An open system exchanges both matter and energy with its surroundings. A closed system exchanges energy but not matter, while an isolated system exchanges neither matter nor energy.
The first law states that energy can neither be created nor destroyed. It can only change from one form to another.
State functions depend only on the initial and final states of a system, not on the process followed. Path functions depend on the specific process used to move between these states. Internal energy and enthalpy are state functions, while heat and work are path functions.
Hess’s law states that the total enthalpy change of a reaction remains the same whether the reaction takes place in one step or several steps, provided the initial and final states are the same.
Exothermic reactions release heat into the surroundings, whereas endothermic reactions absorb heat from the surroundings. For example, the combustion of fuel is exothermic, while the thermal decomposition of calcium carbonate is endothermic.
Students can access NCERT Solutions Class 11 Chemistry Chapter 5 on the ALLEN website. The chapter-wise solutions help them understand thermodynamics concepts, review textbook questions, and prepare for CBSE exams and competitive exams such as JEE.
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NCERT Solutions Class 11 Chemistry Chapter 5 - Chemical Thermodynamics
NCERT Solutions for Class 11 Chemistry Chapter 5 (Chemical Thermodynamics) help students understand energy changes, heat transfer, and why some chemical reactions happen on their own. These solutions explain important topics in simple English and help students learn formulas, solve numerical questions, and prepare for CBSE exams, JEE, and NEET.
NCERT Solutions for Class 11 Chemistry Chapter 5 cover the First, Second, and Third Laws of Thermodynamics with answers to textbook questions. Students can use these solutions to learn the laws, solve numerical problems, revise important topics, and prepare for school exams and competitive exams.
1.0Class 11 Chemistry Chapter 5 Chemical Thermodynamics: Key Concepts
This chapter explains how energy changes during chemical reactions and physical processes. The key concepts include:
Thermodynamic Terms: Understanding systems (open, closed, and isolated), surroundings, state functions, and extensive and intensive properties.
Internal Energy (U): Learning how heat and work transfer energy between a system and its surroundings.
First Law of Thermodynamics: Understanding the law of energy conservation using the equation ΔU=q+w.
Enthalpy (H): Learning how to calculate heat changes at constant pressure and understand the relationship between ΔH and ΔU
Enthalpy Changes of Reactions: Studying standard enthalpy of formation, combustion, atomisation, and solution.
Hess’s Law of Constant Heat Summation: Calculating the enthalpy change of a reaction using different steps.
Entropy (S): Understanding entropy, changes in molecular arrangement, and the Second Law of Thermodynamics.
Gibbs Free Energy (G): Using the equation ΔG=ΔH−TΔS to determine whether a process is spontaneous under given conditions.
Third Law of Thermodynamics: Learning that a perfectly crystalline substance has zero entropy at absolute zero.
2.0Detailed NCERT Textbook Solutions : Class 11 Chemistry Chapter 5
NCERT EXERCISE
Choose the correct answer. A thermodynamic state function is a quantity (i) used to determine heat changes (ii) whose value is independent of path (iii) used to determine pressure volume work (iv) whose value depends on temperature only. Ans. A thermodynamic state function is a quantity whose value is independent of a path. Functions like p, V, T etc. depend only on the state of a system and not on the path. Hence, alternative is correct.
For the process to occur under adiabatic conditions, the correct condition is: (i) ΔT=0 (ii) Δp=0 (iii) q=0 (iv) w=0 Ans. A system is said to be under adiabatic conditions if there is no exchange of heat between the system and its surroundings. Hence, under adiabatic conditions, q=0. Therefore, alternative (iii) is correct.
The enthalpies of all elements in their standard states are: (i) unity (ii) zero (iii) < 0 (iv) different for each element Ans. The enthalpy of all elements in their standard state is zero. Therefore, alternative (ii) is correct.
ΔUθ of combustion of methane is −XkJmol−1. The value of ΔHθ is : (i) =ΔUθ (ii) >ΔUθ (iii) <ΔUθ (iv) =0 Ans. Since ΔHθ=ΔUθ+ΔngRT and ΔUθ=−XkJmol−1,ΔHθ=(−X)+ΔngRT.
⇒ΔHθ<ΔUθ
Therefore, alternative (iii) is correct.
The enthalpy of combustion of methane, graphite and dihydrogen at 298 K are, −890.3kJmol−1−393.5kJmol−1, and -285.8 kJ mol−1 respectively. Enthalpy of formation of CH4(g) will be : (i) −74.8kJmol−1 (ii) -52.27 kJ mol-1 (iii) +74.8kJmol−1 (iv) +52.26kJmol−1
Hence, the enthalpy change involved in the transformation is –7.151kJmol−1
Enthalpy of combustion of carbon to CO2is–393.5kJmol−1 .Calculate the heat released upon formation of 35.2 g of CO2 from carbon and dioxygen gas.
Ans. Formation of CO2 from carbon and dioxygen gas can be represented as:
C(s)+O2(g)→C2(g)ΔfH=−393.5kJmol−1(1mole=44g)Heat released on formation of 44gCO2=−393.5kJmol−1∴Heat released on formation of 35.2gCO2=−314.8kJmol−1=44g−393.5kLmol−1×35.2g=−314.8kJmol−1
12.Enthalpies of formation of CO(g),CO2(g),N2O(g) and N2O4(g)are −110kJmol−1,−393kJmol−1,81kJmol−1 and 9.7kJmol−1respectively. Find the value of ΔrH for the reaction:N2O4(g)+3CO(g)→N2O(g)+3CO2(g)
Ans. ΔrH for a reaction is defined as the difference between ΔfH value of products and ΔfH value of reactants.
What is the standard enthalpy of formation of NH3 gas?
Ans. Standard enthalpy of formation of a compound is the change in enthalpy that takes place during the formation of 1 mole of a substance in its standard form from its constituent elements in their standard state. Re-writing the given equation for 1 mole of
NH3(g),21N2(g)+23H2(g)→NH3(g)∴Standard enthalpy of formation of NH3(g)=21ΔrHθ=21(−92.4kJmol−1)=−46.2kJmol−1
Calculate the standard enthalpy of formation of CH3OH(1) from the following data:
Enthalpy change for the given process CCl4(g)→C(g)+4Cl(g), can be calculated using the following algebraic calculations as: Equation (ii) + 2 × Equation (iii) – Equation (i) – Equation
For an isolated system, ΔU=0, what will be ΔS ? Ans.ΔS will be positive i.e., greater than zero Since ΔU=0,ΔS will be positive and the reaction will be spontaneous.
For the reaction at 298 K,
2A+B→CΔH=400kJmol−1 and ΔS=0.2kJK−1mol−1
At what temperature will the reaction become spontaneous considering ΔH and ΔS to be constant over the temperature range? Ans. From the expression, ΔG=ΔH−TΔS Assuming the reaction at equilibrium, ΔT for the reaction would be:
T=(ΔH−ΔG)ΔS1=ΔSΔH(ΔG=0 at equilibrium) T=2000K
For the reaction to be spontaneous, ΔG must be negative. Hence, for the given reaction to be spontaneous, T should be greater than 2000 K.
For the reaction, 2Cl(g)→Cl2(g), what are the signs of ΔH and ΔS ? Ans. ΔH and ΔS are negative The given reaction represents the formation of chlorine molecule from chlorine atoms. Here, bond formation is taking place. Therefore, energy is being released. Hence, ΔH is negative. Also, two moles of atoms have more randomness than one mole of a molecule. Since spontaneity is decreased, ΔS is negative for the given reaction.
For the reaction
2A(g)+B(g)→2D(g)ΔUθ=−10.5kJ and ΔSθ=−44.1JK−1.
Calculate ΔGθ for the reaction, and predict whether the reaction may occur spontaneously. Ans. For the given reaction,
2A(g)+B(g)→2D(g)Δng=2−(3)=−1 mole
Substituting the value of ΔUθ in the expression of ΔH :
This means that NO(g) has higher energy than the reactants (N2 and O2). Hence, NO(g) is unstable. The negative value of ΔrH indicates that heat is evolved during the formation of NO2(g) from NO(g) and O2(g). The product, NO2(g) is stabilized with minimum energy. Hence, unstable NO(g) changes to stable NO2(g).
Calculate the entropy change in surroundings when 1.00 mol of H2O(l) is formed under standard conditions. ΔfHθ=−286kJmol−1. Ans. It is given that 286kJmol−1 of heat is evolved on the formation of 1 mol of H2O(l). Thus, an equal amount of heat will be absorbed by the surroundings.
NCERT Class 11 Biology – Chapter-Wise Solutions & Explanations
Get Class 11 Chemistry NCERT Solutions with chapter-wise answers to textbook questions. Understand the concepts, revise important topics, and prepare for your exams.