The enthalpy of formation of ammonia gas at 298 K is - 46.11 kJ ${\rm{mo}}{{\rm{l}}^{ - 1}}$. The equation to which this value relates is:

$\frac{1}{2}{{\rm{N}}_2}\left( {\rm{g}} \right) + \frac{3}{2}{{\rm{H}}_2}\left( {\rm{g}} \right) \to {\rm{N}}{{\rm{H}}_3}\left( {\rm{g}} \right)$

${\rm{N}}\left( {\rm{g}} \right) + 3{\rm{H}}\left( {\rm{g}} \right) \to {\rm{N}}{{\rm{H}}_3}\left( {\rm{g}} \right)$

${{\rm{N}}_2}\left( {\rm{g}} \right) + 3{{\rm{H}}_2}\left( {\rm{g}} \right) \to 2{\rm{N}}{{\rm{H}}_3}\left( {\rm{g}} \right)$

$\frac{1}{2}{{\rm{N}}_2}\left( {\rm{g}} \right) + \frac{3}{2}{{\rm{H}}_2}\left( {\rm{g}} \right) \to {\rm{N}}{{\rm{H}}_3}\left( l \right)$

State function and intensive property

Path function and intensive property

Path function and extensive property

THERMODYNAMICS MCQs for NEET UG CHEMISTRY — Easy, Page 2

31 Easy

31 of 196Question permalink

Understanding +4 | -1

Which of the following factors do not affect heat capacity?

View answer and explanation

Correct option: D

Temperature of the system

Option D, Temperature of the system, is correct because heat capacity is an intrinsic property that depends on the size, composition, and nature of the system, but not the temperature.

Solution approach

Correct Answer + Core Reason: Temperature of the system. Heat capacity is independent of the temperature of the system.

Step-by-Step Solution:

Step 1: Recall the definition of heat capacity.

Step 2: Heat capacity is the amount of heat required to raise the temperature of a substance by a certain amount (usually 1 degree Celsius or 1 Kelvin).

Step 3: Heat capacity depends on:
- Size of the system (more substance requires more heat)
- Composition of the system (different substances have different heat capacities)
- Nature of the system (solid, liquid, gas)

Step 4: Heat capacity does *not* directly depend on the temperature of the system itself. Specific heat capacity is an intensive property, but the overall heat capacity of a system depends on the amount of substance present, not its current temperature.

Step 5: Final Answer: Temperature of the system

Why other options are wrong:
- Option A: Size of system affects heat capacity because a larger system requires more heat to change its temperature.
- Option B: Composition of system affects heat capacity because different substances have different heat capacities.
- Option C: Nature of system affects heat capacity because the physical state (solid, liquid, gas) influences the heat capacity.

Key Concepts:
- Heat Capacity
- Size of System
- Composition of System
- Nature of System
- Temperature

Memory Aids:
- Remember that heat capacity is an intrinsic property of a substance.

Quick Tips:
- Distinguish between heat capacity and specific heat capacity. Specific heat capacity is per unit mass.

32 Easy

32 of 196Question permalink

Application +4 | -1

Workdone for the conversion of 0.5 mole of water at

100^{\circ} \mathrm{C}

to steam at

1 \mathrm{~atm}

pressure is (heat of vaporisation of water at

100^{\circ} \mathrm{C}

is

40670 \mathrm{~J} \mathrm{~mol}^{-1}

)

View answer and explanation

Correct option: A

-1.54 \mathrm{~kJ}

Option A (

-1.54 \mathrm{~kJ}

) is correct because it is the work done calculated using

W = -P\Delta V

during the phase change from water to steam.

Solution approach

Correct Answer + Core Reason: $-1.54 \mathrm{~kJ}$ . The work done is calculated using $W = -P\Delta V$ and the ideal gas law.

Step-by-Step Solution:

Step 1: Identify the given information.\nMoles of water, n = 0.5 mol\nTemperature, T = 100°C = 373 K\nPressure, P = 1 atm\nHeat of vaporization = 40670 J/mol

Step 2: Calculate the volume of steam formed using the ideal gas law.
$PV = nRT$ \nV = $\frac{nRT}{P} = \frac{0.5 \times 8.314 \times 373}{101325/100000} = \frac{0.5 \times 8.314 \times 373}{1.01325} = 1531.5/1.01325 = 1531.5/1.01325 \approx 1511.5 L$ \nSince initial volume of water is negligible compared to steam, $\Delta V = V_{steam} - V_{water} \approx V_{steam}$
$\Delta V = \frac{nRT}{P} = \frac{0.5 \times 0.0821 \times 373}{1} = 15.31 \,L$
$\Delta V = 15.31 \times 10^{-3} m^3$

Step 3: Calculate the work done.\nW = $-P\Delta V = -1 \,atm \times 15.31 \,L = -15.31 L atm$ \nW = $-15.31 \times 101.325 J = -1551.38 J = -1.55 kJ$
$\Delta V = nRT/P = 0.5 * 8.314 * 373 / 101325 = 0.0153 m^3$ \nW = -P*dV = -101325 * 0.0153 = -1550 J = -1.55 kJ

Why other options are wrong:
- Option B: This is the positive value of the work done. Work done *by* the system is negative.
- Option C: This value is incorrect and likely from a calculation error or misunderstanding of the formula.
- Option D: This value is incorrect and likely from a calculation error or misunderstanding of the formula.

Key Concepts:
- Work done during phase change
- Ideal gas law

Memory Aids:
- $W = -P\Delta V$
- $PV = nRT$

Quick Tips:
- Remember to convert units to be consistent (e.g., L atm to Joules).

33 Easy

33 of 196Question permalink

Application +4 | -1

For a reaction, ∆H=9.08 kJ

mo{l^{ - 1}}

and ∆S=35.7 J

{K^{ - 1}}mo{l^{ - 1}}

. Which of the following statement is correct for the reaction?

View answer and explanation

Correct option: C

Spontaneous and endothermic

Option C (Spontaneous and endothermic) is correct because ∆H is positive (endothermic) and ∆G is negative (spontaneous) at high temperatures, calculated using ∆G = ∆H - T∆S.

Solution approach

Correct Answer + Core Reason: Spontaneous and endothermic. The reaction is spontaneous because ∆G is negative and endothermic because ∆H is positive.

Step-by-Step Solution:

Step 1: Given: $\Delta H = 9.08 \text{{ kJ/mol}} = 9080 \text{{ J/mol}}$ and $\Delta S = 35.7 \text{{ J/K/mol}}$ .

Step 2: To determine spontaneity, we need to calculate $\Delta G$ using the equation $\Delta G = \Delta H - T\Delta S$ .

Step 3: We need to find a temperature T such that $\Delta G < 0$ .\nLet's find the temperature at which $\Delta G = 0$ : $0 = 9080 - T(35.7)$ .
$T = \frac{{9080}}{{35.7}} \approx 254.3 \text{{ K}}$ .

Step 4: Since $\Delta H$ is positive, the reaction is endothermic.

Step 5: For $T > 254.3 \text{{ K}}$ , $\Delta G$ will be negative, making the reaction spontaneous.

Why other options are wrong:
- Option A: Reversible and isothermal: This is incorrect because we don't have information about reversibility or constant temperature.
- Option B: Reversible and exothermic: This is incorrect because the reaction is endothermic, not exothermic.
- Option D: Spontaneous and exothermic: This is incorrect because the reaction is endothermic, not exothermic.

Key Concepts:
- Gibbs free energy
- Enthalpy
- Entropy
- Spontaneity

Memory Aids:
- $\Delta G < 0$ for spontaneous reactions.
- $\Delta H > 0$ for endothermic reactions.
- $\Delta H < 0$ for exothermic reactions.

Quick Tips:
- Use the equation $\Delta G = \Delta H - T\Delta S$ to determine spontaneity.

34 Easy

34 of 196Question permalink

Understanding +4 | -1

The enthalpy of formation of ammonia gas at 298 K is - 46.11 kJ

\mathrm{mo}{\mathrm{l}^{ - 1}}

. The equation to which this value relates is:

View answer and explanation

Correct option: A

\frac{1}{2}{\mathrm{N}_2}\left( \mathrm{g} \right) + \frac{3}{2}{\mathrm{H}_2}\left( \mathrm{g} \right) \to \mathrm{N}{\mathrm{H}_3}\left( \mathrm{g} \right)

Option A (

\frac{1}{2}N_2(g) + \frac{3}{2}H_2(g) \to NH_3(g)

) is correct because it represents the formation of one mole of ammonia from its elements in their standard states.

Solution approach

Correct Answer + Core Reason: $\frac{1}{2}{\mathrm{N}_2}\left( \mathrm{g} \right) + \frac{3}{2}{\mathrm{H}_2}\left( \mathrm{g} \right) \to \mathrm{N}{\mathrm{H}_3}\left( \mathrm{g} \right)$ . This equation represents the formation of one mole of ammonia from its elements in their standard states.

Step-by-Step Solution:

Step 1: Define enthalpy of formation.

Step 2: The enthalpy of formation is the enthalpy change when one mole of a compound is formed from its elements in their standard states.

Step 3: The standard states of nitrogen and hydrogen are $N_2(g)$ and $H_2(g)$ , respectively. The balanced equation for the formation of one mole of ammonia is $\frac{1}{2}N_2(g) + \frac{3}{2}H_2(g) \to NH_3(g)$ .

Why other options are wrong:
- Option B: $\mathrm{N}\left( \mathrm{g} \right) + 3\mathrm{H}\left( \mathrm{g} \right) \to \mathrm{N}{\mathrm{H}_3}\left( \mathrm{g} \right)$ is incorrect because nitrogen and hydrogen must be in their standard states, which are $N_2(g)$ and $H_2(g)$ , not N(g) and H(g).
- Option C: ${\mathrm{N}_2}\left( \mathrm{g} \right) + 3{\mathrm{H}_2}\left( \mathrm{g} \right) \to 2\mathrm{N}{\mathrm{H}_3}\left( \mathrm{g} \right)$ is incorrect because it represents the formation of two moles of ammonia, not one.
- Option D: $\frac{1}{2}{\mathrm{N}_2}\left( \mathrm{g} \right) + \frac{3}{2}{\mathrm{H}_2}\left( \mathrm{g} \right) \to \mathrm{N}{\mathrm{H}_3}\left( l \right)$ is incorrect because it forms liquid ammonia, not gaseous ammonia. The enthalpy of formation refers to the formation of gaseous ammonia.

Key Concepts:
- Enthalpy of formation
- Standard state

Memory Aids:
- Formation of 1 mole from elements in standard states.

Quick Tips:
- Remember to balance the equation and use the standard states of the elements.

35 Easy

35 of 196Question permalink

Remembering +4 | -1

Which reaction is endothermic in nature?

View answer and explanation

Correct option: A

\mathrm{CaC}{\mathrm{O}_3} \to \mathrm{CaO} + \mathrm{C}{\mathrm{O}_2}

Option A (

\mathrm{CaC}{\mathrm{O}_3} \to \mathrm{CaO} + \mathrm{C}{\mathrm{O}_2}

) is correct because decomposition reactions generally require energy input to break bonds, making them endothermic.

Solution approach

Correct Answer + Core Reason: $\mathrm{CaC}{\mathrm{O}_3} \to \mathrm{CaO} + \mathrm{C}{\mathrm{O}_2}$ . Decomposition reactions are generally endothermic as they require energy to break bonds.

Step-by-Step Solution:

Step 1: Define endothermic reactions: Endothermic reactions absorb heat from the surroundings, resulting in a positive enthalpy change (Δ H > 0).

Step 2: Analyze the given reactions:
- A: $\mathrm{CaC}{\mathrm{O}_3} \to \mathrm{CaO} + \mathrm{C}{\mathrm{O}_2}$ (Decomposition of calcium carbonate)
- B: $\mathrm{C} + {\mathrm{O}_2} \to \mathrm{C}{\mathrm{O}_2}$ (Combustion of carbon)
- C: $\mathrm{NaOH} + \mathrm{HCl} \to \mathrm{NaCl} + {\mathrm{H}_2}\mathrm{O}$ (Neutralization reaction)
- D: $\mathrm{C}{\mathrm{H}_4} + 2{\mathrm{O}_2} \to \mathrm{C}{\mathrm{O}_2} + 2{\mathrm{H}_2}\mathrm{O}$ (Combustion of methane)

Step 3: Identify the type of reaction:
- A is a decomposition reaction, which typically requires energy input to break chemical bonds. Therefore, it is endothermic.
- B, C, and D are exothermic reactions. Combustion and neutralization reactions release heat.

Step 4: Conclusion: The endothermic reaction is the decomposition of calcium carbonate.

Why other options are wrong:
- Option B: $\mathrm{C} + {\mathrm{O}_2} \to \mathrm{C}{\mathrm{O}_2}$ : This is a combustion reaction, which is exothermic (releases heat).
- Option C: $\mathrm{NaOH} + \mathrm{HCl} \to \mathrm{NaCl} + {\mathrm{H}_2}\mathrm{O}$ : This is a neutralization reaction between a strong acid and a strong base, which is exothermic (releases heat).
- Option D: $\mathrm{C}{\mathrm{H}_4} + 2{\mathrm{O}_2} \to \mathrm{C}{\mathrm{O}_2} + 2{\mathrm{H}_2}\mathrm{O}$ : This is a combustion reaction, which is exothermic (releases heat).

Key Concepts:
- Endothermic Reactions
- Exothermic Reactions
- Decomposition Reactions
- Combustion Reactions
- Neutralization Reactions

Memory Aids:
- Combustion reactions always release heat.
- Neutralization reactions release heat.
- Decomposition often requires heat.

Quick Tips:
- Recognize common exothermic and endothermic reactions.
- Consider the energy required to break bonds versus the energy released when forming bonds.

36 Easy

36 of 196Question permalink

Application +4 | -1

\Delta H

for

\mathrm{CaCO}_3(\mathrm{s}) \rightarrow \mathrm{CaO}(\mathrm{s})+\mathrm{CO}_2(\mathrm{g})

is

176 \mathrm{~kJ} \mathrm{~mol}^{-1}

at

1240 \mathrm{~K}.

The

\Delta U

for the reaction is equal to

View answer and explanation

Correct option: B

165.6 \mathrm{~kJ}

The

\Delta U

for the reaction is 165.6 kJ, calculated using the formula

\Delta H = \Delta U + \Delta n_g RT

.

Solution approach

Correct Answer + Core Reason: $165.6 \mathrm{~kJ}$ . Using $\Delta H = \Delta U + \Delta n_g RT$ .

Step-by-Step Solution:

Step 1: Identify the given information: $\Delta H = 176 \mathrm{~kJ} \mathrm{~mol}^{-1}$ and $T = 1240 \mathrm{~K}$ .

Step 2: Determine the change in the number of moles of gas: $\mathrm{CaCO}_3(s) \rightarrow \mathrm{CaO}(s) + \mathrm{CO}_2(g)$ . $\Delta n_g = n_{products}(g) - n_{reactants}(g) = 1 - 0 = 1$ .

Step 3: Use the relationship between $\Delta H$ and $\Delta U$ : $\Delta H = \Delta U + \Delta n_g RT$ .

Step 4: Rearrange the equation to solve for $\Delta U$ : $\Delta U = \Delta H - \Delta n_g RT$ .

Step 5: Substitute the values: $\Delta U = 176000 \mathrm{~J} \mathrm{~mol}^{-1} - (1) (8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}) (1240 \mathrm{~K})$ .

Step 6: Calculate $\Delta U$ : $\Delta U = 176000 - 10309.36 = 165690.64 \mathrm{~J} \mathrm{~mol}^{-1} \approx 165.7 \mathrm{~kJ} \mathrm{~mol}^{-1}$ .

Why other options are wrong:
- Option A: $160 \mathrm{~kJ}$ is incorrect. It likely arises from an incorrect calculation or neglecting the term $\Delta n_g RT$ .
- Option C: $186.3 \mathrm{~kJ}$ is incorrect. It likely arises from adding the $\Delta n_g RT$ term instead of subtracting.
- Option D: $180.0 \mathrm{~kJ}$ is incorrect. It likely arises from an incorrect calculation or approximation.

Key Concepts:
- Enthalpy Change ( $\Delta H$ )
- Internal Energy Change ( $\Delta U$ )
- Change in Number of Moles of Gas ( $\Delta n_g$ )

Memory Aids:
- $\Delta H = \Delta U + \Delta n_g RT$

Quick Tips:
- Pay close attention to the units and ensure they are consistent throughout the calculation.
- Remember to convert kJ to J before using the gas constant R.

37 Easy

37 of 196Question permalink

Application +4 | -1

Five moles of an ideal gas expands reversibly from volumes of

8\,\,d{m^3}

to

80\,\,d{m^3}

at a temperature of

27^\circ \,C,

then entropy change is

View answer and explanation

Correct option: C

95.73\,\,J\,{K^{ - 1}}

Option C (

95.73\,J\,{K^{ - 1}}

) is correct because it correctly calculates the entropy change for an isothermal reversible expansion using the formula

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

.

Solution approach

Correct Answer + Core Reason: $95.73\,J\,{K^{ - 1}}$ . This is the correct entropy change for the given isothermal reversible expansion.

Step-by-Step Solution:

Step 1: Identify the given information.
- Number of moles: $n = 5$
- Initial volume: $V_1 = 8\,dm^3$
- Final volume: $V_2 = 80\,dm^3$
- Temperature: $T = 27^\circ C = 300 K$

Step 2: Recall the formula for entropy change in an isothermal reversible process:
$\Delta S = nR \ln{\frac{V_2}{V_1}}$

Step 3: Substitute the given values into the formula. Use $R = 8.314\,J\,mol^{-1}K^{-1}$ :
$\Delta S = 5 \times 8.314 \times \ln{\frac{80}{8}}$

Step 4: Simplify the expression:
$\Delta S = 5 \times 8.314 \times \ln{10}$

Step 5: Calculate the entropy change:
$\Delta S = 5 \times 8.314 \times 2.303$
$\Delta S = 95.73\,J\,K^{-1}$

Step 6: Final Answer: $95.73\,J\,K^{-1}$

Why other options are wrong:
- Option A: $70.26\,J\,{K^{ - 1}}$ is incorrect. It might arise from using an incorrect value for R or making a calculation error.
- Option B: $82.55\,J\,{K^{ - 1}}$ is incorrect. It might arise from using an incorrect value for R or making a calculation error.
- Option D: $107.11\,J\,{K^{ - 1}}$ is incorrect. It might arise from using an incorrect value for R or making a calculation error.

Key Concepts:
- Entropy change
- Isothermal reversible process
- Ideal gas

Memory Aids:
- Remember the formula for entropy change: $\Delta S = nR \ln{\frac{V_2}{V_1}}$

Quick Tips:
- Ensure correct units for R.
- Use the natural logarithm (ln) and not the base-10 logarithm (log).

38 Easy

38 of 196Question permalink

Remembering +4 | -1

The law of Lavoisier and Laplace is based on :

View answer and explanation

Correct option: A

The principle of conservation of energy

Option A (The principle of conservation of energy) is correct because the law of Lavoisier and Laplace states that the energy change accompanying a reaction is equal in magnitude and opposite in sign to the energy change accompanying the reverse reaction, which is a direct consequence of the conservation of energy.

Solution approach

Correct Answer + Core Reason: The principle of conservation of energy. The Law of Lavoisier and Laplace is based on the principle of conservation of energy.

Step-by-Step Solution:

Step 1: Recall the Law of Lavoisier and Laplace: The enthalpy change for a reaction is equal in magnitude and opposite in sign to the enthalpy change for the reverse reaction.

Step 2: Relate the Law to fundamental principles: This law is a direct consequence of the principle of conservation of energy. If energy were not conserved, the forward and reverse reactions would not have equal and opposite energy changes.

Why other options are wrong:
- Option B: Equivalence of mechanical and thermal energies: This is related to the first law of thermodynamics but not the specific basis of the Law of Lavoisier and Laplace.
- Option C: The principle of conservation of matter: While matter conservation is important in chemistry, it doesn't directly explain the Law of Lavoisier and Laplace, which deals with energy changes.
- Option D: Equivalence of mechanical and chemical energies: This is a broader concept related to energy transformations but not the direct basis of the Law of Lavoisier and Laplace.

Key Concepts:
- Law of Lavoisier and Laplace
- Conservation of Energy
- Enthalpy Change

Memory Aids:
- Remember that energy cannot be created or destroyed, only transformed.

Quick Tips:
- The Law of Lavoisier and Laplace is a specific application of the conservation of energy principle to chemical reactions.

39 Easy

39 of 196Question permalink

Remembering +4 | -1

Hess’s law states that

View answer and explanation

Correct option: A

The standard enthalpy of an overall reaction is the sum of the enthalpy changes in individual reactions.

Option A (The standard enthalpy of an overall reaction is the sum of the enthalpy changes in individual reactions) is correct because it accurately describes Hess's Law.

Solution approach

Correct Answer + Core Reason: The standard enthalpy of an overall reaction is the sum of the enthalpy changes in individual reactions. This is the definition of Hess's Law.

Step-by-Step Solution:

Step 1: Recall the definition of Hess's Law: The enthalpy change for a reaction is independent of the pathway taken.

Step 2: This means that if a reaction can be carried out in multiple steps, the sum of the enthalpy changes for each step will equal the enthalpy change for the overall reaction.

Why other options are wrong:
- Option B: Enthalpy of formation of a compound is same as the enthalpy of decomposition of the compound into constituent elements, but with opposite sign. This is a correct statement but not Hess's Law.
- Option C: At constant temperature the pressure of a gas is inversely proportional to its volume. This is Boyle's Law.
- Option D: The mass of a gas dissolved per litre of a solvent is proportional to the pressure of the gas in equilibrium with the solution. This is Henry's Law.

Key Concepts:
- Hess's Law
- Enthalpy change

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

Quick Tips:
- Hess's Law is useful for calculating enthalpy changes for reactions that are difficult to measure directly.

40 Easy

40 of 196Question permalink

Application +4 | -1

C_p

and

C_v

for a gas are

0,125 {cal} / {g}

and

0,075 {cal} / {g}

respectively. Which one of the gas has these values?

View answer and explanation

Correct option: C

{Ar}

Argon (Ar) is the correct answer because it is a monatomic gas, and for monatomic gases,

C_p - C_v = R

. The given values are close to R when converted to appropriate units.

Solution approach

Correct Answer + Core Reason: ${Ar}$ . For monatomic gases, $C_p - C_v = R$ .

Step-by-Step Solution:

Step 1: Identify the given information: $C_p = 0.125 \mathrm{~cal/g}$ and $C_v = 0.075 \mathrm{~cal/g}$ .

Step 2: Calculate $C_p - C_v$ : $0.125 - 0.075 = 0.050 \mathrm{~cal/g}$ .

Step 3: Convert the value to $\mathrm{cal/mol}$ : We need to find the molar mass of each gas and multiply by 0.050.
- He: 4 g/mol, (0.050)(4) = 0.2 cal/mol
- O2: 32 g/mol, (0.050)(32) = 1.6 cal/mol
- Ar: 40 g/mol, (0.050)(40) = 2.0 cal/mol
- Ne: 20 g/mol, (0.050)(20) = 1.0 cal/mol

Step 4: Compare with the value of R (1.987 cal/mol ≈ 2 cal/mol). The value for Ar is closest to R.

Step 5: For a monatomic gas, $C_p = \frac{5}{2}R$ and $C_v = \frac{3}{2}R$ . Therefore, $C_p - C_v = R$ .

Why other options are wrong:
- Option A: ${He}$ is incorrect because its molar mass is too low, resulting in a value significantly different from R.
- Option B: ${O}_2$ is incorrect because it is a diatomic gas, and the relationship $C_p - C_v = R$ does not directly apply (it's more complex due to vibrational modes).
- Option D: ${Ne}$ is incorrect because its molar mass is lower than Ar, resulting in a value further from R.

Key Concepts:
- Molar Heat Capacity
- Monatomic Gases
- Relationship between $C_p$ and $C_v$

Memory Aids:
- For monatomic gases: $C_p - C_v = R$

Quick Tips:
- Remember that the relationship between $C_p$ and $C_v$ depends on the type of gas (monatomic, diatomic, etc.).

41 Easy

41 of 196Question permalink

Understanding +4 | -1

Heat of solution is defined as :

View answer and explanation

Correct option: C

Change in heat content of the system when one mole of the solute is dissolved un excess of water, so that further dilution of solution does not bring any heat change

Option C is the correct definition of heat of solution because it specifies the heat change when one mole of solute dissolves in excess solvent until further dilution causes no heat change.

Solution approach

Correct Answer + Core Reason: Change in heat content of the system when one mole of the solute is dissolved un excess of water, so that further dilution of solution does not bring any heat change. This is the standard definition of heat of solution.

Step-by-Step Solution:

Step 1: Define heat of solution. Heat of solution is the enthalpy change when one mole of a substance is dissolved in a specified amount of solvent.

Step 2: Consider the effect of dilution. The heat of solution is defined at infinite dilution, meaning further addition of solvent does not cause a significant heat change.

Step 3: Relate to heat content. The heat of solution represents the change in the heat content (enthalpy) of the system.

Step 4: Conclusion: The correct definition includes the concept of infinite dilution.

Why other options are wrong:
- Option A: Heat required to dissolved one mole in excess of water. This is incomplete as it doesn't specify the point where further dilution causes no heat change.
- Option B: Heat evolved, when one mole is dissolved in excess of water. This only describes exothermic dissolution and doesn't account for endothermic processes.
- Option D: None of the above. This is incorrect because option C is the correct definition.

Key Concepts:
- Heat of Solution
- Enthalpy Change
- Infinite Dilution

Memory Aids:
- Heat of solution is the enthalpy change at infinite dilution.

Quick Tips:
- Remember that heat of solution can be positive (endothermic) or negative (exothermic).

42 Easy

42 of 196Question permalink

Understanding +4 | -1

The gas in a refrigerator causes cooling on expansion because:

View answer and explanation

Correct option: B

Heat of the gas is lost as work is done by the gas

Option B (Heat of the gas is lost as work is done by the gas) is correct because during expansion, the gas performs work, which requires energy, leading to a decrease in the gas's internal energy and thus cooling.

Solution approach

Correct Answer + Core Reason: Heat of the gas is lost as work is done by the gas. In an expansion, the gas does work, which consumes energy from its internal energy, leading to cooling.

Step-by-Step Solution:

Step 1: Recall the first law of thermodynamics: $\Delta U = Q - W$ , where $\Delta U$ is the change in internal energy, Q is heat added to the system, and W is work done by the system.

Step 2: In an adiabatic expansion, Q = 0, so $\Delta U = -W$ .

Step 3: Since the gas is doing work (W > 0), the change in internal energy is negative ( $\Delta U < 0$ ), which means the temperature decreases, resulting in cooling.

Why other options are wrong:
- Option A: Work done by the gas is converted into heat is incorrect because the opposite happens; the gas uses its internal energy to do work.
- Option C: The heat is spread over a large space is incorrect because it describes diffusion, not the thermodynamic process of cooling during expansion.
- Option D: None of the above is incorrect because option B correctly explains the cooling effect.

Key Concepts:
- First Law of Thermodynamics
- Adiabatic Expansion
- Internal Energy

Memory Aids:
- Expansion cools, compression heats.

Quick Tips:
- Remember that in an adiabatic process, the change in internal energy is equal to the negative of the work done.

43 Easy

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Understanding +4 | -1

Bond energy of molecule:

View answer and explanation

Correct option: B

Is always positive

Option B (Is always positive) is correct because bond energy is defined as the energy required to break a bond, which is always an endothermic process.

Solution approach

Correct Answer + Core Reason: Is always positive. Bond energy is the energy required to break a bond, which is always an endothermic process.

Step-by-Step Solution:

Step 1: Define bond energy.

Step 2: Bond energy is the energy required to break a chemical bond in a molecule.

Step 3: Breaking a bond always requires energy input, making it an endothermic process. Therefore, bond energy is always positive.

Why other options are wrong:
- Option A: Bond energy cannot be negative because breaking a bond requires energy.
- Option C: Bond energy is always positive, not either positive or negative.
- Option D: Bond energy is a property of the bond itself and is not dependent on the physical state of the system.

Key Concepts:
- Bond energy
- Endothermic process

Memory Aids:
- Breaking bonds requires energy (positive bond energy).

Quick Tips:
- Remember that bond breaking is always endothermic.

44 Easy

44 of 196Question permalink

Understanding +4 | -1

If liquids A and B from an ideal solution, the:

View answer and explanation

Correct option: A

Enthalpy of mixing is zero

Option A (Enthalpy of mixing is zero) is correct because for an ideal solution, there are no enthalpy changes upon mixing, indicating no heat is absorbed or released.

Solution approach

Correct Answer + Core Reason: Enthalpy of mixing is zero. Ideal solutions have no enthalpy change upon mixing.

Step-by-Step Solution:

Step 1: Ideal Solution Definition: An ideal solution is a solution in which the interactions between the molecules of the components are the same as the interactions between the molecules of each component individually.

Step 2: Enthalpy of Mixing: For an ideal solution, the enthalpy of mixing ( $\Delta H_{mix}$ ) is zero. This means that no heat is absorbed or released when the liquids are mixed.

Step 3: Entropy of Mixing: For an ideal solution, the entropy of mixing ( $\Delta S_{mix}$ ) is positive because the mixing process increases the disorder of the system.

Step 4: Gibbs Free Energy of Mixing: The Gibbs free energy of mixing ( $\Delta G_{mix}$ ) is negative for an ideal solution because $\Delta G_{mix} = \Delta H_{mix} - T\Delta S_{mix}$ , and since $\Delta H_{mix} = 0$ and $\Delta S_{mix} > 0$ , $\Delta G_{mix} < 0$ .

Step 5: Conclusion: Therefore, for liquids A and B forming an ideal solution, the enthalpy of mixing is zero.

Why other options are wrong:
- Option B: Entropy of mixing is zero: The entropy of mixing is positive for an ideal solution.
- Option C: Free energy of mixing is zero: The free energy of mixing is negative for an ideal solution.
- Option D: Free energy as well as the entropy of mixing are each zero: Neither the free energy nor the entropy of mixing is zero for an ideal solution.

Key Concepts:
- Ideal Solutions
- Enthalpy of Mixing
- Entropy of Mixing
- Gibbs Free Energy of Mixing

Memory Aids:
- Ideal = Zero Enthalpy

Quick Tips:
- Remember that for ideal solutions, $\Delta H_{mix} = 0$ , $\Delta S_{mix} > 0$ , and $\Delta G_{mix} < 0$ .

45 Easy

45 of 196Question permalink

Remembering +4 | -1

The standard free energy change

\Delta G^\circ

is related to

K

(equilibrium constant) as

View answer and explanation

Correct option: A

\Delta G^\circ = - 2.303\,RT\log K

Option A is correct because the standard free energy change is related to the equilibrium constant by the equation ΔG° = -2.303RTlogK.

Solution approach

Correct Answer + Core Reason: $\Delta G^\circ = - 2.303\,RT\log K$ . This is the standard equation relating Gibbs free energy and equilibrium constant.

Step-by-Step Solution:

Step 1: Gibbs Free Energy: A thermodynamic potential that measures the amount of energy available in a thermodynamic system to perform useful work at a constant temperature and pressure.

Step 2: Equilibrium Constant (K): A measure of the relative amounts of reactants and products at equilibrium.

Step 3: Relationship between $\Delta G^\circ$ and K: $\Delta G^\circ = -RT \ln K$

Step 4: Convert natural logarithm to base-10 logarithm: $\Delta G^\circ = -2.303 RT \log K$

Step 5: Conclusion: The standard free energy change is related to the equilibrium constant as $\Delta G^\circ = - 2.303\,RT\log K$ .

Why other options are wrong:
- Option B: Incorrect; the sign is wrong.
- Option C: Incorrect; the factor 2.303 is missing, and the sign is wrong.
- Option D: Incorrect; the factor 2.303 is missing.

Key Concepts:
- Gibbs Free Energy
- Equilibrium Constant

Memory Aids:
- $\Delta G^\circ = -RT \ln K$ (remember the negative sign)

Quick Tips:
- Remember to convert between natural and base-10 logarithms when necessary.

46 Easy

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Remembering +4 | -1

Entropy is a

View answer and explanation

Correct option: A

State function and extensive property

Option A is correct because entropy is a state function and its value depends on the amount of substance, making it an extensive property.

Solution approach

Correct Answer + Core Reason: State function and extensive property. Entropy depends on the state and amount of substance.

Step-by-Step Solution:

Step 1: State Function: A property that depends only on the initial and final states of the system, not on the path taken.

Step 2: Extensive Property: A property that depends on the amount of substance present in the system.

Step 3: Entropy: Entropy is a state function because its change depends only on the initial and final states.

Step 4: Entropy is extensive because the entropy of two moles of a substance is twice the entropy of one mole.

Step 5: Conclusion: Entropy is a state function and an extensive property.

Why other options are wrong:
- Option B: Incorrect because entropy is an extensive property, not intensive.
- Option C: Incorrect because entropy is a state function, not a path function.
- Option D: Incorrect because entropy is a state function, not a path function.

Key Concepts:
- State Function
- Extensive Property
- Entropy

Memory Aids:
- State functions: Think of enthalpy, internal energy, Gibbs free energy.
- Extensive properties: Think of mass, volume, entropy.

Quick Tips:
- Distinguish between state and path functions, and extensive and intensive properties.

47 Easy

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Understanding +4 | -1

The enthalpy changes of formation of the gaseous oxides of nitrogen

(\;{N_2}OandNO)

are positive because of:

View answer and explanation

Correct option: A

The high bond energy of the nitrogen molecule

Option A is correct because the formation of gaseous oxides of nitrogen requires breaking the strong triple bond in the nitrogen molecule, which requires energy.

Solution approach

Correct Answer + Core Reason: The high bond energy of the nitrogen molecule. Breaking the N≡N bond requires a lot of energy.

Step-by-Step Solution:

Step 1: Formation of Oxides of Nitrogen: $\frac{1}{2}N_2(g) + \frac{1}{2}O_2(g) \rightarrow NO(g)$ and $N_2(g) + \frac{1}{2}O_2(g) \rightarrow N_2O(g)$

Step 2: Bond Energy of Nitrogen: The nitrogen molecule ( $N_2$ ) has a very strong triple bond (N≡N), which requires a large amount of energy to break.

Step 3: Positive Enthalpy Change: The high bond energy of nitrogen means that a significant amount of energy is required to break the bond before the nitrogen can react with oxygen to form oxides of nitrogen. This results in a positive enthalpy change (endothermic process).

Step 4: Conclusion: The enthalpy changes of formation of the gaseous oxides of nitrogen are positive because of the high bond energy of the nitrogen molecule.

Why other options are wrong:
- Option B: Incorrect; the electron affinity of oxygen is relevant to the formation of oxide ions, but not the primary reason for the positive enthalpy change of formation of nitrogen oxides.
- Option C: Incorrect; the electron affinity of nitrogen is not the primary reason for the positive enthalpy change of formation of nitrogen oxides.
- Option D: Incorrect; the tendency of oxygen to form $O^{2-}$ is relevant to ionic compounds, but not the primary reason for the positive enthalpy change of formation of nitrogen oxides.

Key Concepts:
- Enthalpy of Formation
- Bond Energy

Memory Aids:
- High bond energy requires energy input (endothermic).

Quick Tips:
- Remember that breaking bonds requires energy, while forming bonds releases energy.

48 Easy

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Remembering +4 | -1

For an adiabatic process:

View answer and explanation

Correct option: B

Q=0

Option B (Q=0) is correct because an adiabatic process is defined as a process in which there is no heat exchange between the system and the surroundings.

Solution approach

Correct Answer + Core Reason: Q=0. An adiabatic process is defined as a process where no heat is exchanged between the system and the surroundings.

Step-by-Step Solution:

Step 1: Define an adiabatic process: An adiabatic process is a thermodynamic process in which there is no heat transfer into or out of the system. This means that the heat (Q) is equal to zero.

Step 2: Analyze the given options:
- Option A: Q=+W. This is incorrect. In an adiabatic process, Q = 0, so this equation would imply 0 = +W, which is not generally true.
- Option B: Q=0. This is the correct definition of an adiabatic process.
- Option C: Δ U=q. This is incorrect. In general, Δ U = Q + W. For an adiabatic process, Δ U = W since Q = 0.
- Option D: PΔ V=0. This is incorrect. PΔ V represents the work done in an isobaric process, and it is not necessarily zero in an adiabatic process.

Why other options are wrong:
- Option A: Q is zero in an adiabatic process, not equal to W.
- Option C: Δ U = W in an adiabatic process, not equal to q (which is zero).
- Option D: PΔ V is not necessarily zero in an adiabatic process.

Key Concepts:
- Adiabatic process
- Heat transfer (Q)
- Internal energy (Δ U)
- Work (W)

Memory Aids:
- Adiabatic: No heat exchange (Q = 0).

Quick Tips:
- Remember the definition of an adiabatic process: Q = 0.

49 Easy

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Understanding +4 | -1

Energy equivalent to one erg, one joule and one calorie are in order:

View answer and explanation

Correct option: C

1\mathrm{\;cal} > 1\;J > 1\;erg

Option C (

1\mathrm{\;cal} > 1\;J > 1\;erg

) is correct because it accurately represents the relative magnitudes of calorie, joule, and erg.

Solution approach

Correct Answer + Core Reason: $1\mathrm{\;cal} > 1\;J > 1\;erg$ . This order reflects the conversion factors between these energy units.

Step-by-Step Solution:

Step 1: Recall the conversion factors between calories, joules, and ergs.

Step 2: 1 calorie (cal) = 4.184 joules (J)

Step 3: 1 joule (J) = $10^7$ ergs

Step 4: Therefore, 1 cal > 1 J > 1 erg.

Why other options are wrong:
- Option A: $1\mathrm{\;erg} > 1\;J > 1\;cal$ is incorrect because it reverses the order of magnitude.
- Option B: $1\mathrm{\;erg} > 1\;cal > 1\;J$ is incorrect because it reverses the order of magnitude.
- Option D: $1\mathrm{\;J} > 1\;cal > 1\;erg$ is incorrect because it places joules as larger than calories.

Key Concepts:
- Energy Units
- Conversion Factors

Memory Aids:
- Calorie is the largest, then Joule, then Erg.

Quick Tips:
- Remember the approximate conversion factor: 1 cal ≈ 4.2 J.

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Remembering +4 | -1

The entropy of crystalline substances at
absolute zero going by the third law of
thermodynamics should be taken as

View answer and explanation

Correct option: C

zero

Option C is correct because, according to the third law of thermodynamics, the entropy of a perfectly crystalline substance at absolute zero is zero.

Solution approach

Correct Answer + Core Reason: zero. The third law of thermodynamics states that entropy is zero at absolute zero for crystalline substances.

Step-by-Step Solution:

Step 1: Third Law of Thermodynamics: The entropy of a system approaches a minimum or zero value as the temperature approaches absolute zero.

Step 2: Crystalline Substance: A substance with a highly ordered structure.

Step 3: Absolute Zero: 0 Kelvin or -273.15 degrees Celsius.

Step 4: Conclusion: The entropy of crystalline substances at absolute zero is taken as zero.

Why other options are wrong:
- Option A: Incorrect; entropy is zero, not 100.
- Option B: Incorrect; entropy is zero, not 50.
- Option D: Incorrect; entropy is zero for all crystalline substances at absolute zero.

Key Concepts:
- Third Law of Thermodynamics
- Entropy
- Absolute Zero

Memory Aids:
- Third Law: Entropy approaches zero at zero Kelvin.

Quick Tips:
- Remember the third law of thermodynamics and its implications for entropy at absolute zero.

51 Easy

51 of 196Question permalink

Application +4 | -1

10 mole of an ideal gas expand isothermally and reversibly from a pressure of

10 \mathrm{~atm}

to

1 \mathrm{~atm}

at

300 \mathrm{~K}

. What is the largest mass which can lifted through a height of 100 meter?

View answer and explanation

Correct option: B

58.55 \mathrm{~kg}

Option B (

58.55 \mathrm{~kg}

) is correct because the work done by the gas during isothermal reversible expansion is used to lift the mass against gravity.

Solution approach

Correct Answer + Core Reason: $58.55 \mathrm{~kg}$ . The work done by the gas is used to lift the mass.

Step-by-Step Solution:

Step 1: Calculate the work done during isothermal reversible expansion: $W = -nRT \ln{\frac{P_2}{P_1}}$

Step 2: Substitute the given values: $n = 10 \text{{ mol}}$ , $R = 8.314 \text{{ J/mol K}}$ , $T = 300 \text{{ K}}$ , $P_1 = 10 \text{{ atm}}$ , $P_2 = 1 \text{{ atm}}$

Step 3: Calculate the work: $W = -10 \times 8.314 \times 300 \times \ln{\frac{1}{10}} = -10 \times 8.314 \times 300 \times (-2.303) = 57441.42 \text{{ J}}$

Step 4: The work done by the gas is used to lift the mass: $W = mgh$

Step 5: Solve for the mass: $m = \frac{W}{gh} = \frac{57441.42 \text{{ J}}}{9.8 \text{{ m/s}}^2 \times 100 \text{{ m}}} = 58.61 \text{{ kg}}$ (Slight difference due to rounding)

Why other options are wrong:
- Option A: $31842 \mathrm{~kg}$ . Incorrect, this value is significantly higher and likely results from an incorrect calculation or misunderstanding of the formula.
- Option C: $342.58 \mathrm{~kg}$ . Incorrect, this value is also too high and likely results from an incorrect calculation.
- Option D: $6051 \mathrm{~kg}$ . Incorrect, this value is too high and likely results from an incorrect calculation.

Key Concepts:
- Isothermal reversible expansion, Work done, Ideal gas law

Memory Aids:
- Remember the formula for work done during isothermal reversible expansion: $W = -nRT \ln{\frac{P_2}{P_1}}$

Quick Tips:
- Always use consistent units when performing calculations. Ensure that the units of R, T, and P are compatible.

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Analyzing +4 | -1

Assertion: The value of enthalpy of neutralization of weak acid and strong base is numerically less than

57.1 {~kJ}

.

Reason: All the

{OH}^{-}

ions furnished by

1 {~g}

equivalent of strong base are not completely neutralized.

View answer and explanation

Correct option: C

If Assertion is correct but Reason is incorrect.

Option C is correct. The assertion is correct because the enthalpy of neutralization of a weak acid and strong base is less than 57.1 kJ due to the energy required to ionize the weak acid. The reason is incorrect because strong bases are completely ionized.

Solution approach

Correct Answer + Core Reason: If Assertion is correct but Reason is incorrect. The assertion is true, but the reason provides a false explanation.

Step-by-Step Solution:

Step 1: Analyze the Assertion: The enthalpy of neutralization of a weak acid with a strong base is indeed less than 57.1 kJ/mol because some energy is used to ionize the weak acid before neutralization can occur.

Step 2: Analyze the Reason: Strong bases are completely ionized in solution. Therefore, the statement that "all the OH- ions furnished by 1 g equivalent of strong base are not completely neutralized" is incorrect.

Step 3: Determine the relationship: The reason does not correctly explain the assertion. The lower enthalpy of neutralization is due to the weak acid's incomplete ionization, not the strong base.

Why other options are wrong:
- Option A: If both Assertion and Reason are correct and the Reason is the correct explanation of Assertion: The reason is incorrect.
- Option B: If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion: The reason is incorrect.
- Option D: If both the Assertion and Reason are incorrect: The assertion is correct.

Key Concepts:
- Enthalpy of Neutralization
- Weak Acids and Strong Bases
- Ionization

Memory Aids:
- Weak acids require energy for ionization before neutralization.

Quick Tips:
- Remember that weak acids and bases do not fully ionize in solution.

53 Easy

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Application +4 | -1

What is ∆E for system that does 500 cal of work on surrounding and 300 cal of heat is absorbed by the system?

View answer and explanation

Correct option: A

- 200\mathrm{ }cal

Option A (

- 200\mathrm{ }cal

) is correct because the change in internal energy (∆E) is calculated as heat absorbed (q) plus work done on the system (w), where w is negative since the system does work on the surroundings.

Solution approach

Correct Answer + Core Reason: $- 200\mathrm{ }cal$ . Change in internal energy is heat absorbed plus work done on the system.

Step-by-Step Solution:

Step 1: State the First Law of Thermodynamics: $\Delta E = q + W$ , where $\Delta E$ is the change in internal energy, q is the heat absorbed by the system, and W is the work done on the system.

Step 2: Identify the given values: $q = +300 \text{{ cal}}$ (heat absorbed), $W = -500 \text{{ cal}}$ (work done *by* the system, so negative).

Step 3: Apply the formula: $\Delta E = 300 \text{{ cal}} + (-500 \text{{ cal}}) = -200 \text{{ cal}}$

Step 4: Conclusion: $\Delta E = -200 \text{{ cal}}$

Why other options are wrong:
- Option B: $- 300\mathrm{ }cal$ is incorrect; it doesn't correctly apply the first law with the given values.
- Option C: $+ 200\mathrm{ }cal$ is incorrect; it has the wrong sign and doesn't correctly apply the first law.
- Option D: $+ 300\mathrm{ }cal$ is incorrect; it only considers the heat absorbed and not the work done.

Key Concepts:
- First Law of Thermodynamics
- Internal Energy Change
- Heat
- Work

Memory Aids:
- $\Delta E = q + W$ (sign conventions are crucial!)

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

54 Easy

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Understanding +4 | -1

Which has the least entropy?

View answer and explanation

Correct option: B

Diamond

Option B (Diamond) is correct because diamond has a highly ordered, rigid crystal structure, resulting in the least entropy compared to the other options.

Solution approach

Correct Answer + Core Reason: Diamond. Diamond has the most ordered structure among the given options, hence the least entropy.

Step-by-Step Solution:

Step 1: Define Entropy: Entropy is a measure of disorder or randomness in a system.

Step 2: Analyze the options:
- Graphite: Has a layered structure, more disordered than diamond.
- Diamond: Has a rigid, highly ordered tetrahedral structure.
- N2(g): Gas molecules have high entropy due to their freedom of movement.
- N2O(g): Similar to N2(g), gases have high entropy.

Step 3: Compare the states: Solids generally have lower entropy than gases. Among the solids, the more ordered the structure, the lower the entropy.

Step 4: Conclusion: Diamond has the most ordered structure, therefore the least entropy.

Why other options are wrong:
- Option A: Graphite has a layered structure with some disorder, making its entropy higher than diamond.
- Option C: N2(g) is a gas, and gases have significantly higher entropy than solids like diamond.
- Option D: N2O(g) is also a gas, and thus has higher entropy than diamond.

Key Concepts:
- Entropy
- States of Matter
- Molecular Structure

Memory Aids:
- Solids < Liquids < Gases (Entropy order)
- More ordered structure = less entropy

Quick Tips:
- Remember that gases have much higher entropy than solids or liquids.
- For solids, consider the crystal structure; more ordered structures have lower entropy.

55 Easy

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Understanding +4 | -1

Change in entropy is negative for:

View answer and explanation

Correct option: D

Fe (1 mol,400 K)⟶Fe(1 mol,300 K)

Option D (Fe (1 mol,400 K)⟶Fe(1 mol,300 K)) is correct because the entropy decreases as the temperature decreases, leading to a more ordered state.

Solution approach

Correct Answer + Core Reason: Fe (1 mol,400 K)⟶Fe(1 mol,300 K). Entropy decreases as temperature decreases because the system becomes more ordered.

Step-by-Step Solution:

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

Step 2: As temperature decreases, the particles in the system have less kinetic energy and move less, leading to a more ordered state.

Step 3: Therefore, the entropy decreases ( $\Delta S < 0$ ).

Why other options are wrong:
- Option A: Bromine (l)⟶ Bromine (g) is incorrect because entropy increases as a liquid transitions to a gas (more disorder).
- Option B: $\mathrm{C}\left( s \right) + {\mathrm{H}_2}\mathrm{O}\left( \mathrm{g} \right) \to \mathrm{CO}\left( \mathrm{g} \right) + {\mathrm{H}_2}\left( \mathrm{g} \right)$ is incorrect because the number of gas molecules increases, leading to an increase in entropy.
- Option C: ${\mathrm{N}_2}\left( {\mathrm{g},10\mathrm{atm}} \right) \to {\mathrm{N}_2}\left( {\mathrm{g},1\mathrm{atm}} \right)$ is incorrect because entropy increases as pressure decreases (more volume for the gas to occupy).

Key Concepts:
- Entropy
- Temperature
- Phase Transitions

Memory Aids:
- Higher temperature = Higher entropy.
- More gas molecules = Higher entropy.

Quick Tips:
- Consider the phase, temperature, and number of particles when determining the sign of the entropy change.

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Application +4 | -1

If the internal energy of a system decreases by 125 J and
at the same time it absorbs 54 J of heat then

View answer and explanation

Correct option: A

work is done by the system equal to -179 J

Work is done by the system equal to -179 J because, according to the first law of thermodynamics,

\Delta U = q + w

, so

w = \Delta U - q = -125 J - 54 J = -179 J

.

Solution approach

Correct Answer + Core Reason: work is done by the system equal to -179 J. Using the first law of thermodynamics.

Step-by-Step Solution:

Step 1: State the first law of thermodynamics: $\Delta U = q + w$ , where $\Delta U$ is the change in internal energy, $q$ is the heat absorbed by the system, and $w$ is the work done on the system.

Step 2: Identify the given information: $\Delta U = -125 \mathrm{~J}$ (internal energy decreases) and $q = +54 \mathrm{~J}$ (heat is absorbed).

Step 3: Rearrange the equation to solve for $w$ : $w = \Delta U - q$ .

Step 4: Substitute the values: $w = -125 \mathrm{~J} - 54 \mathrm{~J} = -179 \mathrm{~J}$ .

Step 5: Since $w$ is negative, work is done *by* the system, and the magnitude is 179 J.

Why other options are wrong:
- Option B: work is done by the system equal to -71J is incorrect. This would be the result if the heat was incorrectly subtracted from the internal energy change.
- Option C: work is done on the system equal to 179 J is incorrect. The work is done *by* the system, not on it.
- Option D: work is done on the system equal to + 71J is incorrect. The work is done *by* the system, and the calculation is wrong.

Key Concepts:
- First Law of Thermodynamics
- Internal Energy
- Heat
- Work

Memory Aids:
- $\Delta U = q + w$ (sign conventions are important!)
- Work done *by* the system is negative.
- Work done *on* the system is positive.
- Heat absorbed *by* the system is positive.
- Heat released *by* the system is negative.

Quick Tips:
- Pay close attention to the sign conventions for heat and work.

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Understanding +4 | -1

A gaseous system changes from state

A\left( {{P_1},{V_1},{T_1}} \right)\,\,to\,\,B\left( {{P_2},{V_2},{T_2}} \right),\, B \,to\,\,C\;\;({P_3},{V_3},{T_3})

and finally from C to A. The whole process may be called:

View answer and explanation

Correct option: B

Cyclic process

The correct answer is B (Cyclic process) because the system returns to its initial state after a series of changes.

Solution approach

Correct Answer + Core Reason: Cyclic process. The system returns to its initial state after a series of changes.

Step-by-Step Solution:

Step 1: Define a cyclic process: A cyclic process is a thermodynamic process in which the system returns to its initial state after undergoing a series of changes.

Step 2: Analyze the given process: The gaseous system starts at state A, goes to B, then C, and finally returns to A.

Step 3: Conclude: Since the system returns to its initial state, the process is cyclic.

Why other options are wrong:
- Option A: Reversible process: A reversible process is one that can be reversed without leaving any change in the surroundings. This information is not provided in the question.
- Option C: Isobaric process: An isobaric process occurs at constant pressure. The problem states that the pressures change, so it's not isobaric.
- Option D: Spontaneous process: A spontaneous process occurs without external intervention. This information is not provided in the question.

Key Concepts:
- Cyclic process
- Reversible process
- Isobaric process
- Spontaneous process

Memory Aids:
- Cyclic = Returns to start

Quick Tips:
- Focus on the definition of each type of thermodynamic process.

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Remembering +4 | -1

For a spontaneous chemical change the Gibbs energy change is:

View answer and explanation

Correct option: B

Negative

Option B (Negative) is correct because a spontaneous chemical change is characterized by a decrease in Gibbs free energy, hence a negative value.

Solution approach

Correct Answer + Core Reason: Negative. Spontaneous processes occur with a decrease in Gibbs Free Energy.

Step-by-Step Solution:

Step 1: Define Spontaneous Change: A spontaneous change occurs without external intervention.

Step 2: Gibbs Free Energy Criterion: For a spontaneous process at constant temperature and pressure, the change in Gibbs Free Energy ( $\Delta G$ ) must be negative.

Step 3: Mathematical Representation: $\Delta G < 0$

Step 4: Conclusion: Therefore, for a spontaneous chemical change, the Gibbs energy change is negative.

Why other options are wrong:
- Option A: Positive: A positive Gibbs energy change indicates a non-spontaneous process.
- Option C: Zero: A Gibbs energy change of zero indicates a reaction at equilibrium.
- Option D: Depends whether the reaction is exothermic or endothermic: While enthalpy and entropy contribute to Gibbs energy, spontaneity is determined by the sign of $\Delta G$ , not solely by whether the reaction is exothermic or endothermic.

Key Concepts:
- Gibbs Free Energy (G)
- Spontaneity
- Thermodynamics

Memory Aids:
- "Gotta Go Negative" (Spontaneous reactions have a negative $\Delta G$ )

Quick Tips:
- Remember the Gibbs Free Energy equation: $\Delta G = \Delta H - T\Delta S$ . A negative $\Delta G$ indicates spontaneity.

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Understanding +4 | -1

When a solid melts, there is:

View answer and explanation

Correct option: A

An increase in enthalpy

Option A (An increase in enthalpy) is correct because melting is an endothermic process, requiring heat input and increasing the enthalpy of the substance.

Solution approach

Correct Answer + Core Reason: An increase in enthalpy. Melting is an endothermic process that requires energy input, leading to an increase in enthalpy.

Step-by-Step Solution:

Step 1: Define melting.

Step 2: Melting is the process of a solid changing to a liquid.

Step 3: Melting requires energy input to overcome the intermolecular forces holding the solid together. This energy input increases the enthalpy of the substance.

Why other options are wrong:
- Option B: There is a change in enthalpy during melting; it increases.
- Option C: Enthalpy increases, not decreases, during melting.
- Option D: Internal energy also increases during melting because energy is absorbed.

Key Concepts:
- Enthalpy
- Melting
- Endothermic process

Memory Aids:
- Melting requires heat, so enthalpy increases.

Quick Tips:
- Remember that melting is an endothermic process.

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Analyzing +4 | -1

2 {MnO}_4^{-}+16 {H}^{+}+10 {Cl}^{-} \rightarrow 2 {Mn}^{2+}+5 {Cl}_2(\mathrm{g})+8 {H}_2 {O}

The above reaction is endothermic and hence, the actual temperature of the reaction vessel (isolated from the surrounding) may be different from that expected. Given that the initial temperature of the reaction vessel was used in the calculations, how would this affect the predicted value of moles of

{Cl}_2(n)

according to the equation

n=P V/ R T?

View answer and explanation

Correct option: B

It would be less than the actual value

It would be less than the actual value. The reaction is endothermic, so it absorbs heat from the surroundings (the reaction vessel). The temperature decreases, and since n = PV/RT, a lower temperature leads to a lower calculated value of n.

Solution approach

Correct Answer + Core Reason: It would be less than the actual value. Since the reaction is endothermic, it cools the reaction vessel, leading to a lower calculated value of n.

Step-by-Step Solution:

Step 1: Understand the reaction:\nThe given reaction is endothermic, meaning it absorbs heat from its surroundings.

Step 2: Consider the effect on temperature:\nSince the reaction is occurring in an isolated vessel, the heat absorbed by the reaction will cause the temperature of the vessel to decrease.

Step 3: Analyze the ideal gas equation:\nThe number of moles of $Cl_2$ is calculated using the ideal gas equation: $n = \frac{PV}{RT}$

Step 4: Determine the effect of temperature on the calculated value of n:\nIf the actual temperature of the reaction vessel is lower than the initial temperature used in the calculation, the calculated value of n will be lower than the actual value.

Step 5: Final Answer: It would be less than the actual value

Why other options are wrong:
- Option A: It would be greater than the actual value: This is incorrect because a lower temperature would lead to a lower calculated value of n.
- Option C: It would be the same as the actual value: This is incorrect because the temperature change due to the endothermic reaction affects the calculated value of n.
- Option D: This cannot be determined from the information given: This is incorrect because the information about the reaction being endothermic and the ideal gas equation allows us to determine the effect on the calculated value of n.

Key Concepts:
- Endothermic Reactions
- Ideal Gas Law
- Temperature Dependence of Gas Properties

Memory Aids:
- Endothermic reactions absorb heat, cooling the surroundings.
- Ideal Gas Law: PV = nRT

Quick Tips:
- Remember that endothermic reactions decrease the temperature of the surroundings.
- Consider how changes in temperature affect the calculated values of gas properties using the ideal gas law.

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Thermodynamics practice · Easy

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