CBSE Class 11 Chemistry Chapter 6: Thermodynamics NCERT Solutions

Q: What is a thermodynamic state function?

A thermodynamic state function is a property of a system whose value depends only on the current state of the system and not on the path taken to reach that state. Examples include pressure, temperature, volume, and internal energy.

Q: What does it mean for a process to be adiabatic?

An adiabatic process is one where there is no heat transfer between the system and its surroundings. This is represented by the condition q = 0.

Q: What is the standard enthalpy of an element?

The standard enthalpy of an element in its most stable form at standard conditions (298 K and 1 atm pressure) is defined as zero by convention.

Q: How are enthalpy change ($\Delta H$) and internal energy change ($\Delta U$) related?

The relationship is given by $\Delta H = \Delta U + \Delta n_g RT$, where $\Delta n_g$ is the change in the number of moles of gas during the reaction and R is the ideal gas constant.

Q: How can Hess's Law be used in thermodynamics?

Hess's Law states that the total enthalpy change for a reaction is independent of the pathway taken. It is used to calculate the enthalpy of formation or reaction when direct measurement is difficult, by combining the enthalpy changes of known reactions.

Reviewed by NCERT Help Editorial Team · Updated on 29 Jul 2026

NCERT Solutions PDF Class 11 PDF

This resource provides detailed NCERT Solutions for Chapter 6 of the CBSE Class 11 Chemistry syllabus, focusing on Thermodynamics. It covers fundamental concepts such as thermodynamic state functions, their path-independent nature, and the conditions for adiabatic processes where heat exchange is zero. The solutions also clarify the standard enthalpies of elements and the relationship between enthalpy change and internal energy change during combustion reactions. Key topics include calculating the enthalpy of formation using Hess's Law, derived from given enthalpies of combustion. These solutions are designed to help students grasp complex thermodynamic principles, solve numerical problems accurately, and prepare effectively for their board examinations by offering clear explanations and step-by-step problem-solving approaches.

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Board	CBSE
Class	Class 11
Subject	Chemiry
Session	2026
Language	English
Type	NCERT Solutions
Chapter	Chapter 6: Thermodynamics

Chapter summary

Chapter 6 on Thermodynamics for Class 11 Chemistry delves into the fundamental principles governing energy changes in chemical systems. These NCERT Solutions cover the definition and properties of thermodynamic state functions, distinguishing them from path functions. They also explain adiabatic conditions and the standard enthalpy of elements. The solutions provide a clear method for relating enthalpy change to internal energy change and demonstrate how to calculate the enthalpy of formation using Hess's Law with given combustion enthalpies.

Learning outcomes

Understand the definition and properties of thermodynamic state functions.
Identify the conditions for adiabatic processes.
Recall the standard enthalpy of elements.
Differentiate between enthalpy change and internal energy change.
Apply Hess's Law to calculate enthalpy of formation.

Topics covered

Paper topics

Thermodynamic State Functions
Path Independence
Adiabatic Conditions
Heat Exchange (q)
Standard Enthalpy of Elements
Enthalpy of Combustion
Internal Energy Change ($\Delta U$)
Enthalpy Change ($\Delta H$)
Relationship between $\Delta H$ and $\Delta U$
Hess's Law
Enthalpy of Formation

Important topics

Thermodynamic State Functions
Adiabatic Conditions
Relationship between $\Delta H$ and $\Delta U$
Hess's Law
Enthalpy of Formation Calculation

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Questions and Solutions

Question 6.1

Choose the correct answer. A thermodynamic state function is a quantity

used to determine heat changes
whose value is independent of path
used to determine pressure volume work
whose value depends on temperature only.

Solution: A thermodynamic state function is defined as a property of a system whose value depends solely on the initial and final states of the system, irrespective of the path taken to transition between these states. Properties like pressure (p), volume (V), and temperature (T) are examples of state functions because their values are determined by the specific state the system is in. Therefore, the correct characteristic of a thermodynamic state function is that its value is independent of the path. Alternative (ii) accurately describes this property.

Question 6.2

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

$\Delta T = 0$
$\Delta p = 0$
$q = 0$
$w = 0$

Solution: An adiabatic process is a thermodynamic process that occurs without any exchange of heat between the system and its surroundings. This means that the heat absorbed or released by the system is zero. Mathematically, this condition is represented as

q = 0

. While changes in temperature (

\Delta T

) or pressure (

\Delta p

) might occur during an adiabatic process, and work (

w

) can be done, the defining characteristic is the absence of heat transfer. Therefore, alternative (iii) is the correct condition for an adiabatic process.

Question 6.3

The enthalpies of all elements in their standard states are:

unity
zero
$< 0$
different for each element

Solution: In thermodynamics, a reference point is established for enthalpy values. By convention, the enthalpy of any element in its most stable form at standard conditions (defined as 298.15 K or 25°C and 1 atm pressure) is assigned a value of zero. This standard state enthalpy is denoted as

H^{\theta}

. For example, the standard enthalpy of graphite (the standard state of carbon) and gaseous oxygen (the standard state of oxygen) are both zero. Therefore, the enthalpies of all elements in their standard states are zero. Alternative (ii) is correct.

Question 6.4

$\Delta U^{\theta}$ of combustion of methane is – X kJ mol^-1. The value of $\Delta H^{\theta}$ is

= $\Delta U^{\theta}$
$> \Delta U^{\theta}$
$< \Delta U^{\theta}$

$(iv) = 0$

Solution: The relationship between the standard enthalpy change (

\Delta H^{\theta}

) and the standard internal energy change (

\Delta U^{\theta}

) for a reaction is given by the equation:

\Delta H^{\theta} = \Delta U^{\theta} + \Delta n_g RT

Here,

\Delta n_g

represents the change in the number of moles of gaseous reactants and products (

\Delta n_g = \text{moles of gaseous products} - \text{moles of gaseous reactants}

), R is the ideal gas constant, and T is the absolute temperature in Kelvin. For the combustion of methane (

CH_{4(g)} + 2O_{2(g)} \rightarrow CO_{2(g)} + 2H_2O_{(g)}

), the change in the number of moles of gas is

\Delta n_g = (1 + 2) - (1 + 2) = 0

if water is produced as gas. However, if water is produced as liquid,

\Delta n_g = 1 - 3 = -2

. In typical combustion problems where

\Delta U^{\theta}

is given as negative, it implies an exothermic reaction. If

\Delta n_g

is negative (as is common when forming liquid water), the term

\Delta n_g RT

will be negative. Since

\Delta U^{\theta} = -X

kJ mol^-1, and

\Delta n_g RT

is a negative value,

\Delta H^{\theta} = (-X) + (\text{negative value})

. This means

\Delta H^{\theta}

will be more negative than

\Delta U^{\theta}

, hence

\Delta H^{\theta} < \Delta U^{\theta}

. Therefore, alternative (iii) is correct.

Question 6.5

The enthalpies of combustion of methane, graphite and dihydrogen at 298 K are, -890.3 kJ mol^-1, –393.5 kJ mol^-1, and –285.8 kJ mol^-1 respectively. Enthalpy of formation of $CH_{4(g)}$ will be

-74.8 kJ mol^-1
-52.27 kJ mol^-1
+74.8 kJ mol^-1
+52.26 kJ mol^-1

Solution: We are given the enthalpies of combustion for methane, graphite, and dihydrogen, and we need to find the enthalpy of formation of methane (

CH_{4(g)}

). The enthalpy of formation refers to the enthalpy change when one mole of a compound is formed from its constituent elements in their standard states. The given reactions and their enthalpy changes are: 1. Combustion of methane:

CH_{4(g)} + 2O_{2(g)} \longrightarrow CO_{2(g)} + 2H_2O_{(l)} \quad \Delta H_c^{\theta} = -890.3 \text{ kJ mol}^{-1}

(Note: Water is usually formed as liquid in standard combustion enthalpy unless specified otherwise.) 2. Combustion of graphite (standard state of Carbon):

C_{(s, graphite)} + O_{2(g)} \longrightarrow CO_{2(g)} \quad \Delta H_c^{\theta} = -393.5 \text{ kJ mol}^{-1}

3. Combustion of dihydrogen:

H_{2(g)} + \frac{1}{2}O_{2(g)} \longrightarrow H_2O_{(l)} \quad \Delta H_c^{\theta} = -285.8 \text{ kJ mol}^{-1}

We want to find the enthalpy of formation for

CH_{4(g)}

, which corresponds to the reaction:

C_{(s, graphite)} + 2H_{2(g)} \longrightarrow CH_{4(g)} \quad \Delta H_f^{\theta} = ?

We can use Hess's Law to manipulate the given equations to obtain the desired formation equation. Let's denote the given reactions as (1), (2), and (3). We need

C_{(s, graphite)}

as a reactant, which is present in equation (2) as is. So, we use equation (2) as it is:

C_{(s, graphite)} + O_{2(g)} \longrightarrow CO_{2(g)} \quad \Delta H^{\theta} = -393.5 \text{ kJ mol}^{-1} \quad (Equation A)

We need

2H_{2(g)}

as a reactant. Equation (3) has

H_{2(g)}

as a reactant, so we multiply equation (3) by 2:

2H_{2(g)} + O_{2(g)} \longrightarrow 2H_2O_{(l)} \quad \Delta H^{\theta} = 2 \times (-285.8) = -571.6 \text{ kJ mol}^{-1} \quad (Equation B)

We need

CH_{4(g)}

as a product. Equation (1) has

CH_{4(g)}

as a reactant. So, we reverse equation (1) and change the sign of its enthalpy change:

CO_{2(g)} + 2H_2O_{(l)} \longrightarrow CH_{4(g)} + 2O_{2(g)} \quad \Delta H^{\theta} = -(-890.3) = +890.3 \text{ kJ mol}^{-1} \quad (Equation C)

Now, we add equations (A), (B), and (C):

(C_{(s, graphite)} + O_{2(g)}) + (2H_{2(g)} + O_{2(g)}) + (CO_{2(g)} + 2H_2O_{(l)}) \longrightarrow CO_{2(g)} + 2H_2O_{(l)} + (CH_{4(g)} + 2O_{2(g)})

Cancel out the species that appear on both sides:

C_{(s, graphite)} + 2H_{2(g)} \longrightarrow CH_{4(g)}

The enthalpy change for this formation reaction is the sum of the enthalpy changes of equations (A), (B), and (C):

\Delta H_f^{\theta} = \Delta H_A + \Delta H_B + \Delta H_C

\Delta H_f^{\theta} = (-393.5) + (-571.6) + (+890.3)

\Delta H_f^{\theta} = -965.1 + 890.3

\Delta H_f^{\theta} = -74.8 \text{ kJ mol}^{-1}

Thus, the enthalpy of formation of

CH_{4(g)}

is -74.8 kJ mol^-1. Alternative (i) is correct.

Common mistakes

Confusing state functions with path functions.
Incorrectly applying the relationship between $\Delta H$ and $\Delta U$.
Errors in setting up and solving equations for enthalpy of formation using Hess's Law.

Revision tips

Focus on understanding the definition of state functions and why path independence is crucial.
Memorize the standard enthalpy of elements in their standard states.
Practice the formula relating $\Delta H$ and $\Delta U$ for reactions involving gases.
Work through the Hess's Law examples to master enthalpy of formation calculations.

Practice MCQs

Q1. Which of the following is a characteristic of a thermodynamic state function?

Q2. What is the condition for a process to occur under adiabatic conditions?

Q3. The standard enthalpy of all elements in their standard states is defined as:

Q4. For the combustion of methane, if $ $\Delta$ $U^{\theta} = -X$ $ kJ mol$^{-1}$, what is the relationship between $ $\Delta$ $H^{\theta}$ $ and $ $\Delta$ $U^{\theta}$ $?

Frequently asked questions

What is a thermodynamic state function?