THERMODYNAMICS MCQs for NEET UG CHEMISTRY — Easy, Page 7

181 Easy

Understanding +4 | -1

For the isothermal expansion of an ideal gas

View answer and explanation

Correct option: D

E and H are unaltered

The correct answer is D (E and H are unaltered) because for an ideal gas in an isothermal process, internal energy (E) and enthalpy (H) depend only on temperature, which remains constant.

Solution approach

Correct Answer + Core Reason: E and H are unaltered. For an ideal gas, internal energy and enthalpy depend only on temperature in an isothermal process.

Step-by-Step Solution:

Step 1: Define isothermal process: An isothermal process is one that occurs at constant temperature.

Step 2: Recall the dependence of internal energy (E) and enthalpy (H) on temperature for an ideal gas: For an ideal gas, internal energy (E) and enthalpy (H) are functions of temperature only.

Step 3: Apply to the isothermal process: Since the temperature is constant in an isothermal process, the internal energy (E) and enthalpy (H) remain constant.

Why other options are wrong:
- Option A: E and H increases: This is incorrect because temperature is constant, so E and H do not increase.
- Option B: E increases but H decreases: This is incorrect because both E and H depend only on temperature and would remain constant.
- Option C: H increases but E decreases: This is incorrect because both E and H depend only on temperature and would remain constant.

Key Concepts:
- Isothermal process
- Internal energy (E)
- Enthalpy (H)
- Ideal gas

Memory Aids:
- Isothermal = Constant temperature
- Ideal gas: E and H depend only on T

Quick Tips:
- Remember that for ideal gases, internal energy and enthalpy are functions of temperature only.

182 Easy

182 of 196Question permalink

Application +4 | -1

The change in internal energy if

20\,\,\mathrm{J}

of heat is given to the system at constant pressure and

10\,\,\mathrm{J}

of work is done by the system is

View answer and explanation

Correct option: C

10\,\,\mathrm{J}\,\,

Option C (

10\,\mathrm{J}\,\,

) is correct because the change in internal energy is calculated as the heat added to the system minus the work done by the system.

Solution approach

Correct Answer + Core Reason: $10\,\mathrm{J}\,\,$ . The change in internal energy is given by the first law of thermodynamics: $\Delta U = q - w$ .

Step-by-Step Solution:

Step 1: Identify the given information:\nHeat given to the system, $q = 20\,\mathrm{J}$ \nWork done by the system, $w = 10\,\mathrm{J}$

Step 2: Write the formula for the change in internal energy:
$\Delta U = q - w$

Step 3: Substitute the given values:
$\Delta U = 20\,\mathrm{J} - 10\,\mathrm{J}$

Step 4: Calculate the change in internal energy:
$\Delta U = 10\,\mathrm{J}$

Step 5: Final Answer:
$\Delta U = 10\,\mathrm{J}$

Why other options are wrong:
- Option A: This value has the wrong sign and is obtained by adding the heat and work with incorrect signs.
- Option B: This value has the wrong sign and is obtained by subtracting the heat from the work done.
- Option D: This value is obtained by adding the heat and work, which is incorrect according to the first law of thermodynamics.

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

Memory Aids:
- $\Delta U = q - w$ (Internal energy change equals heat added minus work done)

Quick Tips:
- Remember the sign conventions: Heat added to the system is positive, and work done by the system is positive.

183 Easy

183 of 196Question permalink

Understanding +4 | -1

Least random state of water is:

View answer and explanation

Correct option: A

Ice

Option A (Ice) is correct because ice has a highly ordered crystalline structure, making it the least random state of water compared to liquid water and steam.

Solution approach

Correct Answer + Core Reason: Ice. Solid ice has the most ordered structure and therefore the least randomness.

Step-by-Step Solution:

Step 1: Recall the definition of entropy and its relation to randomness.

Step 2: Entropy is a measure of the disorder or randomness of a system. Higher entropy means more randomness.

Step 3: Compare the states of water: ice (solid), liquid water, and steam (gas).

Step 4: Ice has a highly ordered crystalline structure, which means it has the least randomness and therefore the lowest entropy.

Step 5: Liquid water has more disorder than ice, and steam has the most disorder.

Why other options are wrong:
- Option B: Liquid water has more randomness than ice.
- Option C: Steam has the most randomness among the three states.
- Option D: The three states have different levels of randomness.

Key Concepts:
- Entropy
- States of matter

Memory Aids:
- Solid < Liquid < Gas (in terms of entropy)

Quick Tips:
- Remember that solids generally have lower entropy than liquids and gases.

184 Easy

184 of 196Question permalink

Analyzing +4 | -1

If a gas has

2 \mathrm{~atm}

and

5 \mathrm{~atm}

pressure at

30^{\circ} \mathrm{C}

and

27^{\circ} \mathrm{C}

respectively. Then it will

View answer and explanation

Correct option: B

warm on expansion

Option B (warm on expansion) is correct because the gas has a higher pressure than expected for an ideal gas at the given temperature, indicating attractive forces dominate, leading to warming upon expansion (Joule-Thomson effect).

Solution approach

Correct Answer + Core Reason: warm on expansion. The gas warms upon expansion because it deviates from ideal behavior, and attractive forces dominate.

Step-by-Step Solution:

Step 1: Analyze the given information.\nWe have a gas with two different states:\nState 1: $P_1 = 2 \text{ atm}$ , $T_1 = 30^{\circ} \text{C} = 303 \text{ K}$ \nState 2: $P_2 = 5 \text{ atm}$ , $T_2 = 27^{\circ} \text{C} = 300 \text{ K}$

Step 2: Consider the ideal gas behavior.\nFor an ideal gas, if the pressure increases, the temperature should also increase (or volume decrease) to maintain $PV = nRT$ . Let's compare the ratio $\frac{P}{T}$ for both states.
$\frac{P_1}{T_1} = \frac{2}{303} = 0.0066$
$\frac{P_2}{T_2} = \frac{5}{300} = 0.0167$ \nSince $\frac{P_2}{T_2} > \frac{P_1}{T_1}$ , the gas is not behaving ideally. The pressure is higher than expected for an ideal gas at the given temperature.

Step 3: Consider the Joule-Thomson effect.\nThe Joule-Thomson effect describes the temperature change of a real gas when it expands adiabatically through a valve or porous plug. The temperature change depends on the Joule-Thomson coefficient, $\mu_{JT} = \left( \frac{\partial T}{\partial P} \right)_H$ .\nIf $\mu_{JT} > 0$ , the gas cools upon expansion.\nIf $\mu_{JT} < 0$ , the gas warms upon expansion.\nIf $\mu_{JT} = 0$ , the gas temperature remains constant upon expansion (ideal gas behavior).

Step 4: Determine the sign of $\mu_{JT}$ .\nSince the pressure is higher than expected, it suggests that attractive forces between the gas molecules are significant. When the gas expands, these attractive forces have to be overcome, which requires energy. This energy comes from the internal energy of the gas, causing the temperature to decrease. However, the question states that the gas warms upon expansion. This implies that the gas is already in a state where the repulsive forces are dominant, and the expansion allows the gas to release energy, thus warming it.\nAlternatively, if the gas has strong intermolecular attractions, the internal pressure is high. When the gas expands, the molecules move further apart, and the potential energy associated with the intermolecular attractions increases. This increase in potential energy comes at the expense of kinetic energy, causing the temperature to drop. However, in this case, the gas warms upon expansion, suggesting that the gas is already in a state where repulsive forces are dominant, and the expansion allows the gas to release energy, thus warming it.

Step 5: Conclusion.\nSince the gas warms upon expansion, it indicates that repulsive forces are dominant, and the Joule-Thomson coefficient is negative. Therefore, the gas will warm on expansion.

Why other options are wrong:
- Option A: cool on expansion: This would be true if the attractive forces were dominant and the Joule-Thomson coefficient was positive.
- Option C: no change on expansion: This would be true for an ideal gas where the Joule-Thomson coefficient is zero.
- Option D: None of the above: This is incorrect because one of the options is correct.

Key Concepts:
- Ideal gas law
- Joule-Thomson effect
- Intermolecular forces
- Joule-Thomson coefficient

Memory Aids:
- Remember the Joule-Thomson coefficient and its relation to cooling or warming upon expansion.
- Attractive forces lead to cooling, repulsive forces lead to warming.

Quick Tips:
- Analyze the deviation from ideal gas behavior to determine the dominant intermolecular forces.
- Consider the Joule-Thomson effect when dealing with real gas expansion.

185 Easy

185 of 196Question permalink

Remembering +4 | -1

The enthalpy change for the process, C(s) ⟶ C(g) is known as enthalpy of :

View answer and explanation

Correct option: D

Sublimation

Option D (Sublimation) is correct because sublimation is the process of a solid changing directly into a gas.

Solution approach

Correct Answer + Core Reason: Sublimation. Sublimation is the process of a solid directly converting to a gas.

Step-by-Step Solution:

Step 1: Define the process: The process is C(s) ⟶ C(g), which means solid carbon is converting directly into gaseous carbon.

Step 2: Define the terms:
- Fusion: Solid to liquid
- Vaporization: Liquid to gas
- Combustion: Reaction with oxygen, usually producing heat and light
- Sublimation: Solid to gas

Step 3: Match the process to the definition: Since the process is solid to gas, it is sublimation.

Why other options are wrong:
- Option A: Fusion. Fusion is the process of a solid changing into a liquid.
- Option B: Vaporisation. Vaporization is the process of a liquid changing into a gas.
- Option C: Combustion. Combustion is a chemical process that involves rapid reaction between a substance with an oxidant, usually oxygen, to produce heat and light.

Key Concepts:
- Phase transitions
- Sublimation
- Fusion
- Vaporization
- Combustion

Memory Aids:
- Sublimation: SOLID to GAS (S to G)
- Fusion: SOLID to LIQUID (S to L)
- Vaporization: LIQUID to GAS (L to G)

Quick Tips:
- Remember the definitions of the different phase transitions.

186 Easy

186 of 196Question permalink

Analyzing +4 | -1

Consider the following two reactions :

\begin{aligned} & {X} \rightarrow 2 {Y} ; \Delta {H}_{\text {reaction }}=\Delta {H}_1 \\ & \frac{1}{2} {X} \rightarrow {Z} ; \Delta {H}_{\text {reaction }}=\Delta {H}_2 \end{aligned}

What would be the enthalpy change for the process

{Y} \rightarrow {Z}

?

View answer and explanation

Correct option: A

\Delta {H}_2-\frac{1}{2} \Delta {H}_1

Option A (

\Delta {H}_2-\frac{1}{2} \Delta {H}_1

) is correct because it is obtained by manipulating the given reactions to arrive at the desired reaction.

Solution approach

Correct Answer + Core Reason: $\Delta {H}_2-\frac{1}{2} \Delta {H}_1$ . This is obtained by manipulating the given reactions to get the desired reaction Y -> Z.

Step-by-Step Solution:

Step 1: Write down the given reactions and their enthalpy changes:\nReaction 1: $X \rightarrow 2Y$ ; $\Delta H_{reaction} = \Delta H_1$ \nReaction 2: $\frac{1}{2}X \rightarrow Z$ ; $\Delta H_{reaction} = \Delta H_2$

Step 2: We want to find the enthalpy change for the reaction: $Y \rightarrow Z$

Step 3: Manipulate the given reactions to obtain the desired reaction.\nFirst, reverse Reaction 1 and divide by 2:
$-\frac{1}{2} (X \rightarrow 2Y)$ becomes $Y \rightarrow \frac{1}{2}X$ ; $\Delta H = -\frac{1}{2}\Delta H_1$

Step 4: Add the modified Reaction 1 to Reaction 2:
$Y \rightarrow \frac{1}{2}X$ ; $\Delta H = -\frac{1}{2}\Delta H_1$
$\frac{1}{2}X \rightarrow Z$ ; $\Delta H = \Delta H_2$ \nAdding these gives: $Y \rightarrow Z$ ; $\Delta H = \Delta H_2 - \frac{1}{2}\Delta H_1$

Step 5: Final Answer: $\Delta H_2 - \frac{1}{2}\Delta H_1$

Why other options are wrong:
- Option B: $\Delta {H}_2-\Delta {H}_1$ : This is incorrect because it doesn't account for the stoichiometry of the first reaction.
- Option C: $\Delta {H}_1-\frac{1}{2} \Delta {H}_2$ : This is incorrect because it subtracts half of the second reaction's enthalpy from the first, which is not the correct manipulation.
- Option D: $\frac{1}{2}(\Delta {H}_1+\Delta {H}_2)$ : This is incorrect because it adds the enthalpies and divides by 2, which doesn't follow from manipulating the reactions.

Key Concepts:
- Hess's Law
- Enthalpy Change of Reaction
- Manipulating Chemical Equations

Memory Aids:
- Hess's Law: Enthalpy change is independent of the path taken.

Quick Tips:
- Reverse a reaction, change the sign of $\Delta H$ .
- Multiply a reaction by a factor, multiply $\Delta H$ by the same factor.
- Add reactions, add their $\Delta H$ values.

187 Easy

187 of 196Question permalink

Understanding +4 | -1

An ideal gas undergoing expansion in vacuum shows:

View answer and explanation

Correct option: D

All of these

Option D is correct because in a vacuum expansion, no external pressure opposes the expansion, so no work is done (W=0), no heat is exchanged (q=0), and the internal energy remains constant (∆U=0).

Solution approach

Correct Answer + Core Reason: All of these. In vacuum expansion, no work is done, no heat is exchanged, and internal energy remains constant.

Step-by-Step Solution:

Step 1: Define expansion in vacuum. Expansion in vacuum means the external pressure is zero ( $P_{ext} = 0$ ).

Step 2: Calculate work done. Work done during expansion is given by $W = -P_{ext} \Delta V$ . Since $P_{ext} = 0$ , $W = 0$ .

Step 3: Apply the first law of thermodynamics. The first law of thermodynamics states that $\Delta U = q + W$ .

Step 4: Determine heat exchange. Since the gas is expanding into a vacuum, there is no external force to overcome, and the process is adiabatic (no heat exchange), so $q = 0$ .

Step 5: Calculate change in internal energy. Substituting $q = 0$ and $W = 0$ into the first law, we get $\Delta U = 0 + 0 = 0$ .

Why other options are wrong:
- Option A: Internal energy remains constant, so this option is partially correct but not fully.
- Option B: No work is done, so this option is partially correct but not fully.
- Option C: No heat is exchanged, so this option is partially correct but not fully.

Key Concepts:
- Expansion in Vacuum
- Work Done
- First Law of Thermodynamics
- Internal Energy

Memory Aids:
- Vacuum: No external pressure
- No external pressure: No work done

Quick Tips:
- Remember that expansion in vacuum implies zero external pressure and therefore zero work done.

188 Easy

188 of 196Question permalink

Understanding +4 | -1

An example of closed system is:

View answer and explanation

Correct option: C

Some amount of hot water enclosed in a closed container which is not insulated

Option C (Some amount of hot water enclosed in a closed container which is not insulated) is correct because a closed system allows energy exchange but not mass exchange.

Solution approach

Correct Answer + Core Reason: Some amount of hot water enclosed in a closed container which is not insulated. A closed system allows energy exchange but not mass exchange.

Step-by-Step Solution:

Step 1: Define closed system. A closed system is a system that can exchange energy (heat, work) with its surroundings but cannot exchange matter.

Step 2: Analyze the given options.
- Option A: Open beaker allows both mass and energy exchange (open system).
- Option B: Closed and insulated beaker prevents both mass and energy exchange (isolated system).
- Option C: Closed container allows energy exchange but not mass exchange (closed system).

Step 3: Conclusion: Therefore, the example of a closed system is some amount of hot water enclosed in a closed container which is not insulated.

Why other options are wrong:
- Option A: Hot water present in an open beaker: This is an open system because both mass and energy can be exchanged with the surroundings.
- Option B: Some amount of water present in equilibrium with its vapour in a closed and insulated beaker: This is an isolated system because neither mass nor energy can be exchanged with the surroundings.
- Option D: None of the above: Option C is a correct example of a closed system.

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

Memory Aids:
- Closed system: Closed container, energy exchange allowed.

Quick Tips:
- Understand the differences between open, closed, and isolated systems.

189 Easy

189 of 196Question permalink

Understanding +4 | -1

Which of the following are not state functions?
(I) q+w (II) q
(III) w (IV) H-TS

View answer and explanation

Correct option: C

(II) and (III)

Option C ((II) and (III)) is correct because q (heat) and w (work) are path-dependent functions, while q+w (which equals Δ U) and H-TS (which equals Gibbs Free Energy) are state functions.

Solution approach

Correct Answer + Core Reason: (II) and (III). Heat (q) and work (w) are path functions, not state functions.

Step-by-Step Solution:

Step 1: Define a state function as a property that depends only on the initial and final states, not the path taken.

Step 2: Recall that heat (q) and work (w) are path-dependent, meaning their values depend on the process by which the system changes.

Step 3: Recognize that q + w = Δ U (change in internal energy), which is a state function. Also, H - TS = G (Gibbs free energy), which is a state function.

Why other options are wrong:
- Option A: (II),(III) and (IV) is incorrect because (IV) H-TS is a state function.
- Option B: (I),(II) and (III) is incorrect because (I) q+w is a state function.
- Option D: (I) and (IV) is incorrect because (I) q+w and (IV) H-TS are state functions.

Key Concepts:
- State function
- Path function
- Internal energy (U)
- Gibbs free energy (G)

Memory Aids:
- State functions: "I Have Visited Every Place" (Internal energy, Enthalpy, Volume, Entropy, Pressure)
- Heat and work are path functions.

Quick Tips:
- Remember that Δ U is a state function, even though q and w are not.

190 Easy

190 of 196Question permalink

Remembering +4 | -1

Which of the following is a path function?

View answer and explanation

Correct option: D

Heat

Option D (Heat) is correct because heat is a form of energy transfer that depends on the path taken, not just the initial and final states.

Solution approach

Correct Answer + Core Reason: Heat. Heat is a path function because its value depends on the process or path taken to achieve a particular change of state.

Step-by-Step Solution:

Step 1: Define state and path functions.\nA state function is a property of a system that depends only on the current state of the system, not on the path taken to reach that state. Examples include internal energy, enthalpy, and entropy.\nA path function is a property of a system that depends on the path taken to reach a particular state. Examples include heat and work.

Step 2: Analyze each option.\nInternal energy (U) is a state function.\nEntropy (S) is a state function.\nEnthalpy (H) is a state function.\nHeat (q) is a path function.

Why other options are wrong:
- Option A: Internal energy. Internal energy is a state function, not a path function.
- Option B: Entropy. Entropy is a state function, not a path function.
- Option C: Enthalpy. Enthalpy is a state function, not a path function.

Key Concepts:
- State functions
- Path functions
- Thermodynamics

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

Quick Tips:
- Remember that heat and work are path functions, while internal energy, enthalpy, and entropy are state functions.

191 Easy

191 of 196Question permalink

Remembering +4 | -1

\mathrm{C}_{\mathrm{p}} / \mathrm{C}_{\mathrm{v}}

for noble gases is :

View answer and explanation

Correct option: A

1.66

Option A (1.66) is correct because noble gases are monoatomic, and for monoatomic gases, the ratio of specific heats,

\gamma = C_p/C_v = 5/3 \approx 1.66

.

Solution approach

Correct Answer + Core Reason: 1.66. For noble gases (monoatomic), the ratio of specific heats $\mathrm{C}_{\mathrm{p}} / \mathrm{C}_{\mathrm{v}}$ is 1.66.

Step-by-Step Solution:

Step 1: Define $\mathrm{C}_{\mathrm{p}}$ and $\mathrm{C}_{\mathrm{v}}$ :
$\mathrm{C}_{\mathrm{p}}$ is the specific heat at constant pressure.
$\mathrm{C}_{\mathrm{v}}$ is the specific heat at constant volume.

Step 2: Define the ratio $\gamma = \frac{\mathrm{C}_{\mathrm{p}}}{\mathrm{C}_{\mathrm{v}}}$ :\nFor a monoatomic gas (like noble gases), the degrees of freedom are 3 (translational only).
$\mathrm{C}_{\mathrm{v}} = \frac{f}{2}R = \frac{3}{2}R$ , where $f$ is the degrees of freedom and $R$ is the ideal gas constant.
$\mathrm{C}_{\mathrm{p}} = \mathrm{C}_{\mathrm{v}} + R = \frac{3}{2}R + R = \frac{5}{2}R$

Step 3: Calculate the ratio:
$\gamma = \frac{\mathrm{C}_{\mathrm{p}}}{\mathrm{C}_{\mathrm{v}}} = \frac{\frac{5}{2}R}{\frac{3}{2}R} = \frac{5}{3} \approx 1.66$

Step 4: Conclusion: The ratio $\mathrm{C}_{\mathrm{p}} / \mathrm{C}_{\mathrm{v}}$ for noble gases is approximately 1.66.

Why other options are wrong:
- Option B: 1.43: This value is closer to the ratio for diatomic gases (like $N_2$ or $O_2$ ), where $\gamma = 1.4$ .
- Option C: 1.8: This value is not a standard ratio for any common type of gas.
- Option D: 1.33: This value is also not a standard ratio for any common type of gas. It is closer to 4/3, which is not applicable here.

Key Concepts:
- Specific Heat at Constant Pressure ( $\mathrm{C}_{\mathrm{p}}$ )
- Specific Heat at Constant Volume ( $\mathrm{C}_{\mathrm{v}}$ )
- Degrees of Freedom
- Ratio of Specific Heats ( $\gamma$ )

Memory Aids:
- Monoatomic gases have 3 degrees of freedom.
- Diatomic gases have 5 degrees of freedom (at moderate temperatures).

Quick Tips:
- Remember the formula for $\mathrm{C}_{\mathrm{v}}$ and $\mathrm{C}_{\mathrm{p}}$ in terms of degrees of freedom.
- Know the typical values of $\gamma$ for monoatomic and diatomic gases.

192 Easy

192 of 196Question permalink

Application +4 | -1

{C_2}{H_6}\left( g \right) + \frac{7}{2}{O_2}\left( g \right) \to 2C{O_2}\left( g \right) + 3{H_2}O\left( l \right);\,\Delta H = - 1560\,KJ

{C_2}{H_2}\left( g \right) + \frac{5}{2}{O_2}\left( g \right) \to 2C{O_2}\left( g \right) + {H_2}O\left( l \right);\,\Delta H = - 1310\,KJ

{H_2}\left( g \right) + \frac{1}{2}{O_2}\left( g \right) \to {H_2}O\left( l \right);\Delta H = - 290\,KJ

Enthalpy of hydrogenation of ethyne

\left( {{C_2}{H_2}} \right)

is

View answer and explanation

Correct option: C

- 330 KJ

Option C (– 330 KJ) is correct because it is the enthalpy change for the hydrogenation of ethyne to ethane, calculated using Hess's law from the given combustion enthalpies.

Solution approach

Correct Answer + Core Reason: – 330 KJ. The enthalpy of hydrogenation is calculated using Hess's Law.

Step-by-Step Solution:

Step 1: Identify the target reaction: ${C_2}{H_2}(g) + 2{H_2}(g) \to {C_2}{H_6}(g)$

Step 2: Write down the given reactions and their enthalpy changes:\nReaction 1: ${C_2}{H_6}(g) + \frac{7}{2}{O_2}(g) \to 2C{O_2}(g) + 3{H_2}O(l); \Delta H_1 = -1560 kJ$ \nReaction 2: ${C_2}{H_2}(g) + \frac{5}{2}{O_2}(g) \to 2C{O_2}(g) + {H_2}O(l); \Delta H_2 = -1310 kJ$ \nReaction 3: ${H_2}(g) + \frac{1}{2}{O_2}(g) \to {H_2}O(l); \Delta H_3 = -290 kJ$

Step 3: Manipulate the given reactions to obtain the target reaction. Reverse Reaction 1 and multiply Reaction 3 by 2:\nReversed Reaction 1: $2C{O_2}(g) + 3{H_2}O(l) \to {C_2}{H_6}(g) + \frac{7}{2}{O_2}(g); -\Delta H_1 = 1560 kJ$
2 * Reaction 3: $2{H_2}(g) + {O_2}(g) \to 2{H_2}O(l); 2\Delta H_3 = -580 kJ$

Step 4: Add the reversed Reaction 1, Reaction 2, and 2 * Reaction 3:
$2C{O_2}(g) + 3{H_2}O(l) + {C_2}{H_2}(g) + \frac{5}{2}{O_2}(g) + 2{H_2}(g) + {O_2}(g) \to {C_2}{H_6}(g) + \frac{7}{2}{O_2}(g) + 2C{O_2}(g) + {H_2}O(l) + 2{H_2}O(l)$

Step 5: Simplify the equation:
${C_2}{H_2}(g) + 2{H_2}(g) \to {C_2}{H_6}(g)$

Step 6: Calculate the enthalpy change for the target reaction:
$\Delta H = -\Delta H_1 + \Delta H_2 + 2\Delta H_3 = 1560 kJ - 1310 kJ - 580 kJ = -330 kJ$

Why other options are wrong:
- Option A (– 40 KJ): This value is incorrect as it doesn't properly account for the stoichiometry and enthalpy changes of all the reactions involved in the hydrogenation process.
- Option B (– 620 KJ): This value is incorrect, likely arising from an error in applying Hess's Law or incorrect manipulation of the given enthalpy values.
- Option D (– 410 KJ): This value is incorrect, possibly due to a mistake in the arithmetic when combining the enthalpy changes.

Key Concepts:
- Hess's Law
- Enthalpy of Reaction
- Hydrogenation

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

Quick Tips:
- Always write out the target reaction and manipulate the given reactions to match it.
- Pay attention to the signs of enthalpy changes when reversing reactions.

193 Easy

193 of 196Question permalink

Application +4 | -1

When a

60 \, \mathrm{W}

electric heater is immersed in a gas for

100 \, \text{s}

in a constant volume container with adiabatic walls, the temperature of the gas rises by

5^{\circ} \mathrm{C}

. The heat capacity of the given gas is ______

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

(Nearest integer)

View answer and explanation

Correct option: A

1200

Option A (1200) is correct because the heat capacity is calculated by dividing the energy supplied by the temperature rise.

Solution approach

Correct Answer + Core Reason: 1200. The heat capacity is calculated by dividing the energy supplied by the temperature rise.

Step-by-Step Solution:

Step 1: Calculate the energy supplied by the heater: Energy (Q) = Power (P) x Time (t)

Step 2: Substitute the given values: P = 60 W, t = 100 s\nQ = 60 W x 100 s = 6000 J

Step 3: Calculate the heat capacity (C): C = Q / \u0394T

Step 4: Substitute the values: Q = 6000 J, \u0394T = 5 °C\nC = 6000 J / 5 °C = 1200 J/K

Why other options are wrong:
- Option B: 0 is incorrect. It implies no heat capacity, which is not possible for a gas.
- Option C: 0 is incorrect. It implies no heat capacity, which is not possible for a gas.
- Option D: 0 is incorrect. It implies no heat capacity, which is not possible for a gas.

Key Concepts:
- Heat capacity
- Power and energy relationship

Memory Aids:
- Q = mc\u0394T, C = Q/\u0394T

Quick Tips:
- Ensure units are consistent before calculations.

194 Easy

194 of 196Question permalink

Analyzing +4 | -1

A: Work done in an irreversible isothermal process at constant volume is zero.
R: Work is assigned negative sign during expansion and is assigned positive sign during compression.

View answer and explanation

Correct option: B

If both A and R are correct, but R is not the correct explanation of A

Option B is correct. Assertion is correct because at constant volume, no work is done. Reason is also correct about the sign convention for work, but it doesn't explain the assertion.

Solution approach

Correct Answer + Core Reason: If both A and R are correct, but R is not the correct explanation of A. Work done at constant volume is zero, and the sign convention for work is as stated, but they are unrelated.

Step-by-Step Solution:

Step 1: Analyze Assertion (A): Work done in an irreversible isothermal process at constant volume is zero. This is correct because work done is given by $W = -P\Delta V$ , and if the volume is constant ( $\Delta V = 0$ ), then $W = 0$ .

Step 2: Analyze Reason (R): Work is assigned a negative sign during expansion and a positive sign during compression. This is also correct based on the standard sign convention in thermodynamics.

Step 3: Determine the relationship: While both A and R are correct statements, R does not explain A. The reason A is true is because of constant volume, not the sign convention of work.

Why other options are wrong:
- Option A: If both A and R are correct and R is the correct explanation of A: R does not explain A.
- Option C: If A is correct but R is incorrect: R is correct.
- Option D: If both A and R are incorrect: Both A and R are correct.

Key Concepts:
- Work done in thermodynamics
- Constant volume process
- Sign convention for work

Memory Aids:
- No volume change, no work done.

Quick Tips:
- Remember the formula for work: $W = -P\Delta V$ . If $\Delta V = 0$ , then $W = 0$ .

195 Easy

195 of 196Question permalink

Analyzing +4 | -1

Given below are two statements one is labelled as Assertion (A) and the other is labelled as Reason (R)

Assertion (A): Heat of neutralisation of

HF

(aq.) with

{NaOH}

(aq.) is greater than

13.7 \mathrm{~Kcal}

.

Reason (R): Generally some heat is consumed in the ionisation of weak acid.
On the basis of above statements choose the most appropriate option from the following options:

View answer and explanation

Correct option: A

Both (A) and (R) are correct but (R) is not the correct explanation of (A)

Option A is correct. The heat of neutralization of HF with NaOH is greater than 13.7 kcal because of the additional energy released due to the formation of hydrogen bonds. The reason is true, but it explains the lower heat of neutralization for weak acids, not the higher heat for HF.

Solution approach

Correct Answer + Core Reason: Both (A) and (R) are correct but (R) is not the correct explanation of (A). HF's high heat of neutralization is due to hydrogen bond formation, not ionization of a weak acid.

Step-by-Step Solution:

Step 1: Analyze Assertion (A): Heat of neutralization of HF(aq.) with NaOH(aq.) is greater than 13.7 kcal (which is approximately 57.3 kJ). This is true because HF, upon neutralization, forms strong hydrogen bonds, releasing additional energy.

Step 2: Analyze Reason (R): Generally, some heat is consumed in the ionization of a weak acid. This is also true. Weak acids are not fully ionized, so energy is required to ionize them before neutralization.

Step 3: Determine the relationship: While both A and R are correct, R does not explain A. The reason for the higher heat of neutralization of HF is the formation of hydrogen bonds, not the ionization of a weak acid.

Why other options are wrong:
- Option B: (A) is correct but (R) is not correct: R is correct.
- Option C: (A) is not correct but (R) is correct: A is correct.
- Option D: Both (A) and (R) are correct and (R) is the correct explanation of (A): R does not explain A.

Key Concepts:
- Heat of Neutralization
- Hydrogen Bonding
- Weak Acids

Memory Aids:
- HF forms strong hydrogen bonds.

Quick Tips:
- Consider hydrogen bonding when dealing with HF.

196 Easy

196 of 196Question permalink

Remembering +4 | -1

The enthalpies of elements under the following conditions are assumed to be

View answer and explanation

Correct option: A

Zero at 298 K and 1 atm

Option A (Zero at 298 K and 1 atm) is correct because the standard enthalpy of formation of an element in its standard state is defined as zero at 298 K and 1 atm.

Solution approach

Correct Answer + Core Reason: Zero at 298 K and 1 atm. By convention, the standard enthalpy of formation of an element in its standard state is zero at 298 K (25°C) and 1 atm pressure.

Step-by-Step Solution:

Step 1: Define Standard State: The standard state of an element is its most stable form at 298 K (25°C) and 1 atm pressure.

Step 2: Define Standard Enthalpy of Formation: The standard enthalpy of formation (Δ H_f^°) is the enthalpy change when one mole of a substance is formed from its elements in their standard states.

Step 3: Apply the definition to elements: Since elements in their standard states are already in their most stable form, no energy is required to "form" them from themselves. Therefore, their standard enthalpy of formation is zero.

Why other options are wrong:
- Option B: The enthalpy of elements is not unity in their standard state.
- Option C: The standard state is defined at 298 K, not 273 K.
- Option D: The enthalpy of elements is not unity in their standard state, and the standard state is defined at 298 K, not 273 K.

Key Concepts:
- Standard State
- Standard Enthalpy of Formation (Δ H_f^°)

Memory Aids:
- Elements in their standard state are already "formed", so Δ H_f^° = 0

Quick Tips:
- Remember the definition of standard state (298 K and 1 atm).

Your Progress

Thermodynamics practice · Easy

Explore Other Chapters in CHEMISTRY

Continue Your Practice

Your practice