles of $OH^-$ .

Step 3: Since each mole of $H^+$ reacting with each mole of $OH^-$ releases -57 kJ, the reaction of 2 moles of $H^+$ with 2 moles of $OH^-$ will release twice the energy.

Step 4: Calculate the enthalpy change: $\Delta H = 2 \times (-57 \text{ kJ}) = -114 \text{ kJ}$ .

Why other options are wrong:
- Option A: This option is incorrect because it only considers the enthalpy change for one mole of $H^+$ and $OH^-$ reacting.
- Option B: This option is incorrect. There's no clear reason why -76 kJ would be the answer.
- Option D: This option is incorrect. It seems to be multiplying the enthalpy change by 4 instead of 2.

Key Concepts:
- Enthalpy of neutralization: The enthalpy change when one mole of acid is neutralized by a base.
- Strong acid-strong base neutralization: The enthalpy change is approximately constant for strong acid-strong base reactions.

Memory Aids:
- Remember that for diprotic acids and bases, the enthalpy change will be twice that of monoprotic acids and bases.

Quick Tips:
- Always check the stoichiometry of the reaction to determine how many moles of acid and base are reacting.

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Understanding +4 | -1

If a refrigerator door is kept open, then we get:

View answer and explanation

Correct option: B

Room heated

Option B (Room heated) is correct because the refrigerator expels more heat into the room than it removes, due to the energy consumed by its motor.

Solution approach

Correct Answer + Core Reason: Room heated. A refrigerator transfers heat from its inside to the surroundings, consuming energy in the process, which also contributes to the heat in the room.

Step-by-Step Solution:

Step 1: A refrigerator works by extracting heat from its interior and releasing it to the surroundings.

Step 2: The refrigerator's motor consumes electrical energy, which is converted into heat.

Step 3: The total heat released into the room is the sum of the heat extracted from the refrigerator's interior and the heat generated by the motor.

Step 4: Therefore, the net effect of running a refrigerator with its door open is to heat the room.

Why other options are wrong:
- Option A: Room cooled. This is incorrect because the refrigerator releases more heat than it removes.
- Option C: More heat is passed out. While heat is passed out, the net effect is heating the room, not just passing heat.
- Option D: No effect on room. This is incorrect because the refrigerator's operation generates heat.

Key Concepts:
- Thermodynamics
- Heat transfer
- Energy conservation

Memory Aids:
- Refrigerators move heat, but don't eliminate it; they add energy to the system.

Quick Tips:
- Consider the energy consumption of the appliance in addition to its primary function.

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Remembering +4 | -1

An isolated system is that system in which:

View answer and explanation

Correct option: C

There is no exchange of mass and energy with the surroundings

Option C is correct because an isolated system, by definition, does not exchange either mass or energy with its surroundings.

Solution approach

Correct Answer + Core Reason: There is no exchange of mass and energy with the surroundings. An isolated system is defined as one that does not exchange mass or energy with its surroundings.

Step-by-Step Solution:

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

Step 2: Analyze the given options:
- Option A: There is no exchange of energy with the surroundings. This is only partially correct, as an isolated system also does not exchange mass.
- Option B: There is exchange of mass and energy with the surroundings. This describes an open system, not an isolated system.
- Option C: There is no exchange of mass and energy with the surroundings. This is the correct definition of an isolated system.
- Option D: There is exchange of mass with the surroundings. This describes an open system, not an isolated system.

Why other options are wrong:
- Option A: It only mentions energy, but an isolated system also doesn't exchange mass.
- Option B: This describes an open system.
- Option D: This describes an open system.

Key Concepts:
- Isolated system
- Open system
- Closed system
- Thermodynamic system

Memory Aids:
- Isolated: No exchange of mass or energy.

Quick Tips:
- Remember the definitions of open, closed, and isolated systems.

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The law of conservation of energy states that :

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Correct option: C

Energy is neither created nor destroyed

Option C is correct because the law of conservation of energy states that energy cannot be created or destroyed, but can be transformed from one form to another.

Solution approach

Correct Answer + Core Reason: Energy is neither created nor destroyed. The law of conservation of energy states that the total energy of an isolated system remains constant; energy can neither be created nor destroyed, but can transform from one form to another.

Step-by-Step Solution:

Step 1: State the law of conservation of energy: The law of conservation of energy states that the total energy of an isolated system remains constant; it is said to be conserved over time. Energy can neither be created nor destroyed, but can be transformed from one form to another.

Step 2: Analyze the given options:
- Option A: The internal energy of a system is constant. This is not always true; internal energy can change due to heat or work.
- Option B: The heat content of a system is constant. This is not always true; heat can be exchanged with the surroundings.
- Option C: Energy is neither created nor destroyed. This is the correct statement of the law of conservation of energy.
- Option D: There is an equivalence between energy and mass. This is related to Einstein's mass-energy equivalence (E=mc²), but not the primary statement of the law of conservation of energy.

Why other options are wrong:
- Option A: Internal energy is not always constant.
- Option B: Heat content is not always constant.
- Option D: This is related to mass-energy equivalence, not the conservation of energy itself.

Key Concepts:
- Law of conservation of energy
- Internal energy
- Heat
- Mass-energy equivalence

Memory Aids:
- Energy is conserved: neither created nor destroyed.

Quick Tips:
- Remember the different forms of energy and how they can be transformed.

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