Thermodynamics is not concerned about___. (i) energy changes involved in a chemical reaction. (ii) the extent

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Q. Thermodynamics is not concerned about______.

(i) energy changes involved in a chemical reaction.

(ii) the extent to which a chemical reaction proceeds.

(iii) the rate at which a reaction proceeds.

(iv) the feasibility of a chemical reaction.

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Concept understanding — State Function

What is a State Function? The Intuition

Imagine you're standing at the base of a hill. Your height above sea level is a number — say, 100 metres. Now, you walk to the top of the hill. Your height is now 500 metres.

Here's the key question: Does it matter how you got to the top? Did you take the steep path, the winding road, or did you get carried up by a helicopter?

The answer is no. Your height at the top is 500 metres, regardless of the path you took. Height is a state function — it depends only on where you are, not on how you got there.

Now contrast that with distance walked. If you took the winding road, you walked 2 km. If you took the steep path, you walked 500 m. The distance walked depends entirely on the path. That's not a state function — it's a path function.

Note

A state function is a property whose value depends only on the current state of the system, not on the history or path taken to reach that state.

The Precise Statement

In thermodynamics and physics, a state function (or state variable) is any property of a system that is determined entirely by the system's current equilibrium conditions — typically its temperature, pressure, volume, composition, and so on.

If you know the state of the system (say, "1 mole of ideal gas at 300 K and 1 atm"), then every state function has a fixed value. You don't need to know whether the gas was heated slowly, compressed quickly, or cooled then expanded. The value is the same.

Common State Functions in Chemistry & Physics

Property	Symbol	Why it's a state function
Pressure	$P$	Depends only on current $T$ , $V$ , $n$
Volume	$V$	Depends only on current $P$ , $T$ , $n$
Temperature	$T$	A fundamental state variable
Internal Energy	$U$	Depends only on current $P$ , $T$ , composition
Enthalpy	$H$	$H = U + PV$ — a combination of state functions
Entropy	$S$	Depends only on current state
Gibbs Free Energy	$G$	$G = H - TS$ — again, a combination

Common Path Functions (the opposite)

Property	Why it's a path function
Work ( $W$ )	Depends on how you change volume (e.g., fast vs slow)
Heat ( $Q$ )	Depends on how you transfer energy (e.g., conduction vs radiation)

The Mathematical Signature

Here's the crisp, exam-ready way to recognise a state function:

Important

For a state function $f$ , the cyclic integral is zero:

$\oint df = 0$

This means: if you go from state A to state B and back to A by any path, the net change in $f$ is zero. The value of $f$ at A is always the same when you return.

Equivalently, the change in a state function between two states is path-independent:

\Delta f = f_{\text{final}} - f_{\text{initial}}

This is a single number — no integral over a path needed.

A Concrete Example: Internal Energy

Consider a gas in a cylinder. You take it from State 1 ( $T_1 = 300$ K, $P_1 = 1$ atm) to State 2 ( $T_2 = 400$ K, $P_2 = 2$ atm).

Path A: Heat the gas at constant volume, then compress it at constant temperature.
Path B: Compress the gas at constant temperature, then heat it at constant volume.

The work done ( $W$ ) and heat transferred ( $Q$ ) will be different for Path A vs Path B. But the change in internal energy $\Delta U$ will be identical for both paths. That's because $U$ is a state function — it only cares about the starting and ending states.

Watch out

A common mistake: students think "energy is conserved, so $\Delta U$ is always zero." No — $\Delta U$ is zero only for a cyclic process (returning to the same state). For a change between two different states, $\Delta U$ is non-zero but path-independent.

Why This Matters for Exams

When you see a problem asking for $\Delta H$ or $\Delta U$ or $\Delta S$ , you do not need to know the path. You only need the initial and final states. That's why we can use tables of standard enthalpies, entropies, and Gibbs energies — they are state functions, so their values are fixed for a given substance at a given temperature and pressure.

When you see a problem asking for $W$ or $Q$ , you must know the path (isothermal, adiabatic, isobaric, etc.). These are path functions.

Final takeaway: A state function is like your bank account balance at a given moment. It doesn't matter if you earned the money, inherited it, or found it on the street — the balance is what it is. A path function is like the total amount of money that moved in and out of your account over a month — that depends entirely on how you earned and spent.

This topic is commonly searched as "State Function 11 chemistry important questions" or "State Function formula and examples", and it maps cleanly onto the Class 11 Chemistry portion of the NCERT/CBSE syllabus. Because state function shows up repeatedly in JEE Main, NEET and state CET Chemistry papers, mastering the underlying idea (not just the formula) is genuinely worth the extra time.

Thermodynamics focuses on energy changes, spontaneity, and the extent of a reaction at equilibrium, but it does not provide information about how quickly a reaction occurs. The correct option is (iii).

Thermodynamics is a fundamental branch of science that deals with energy and its transformations. In the context of chemical reactions, it helps us understand the overall energy balance, whether a reaction will occur spontaneously under given conditions, and how far it will proceed before reaching equilibrium. It's a macroscopic science, meaning it describes the properties of matter in bulk and focuses on the initial and final states of a system, rather than the microscopic details of the reaction pathway.

Crucially, thermodynamics tells us about the possibility and direction of a process, but it provides no information about the speed or rate at which that process will occur. This distinction is key to understanding the scope of thermodynamics.

Let's analyze each option:

Understanding the core concerns of Thermodynamics:

Thermodynamics primarily addresses:
- Energy changes: Quantifying the heat absorbed or released during a process (e.g., enthalpy change, $\Delta H$ ).
- Spontaneity or Feasibility: Determining if a process will occur on its own under specific conditions (e.g., Gibbs free energy change, $\Delta G$ ).
- Equilibrium: Predicting the final state where no net change occurs (e.g., equilibrium constant, $K$ ).
It explicitly does not deal with the time taken for a process to occur.
Analyzing option (i): energy changes involved in a chemical reaction.
- This is a central concern of thermodynamics. The First Law of Thermodynamics deals with the conservation of energy, and concepts like enthalpy ( $\Delta H$ ) are used to quantify the heat exchanged during a chemical reaction at constant pressure. For instance, determining if a reaction is exothermic (releases heat) or endothermic (absorbs heat) is a direct application of thermodynamics.
- Therefore, thermodynamics is concerned with energy changes.
Analyzing option (ii): the extent to which a chemical reaction proceeds.
- Thermodynamics helps predict the position of equilibrium for a reversible reaction. The Gibbs free energy change ( $\Delta G$ ) is directly related to the equilibrium constant ( $K$ ) by the equation $\Delta G^\circ = -RT \ln K$ .
- A large value of $K$ indicates that the reaction proceeds extensively, forming a high concentration of products at equilibrium. Conversely, a small $K$ means the reaction proceeds to a lesser extent.
- Therefore, thermodynamics is concerned with the extent to which a reaction proceeds.
Analyzing option (iii): the rate at which a reaction proceeds.
- The rate of a reaction refers to how quickly reactants are converted into products over a period of time. This area of study falls under chemical kinetics.
- Chemical kinetics investigates factors such as activation energy, reaction mechanisms, and the influence of temperature, concentration, and catalysts on reaction speed.
- Thermodynamics provides no information about how fast a reaction will occur. A reaction might be thermodynamically favorable (spontaneous, $\Delta G < 0$ ) but kinetically very slow (e.g., the conversion of diamond to graphite, or the rusting of iron).
- Therefore, thermodynamics is not concerned with the rate at which a reaction proceeds. This is the correct answer.
Watch out

It is a common mistake to confuse thermodynamic spontaneity with reaction speed. A reaction can be thermodynamically spontaneous (meaning it can happen) but kinetically very slow (meaning it happens very slowly). Thermodynamics tells you if a reaction is possible; kinetics tells you how fast it will happen.
Analyzing option (iv): the feasibility of a chemical reaction.
- Feasibility, or spontaneity, refers to whether a reaction can occur on its own without continuous external energy input.
- The Gibbs free energy change ( $\Delta G$ $Δ G$ ) is the primary thermodynamic criterion for spontaneity:
  - If $\Delta G < 0$ , the reaction is spontaneous (feasible) under the given conditions.
  - If $\Delta G > 0$ , the reaction is non-spontaneous (not feasible) and requires continuous energy input to proceed.
  - If $\Delta G = 0$ , the reaction is at equilibrium.
- Therefore, thermodynamics is concerned with the feasibility of a chemical reaction.

Based on this analysis, the only aspect that thermodynamics is not concerned about is the rate at which a reaction proceeds.

✓Final answer

Thermodynamics is not concerned about (iii) the rate at which a reaction proceeds.

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

KCET 2026Set D31 markMCQ
$Q. Which of the following is a correct statement for a thermodynamic system? (A) The internal energy changes in all processes (B) Internal energy and entropy are state functions (C) Work is a state function (D) The work done in an adiabatic process is always zero$
›Reveal solutionSolution

Distinguishing state functions (path-independent) from path functions (path-dependent) resolves this question.

Step 1 — Statement (A)

Internal energy does not change in every process — for example, in an isothermal expansion of an ideal gas, $\Delta U = 0$ because internal energy depends only on temperature. This statement is false.

Step 2 — Statement (B)

Internal energy $U$ and entropy $S$ are both state functions — their values depend only on the current state of the system (defined by variables like $T$ , $P$ , $V$ , composition), not on the path taken to reach that state. This is a fundamental and correct thermodynamic principle.

Step 3 — Statements (C) and (D)

Work is explicitly a path function, not a state function, since its value depends on how the process is carried out (e.g. reversibly vs irreversibly), so (C) is false. The work done in an adiabatic process is generally non-zero (e.g. adiabatic compression/expansion does real work; only free expansion into vacuum gives zero work), so (D) is false.

✓Final answer
$The correct option is (B) — internal energy and entropy are state functions.$
COMEDK 2022Set 20221 markMCQ
$-$
›Reveal solutionSolution

The question asks which are NOT path functions: I and IV.

Concept: state functions depend only on the state of the system; path functions depend on how the change is carried out.

I. H - TS = G, the Gibbs free energy - a STATE function (H, T, S are all state functions).

II. W (work) - a PATH function.

III. q (heat) - a PATH function.

IV. q + W = Delta U, the internal energy change - a STATE function (this is exactly the first law: although q and W individually are path-dependent, their sum is not).

The question asks which are NOT path functions: I and IV.

✓Final answer
$The correct option is (D) — I and IV$

ANSWER: D

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