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Match the column I with column II and mark the appropriate choice. $$\begin{array}{|l|l|l|l|} \hline {\text { Column I }} & & {\text { Column II }} \\ \hline \text { (A) } & \text { State function } & \text { (i) } & \text { At constant pressure } \\ \hline \text { (B) } & \Delta H=q & \text { (ii) } & \text { Specific heat } \\\ \hline \text { (C) } & \Delta U=q & \text { (iii) } & \text { Entropy } \\ \hline \text { (D) } & \text { Intensive property } & \text { (iv) } & \text { At constant volume } \\ \hline \end{array}$$

Short Answer

Expert verified
The appropriate matches are: (A)-(iii), (B)-(i), (C)-(iv), (D)-(ii).

Step by step solution

01

Understanding State Function

State function is a property that is independent of the path used to reach that specific value. Some examples of state functions are enthalpy (H), entropy (S), pressure (P), temperature (T), and volume (V).
02

Identifying \( \Delta H=q \) at Constant Pressure

The equation \( \Delta H=q \) defines the enthalpy change (\( \Delta H \) as the heat transferred (q) at constant pressure. Therefore, \( \Delta H=q \) corresponds to a process occurring at constant pressure.
03

Associating \( \Delta U=q \) with a Process

The change in internal energy (\( \Delta U \) is equal to the heat transferred (q) in a process occurring at constant volume, as the work done is zero in such a scenario when considering the first law of thermodynamics.
04

Understanding Intensive Property

An intensive property is one that does not depend on the amount of substance present. Specific heat is an example of an intensive property because it is the heat capacity per unit mass and does not depend on the total mass of the substance.
05

Matching Each Concept with Its Corresponding Description

Now, match each term from Column I to the appropriate description in Column II based on the explanations given in the previous steps. (A) is related to (iii) because state functions include entropy. (B) should be matched with (i) as it pertains to heat transfer at constant pressure. (C) corresponds with (iv) since it is about change in internal energy at constant volume. And (D) pairs with (ii) because it's referring to an intensive property, specific heat.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

State Function
A state function is a fundamental concept in chemical thermodynamics that describes a property whose value does not depend on the path taken to reach that specific value. Imagine hiking to the top of a mountain; whether you take a steep, short path or a longer, winding trail, your elevation at the summit remains the same. Similarly, in thermodynamics, state functions like enthalpy (H), entropy (S), pressure (P), temperature (T), and volume (V) are solely determined by the current state of the system and are indifferent to how the system arrived there.

Understanding state functions is critical because they help provide a snapshot of a system's condition. State functions are particularly useful for calculations involving energy, as they provide a means to quantify changes within a system without needing to know the specific details of the processes that led to those changes.
Enthalpy Change
The term enthalpy change, represented by \( \Delta H \) in thermodynamics, signifies the total heat content change within a system at constant pressure. It's often correlated with the heat absorbed or released during chemical reactions. Imagine making a cup of tea; the warmth you feel when you touch the cup is akin to the enthalpy change of the water as it gains heat on a stove.

Specifically, when a reaction occurs at constant pressure, the heat transfer (\( q \) can be directly equated to \( \Delta H \). This implies that the enthalpy change provides a convenient measure for the energy change as heat, which is especially important for understanding exothermic and endothermic reactions where heat release or absorption is involved.
Internal Energy
The internal energy of a thermodynamic system, symbolized by \( \Delta U \), encompasses all the energy contained within the system. This includes both potential and kinetic energy of the particles that compose the system. Consider a balloon filled with air; the movement and interactions of the gas molecules inside it contribute to the balloon's internal energy.

In the context of thermodynamics, \( \Delta U \) indicates the change in a system's internal energy. At constant volume, according to the first law of thermodynamics, this change is equal to the heat transferred (\( q \) into or out of the system since no work is performed on or by the system. Therefore, internal energy change is a critical component for understanding energy conservation and distribution within chemical systems.
Intensive Property
An intensive property is a physical quantity of a material that does not depend on the amount of substance present. This contrasts with extensive properties, which do vary with the size or extent of the system. A good analogy for an intensive property is the boiling point of water; it doesn't matter if you have a cup or a pool of water, the boiling point remains consistent.

A common example of an intensive property is specific heat, which is the heat capacity per unit mass. It allows the determination of how much energy is required to change the temperature of a substance by a certain amount per unit mass, highlighting that it's independent of the system's scale. Recognizing whether a property is intensive or extensive is vital when measuring or calculating properties that can affect the behavior and state of a system, such as in the study of material properties and phase transitions.

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Most popular questions from this chapter

Match the column I with column II and mark the appropriate choice. $$ \begin{array}{|l|l|l|l|} \hline \text { (A) } & \mathrm{H}_{2(\mathrm{~g})}+\mathrm{Br}_{2(\mathrm{~g})} \rightarrow 2 \mathrm{HBr}_{(g)} & \text { (i) } & \Delta H=\Delta U-2 R T \\ \hline \text { (B) } & \mathrm{PCl}_{5(g)} \rightarrow \mathrm{PCl}_{3(\mathrm{~g})}+\mathrm{Cl}_{2(g)} & \text { (ii) } & \Delta H=\Delta U+3 R T \\ \hline \text { (C) } & \mathrm{N}_{2(g)}+3 \mathrm{H}_{2(g)} \rightarrow 2 \mathrm{NH}_{3(g)} & \text { (iii) } & \Delta H=\Delta U \\ \hline \text { (D) } & \begin{array}{l} 2 \mathrm{~N}_{2} \mathrm{O}_{5(g)} \rightarrow 4 \mathrm{NO}_{2(g)} \\ +\mathrm{O}_{2(g)} \end{array} & \text { (iv) } & \Delta H=\Delta U+R T \\ \hline \end{array} $$ (a) (A) \(\rightarrow\) (iii), (B) \(\rightarrow\) (i), (C) \(\rightarrow\) (ii), (D) \(\rightarrow\) (iv) (b) (A) \(\rightarrow\) (iii), (B) \(\rightarrow\) (iv), (C) \(\rightarrow\) (i), (D) \(\rightarrow\) (ii) (c) (A) \(\rightarrow\) (ii), (B) \(\rightarrow\) (i), (C) \(\rightarrow\) (iv), (D) \(\rightarrow\) (iii) (d) (A) \(\rightarrow\) (iv), (B) \(\rightarrow\) (ii), (C) \(\rightarrow\) (i), (D) \(\rightarrow\) (iii)

Bond dissociation energies of \(\mathrm{H}_{2}, \mathrm{Cl}_{2}\) and \(\mathrm{HCl}_{(g)}\) are 104,58 and \(103 \mathrm{kcal} \mathrm{mol}^{-1}\) respectively. Calculate the enthalpy of formation of \(\mathrm{HCl}\) gas. (a) \(-22 \mathrm{kcal}\) (b) \(+22 \mathrm{kcal}\) (c) \(+184 \mathrm{kcal}\) (d) \(-184 \mathrm{kcal}\)

For the reaction: \(\mathrm{H}_{2(g)}+\mathrm{Cl}_{2(g)} \rightarrow 2 \mathrm{HCl} ;\) \(\Delta H=-44 \mathrm{kcal}\) What is the enthalpy of decomposition of HCl? (a) \(+44 \mathrm{kcal} / \mathrm{mol}\) (b) - \(44 \mathrm{kcal} / \mathrm{mol}\) (c) \(-22 \mathrm{kcal} / \mathrm{mol}\) (d) \(+22 \mathrm{kcal} / \mathrm{mol}\)

At dynamic equilibrium the reaction on both sides occur at the same rate and the mass on both sides of the equilibrium does not undergo any change. This condition can be achieved only when the value of \(\Delta G\) is (a) \(-1\) (b) \(+1\) (c) \(+2\) (d) 0

What will be the work done when one mole of a gas expands isothermally from \(15 \mathrm{~L}\) to \(50 \mathrm{~L}\) against a constant pressure of 1 atm at \(25^{\circ} \mathrm{C}\) ? (a) \(-3542 \mathrm{cal}\) (b) \(-843.3 \mathrm{cal}\) (c) \(-718 \mathrm{cl}\) (d) \(-60.23 \mathrm{cal}\)

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