Chapter 06: Chemical Energetics

Short Questions & Flashcards Study Portal

Short Questions

Enthalpy Change

Q.1

and Endothermic Reactions.

Answer

•Exothermic Reaction: Heat is released; AH is negative ExamPle: Combustion of methane. • Endothermic Reaction: Heat is absorbed; AH is positive. Example: Decomposition of calcium carbonate.
Illustration (added) - Reaction Energy Profile Diagram Reactants Products dH < 0 Exothermic Reactants Products dH > 0 Endothermic

Thermochemistry

Q.2

What do you understand by enthalpy of a system? of a system at constant pressure. It is the sum of the internal energy and the product of pressure and volume. Difference clearly between

Answer

Entropy

Free Energy Change

Q.3

(S) and Gibbs Free Energy (G):

Answer

•Entropy (S): Measure of randomness or disorder in a system. • Gibbs Free Energy (G): Energy available to do useful work; determines spontaneity of a reaction (G = H - TS) between clearly

Q.3

its two applications.

Answer

See Q.7 from theory. Na* (g) → Na*(aq) Example:

Q.4

Distinguish and standard enthalpy of reaction standard enthalpy of Formation.

Answer

•Standard Enthalpy of Reaction (AH,): Enthalpy change when a chemical reaction occurs under standard conditions. • Standard Enthalpy of Formation (AH;): a Enthalpy change when 1 mole of compound is formed from its elements in standard states.

Born-Haber Cycle

Q.4

What is lattice energy? How does Born-Haber cycle help to calculate the lattice energy of NaCl?

Answer

See Q.13 from theory. NUMERICAL PROBLEMS

Illustration (added) - Born-Haber Cycle Na(s) + 1/2 Cl₂(g) Na(g) [Sublimation] Na⁺(g) + e⁻ [Ionization] Na⁺(g) + Cl⁻(g) [Electron Affinity] NaCl(s) [Lattice Energy] Lattice Energy (U)

Q.5

Define the following enthalpies and give one example of each. (i) Standard enthalpy of solution (ii) Standard enthalpy of hydration (iii) Standard enthalpy of atomization (iv) Standard enthalpy of large negative

Answer

(i) Standard Enthalpy of Solution ДН when 1 mole of a substance dissolves in a solvent. Example NaCt) → Na (ag) + Cl (ag) (ii) Standard Enthalpy of Hydration AH when 1 mole of gaseous ions hydrated Nat (aq) (8) → Nat Example: (iii) Standard Enthalpy of Atomization AH when 1 mole of gaseous atoms is formed from an element in standard state. Example: 2H2(g) → H(g) (iv) Standard Enthalpy of Combustion AH when 1 mole of a substance completely burns in oxygen. Example: CH 4(8) + 202(g) CO2(g) + 2H,0(g)

Q.5

When 0.400g NaOH is dissolved in 100,0g of water, the temperature rises from 25.00 to 26.03°C. Calculate: (i). q water, (ii). AH for the solution process.

Answer

Given: • Mass of NaOH = 0.400g • Mass of water = 100.0g • Initial temperature = 25.00°C • Final temperature = 26.03°C • AT = 26.03 - 25.00 = 1.03°C Specific heat capacity of water, c = 4.18J/g°C Step (i): Calculate heat absorbed by water (qwater) 9=m. c. AT q = 100.0g × 4.18J/g°C × 1.03°C q = 430.54J Answer (i): Qwater = 430.54J (ii): Calculate AH for the solution process (in kJ/mol) 1. Molar mass of NaOH = 23 +16 + 1 = 40 g/mol 2. Moles of NaOH 0.400g - = 0.010mol n= ole 40g/ mol Now use: . 430.54J 9 = 43054J / mol = 43.05kJ. / mol AH = n 0.010mol Answer (ii): AH(solution) = -43.05kJ/mol

Q.6

Explain why the lattice enthalpy of an ionic compound is typically a large negative value.

Answer

It is the energy released when gaseous ions form an ionic solid: The strong electrostatic attraction between oppositely charged ions results in a large release of energy. influence the

Hess's Law

Q.6

By applying Hess' law, calculate the enthalpy change for the formation of an aqueous solution of NH4Cl from NH3 gas and HCl gas. The results for the various reactions are as follows. (i) NH 3(g) + aq NH NH AH = =-35.16kJmol" → HC (a) AH = -72.41kJmol (ii) HC (g) + aq (i) NH 3(a) + HC (ag) → NHLag) AH= -51.48kJmol

Answer

Target reaction: NH3(8) + HCLg) → NH, Claq) Given: AH = -35.16kJ/ mol 1. NH3(g) → NH 3(ag) AH= -72.41kJ/ mol 2. HC (g) HC (aq) AH = -51.48kJ/ mol 3. NH 3(ag) + HC (ag) Using Hess's Law: Add all three steps to get the overall reaction: NH 3(8) + HC g) → NH 4Ch (4) AH = (-35.16) + (-72.41) + (-51.48) AH = -159.05kJ/ mol Answer: AH=-159.05kJ/ mol Calculate the heat of formation of ethyl alcohol from the following information.

Illustration (added) - Hess's Law Cycle Reactants A Products B dH₁ (Direct Path) Intermediates C dH₂ dH₃ dH₁ = dH₂ + dH₃

Q.7

What factors magnitude of the lattice enthalpy?

Answer

1. Ionic Charges: Higher charge → stronger attraction → more negative lattice enthalpy. 2. Ionic Radii: Smaller size → stronger attraction → more negative lattice enthalpy.

Q.7

(i) Heat of combustion of ethyl alcohol is -1367kJmol 1 (ii) Heat of formation of carbon dioxide is -393.kJ mol (iii) Heat of formation of water is -285.8 kJmol. Given information: 1. Heat of combustion of ethyl alcohol: CH2OH, + 302(g) 2C02(g) +3,0) AH=- 1367 kJ/ mol AH; of CO2(g) = -393.7kJ/mol AH; of H20(1) = -285.8kJ/mal Step 1: Use Hess's Law Equation According to Hess's Law: = AH® AH ZAH; f. (products) combustion (reactants) | Rewriting to find the AHt of C2HsOH(L): АН, (С, НОН) = (products) - AH Lee. Step 2: Products • 2 mol CO2: 2 X (-393.7) = -787.4kJ • 3 mol H2O: 3 X (-285.8) = -857.4kJ Total AH; of products = - 787.4 + (-85.7.4) = - 1644.8kJ Now calculate AH; of ethanol: дН; (С,Н,ОН)=-1644.8- (-1367) = - 1644.8+ 1367 = -277.8kJ / mol

Answer

wer: AH; (CH2OH) = -277.8kJ / mol

Q.8

Explain why the enthalpy of hydration is always an exothermic process for gaseous ions. What are the main interactions responsible for the release of energy during hydration?

Answer

Because energy is released when water molecules attract and surround gaseous ions. Main interactions: Ion-dipole interactions between ions and water molecules.

Q.8

Using the information given in the table below, calculate the lattice energy of potassium bromide. Reactions →K Br A (s) +1/2B2(1) - K(.) → K(g) K (g) K(g) te Br +é → Br AH/kJmol-' -392 +90 +420 +112 -342 Given: Reaction AH(kJ/mol) K(s) +1/2B*21) → KBr(s) = 392 +90 K(s) → K(g) +420 K (s) → K(g) té +112 1/2B 2(1) → Br(g) -324 > Br (g) Brig) te Step 1: Understand the overall process We want to find the lattice energy (U) of KBr, which is the energy released when gaseous ions combine to form solid ionic lattice K+ Lattice energy = AH lau *(g) + Br (g) → KBr (s) Step2: Write the Born Haber cycle The formation of KBr from its element in their standard states can be broken down into these steps: 1. Sublimation of K(s) to K(g): K(s) K(g) AH= +90kJ/ mol 2. Ionization of K(g) to K+ . K(g) →K(g)te AH=+420kJ/ mol 3. Dissociation of Br2(1) to Br (g): 1 Br2(1) → Br(g) AH=+112kJ| mol 2 4. Electron affinity of Br(g) to Br (g): Br(g) +e-→ Brg) =-324kJ mol 5. Formation of KBr(s) from gaseous ions (Lattice énergy): K+ AH eat =? (8) → KBr) (8) + Br Step 3: Write the enthalpy cycle equation The total enthalpy change for the formation of KBr(s) from K(s) and ½ Br2(1) is: AH, = Sublimation + Ionization + Dissociation + Electron affinity + Lattice energy -392 = (90 + 420 + 112 - 324) + AH eart Calculate the sum inside the parentheses: -392 = 598 + AH eart AH eatr=-392-298=-690kJ/mol Final

Answer

wer: The lattice energy of potassium bromide is:-690kJ/mol

Q.9

For the reaction CH4(8) + 202(g) → CO2(g) +2H0(8)) identify all the bonds that need to be broken and all the bonds that need to be formed to carry out a bond energy calculation of AH.

Answer

Bonds broken: 4C - H bonds in CH4 20 = 0 double bonds Bonds formed: 2C= C=0 bonds in CO2 40-H bonds in H2O
Illustration (added) - Ethene Carbon-Carbon Sigma & Pi Bonding C C Sigma Bond (sp2-sp2) Pi Bond Overlap

Q.9
Calculate the entropy of the surrounding Surrounding 2Ca(s) + 02(g) → 2CaO (s) Given:

Answer

Reaction 2Cа(s) +02(g) - • AH reaction = -1270.2kJmol Temperature, T = 298K Step 1: Understand the relationship The entropy change of the surroundings is related to the enthalpy change of the system by the formula: АН® reaction = ASO surroundings Here: • ДН° reaction should be in Joules (not kJ) for consistency with entropy units. • Temperature in Kelvin. Step 2: Convert enthalpy change to Joules Traction =- 1270.2kJ/ mol = -1270.2×10°= -1,270,200J / mol Step 3: Calculate AS surrounding® -1,270, 200 1,270,200 = AS° surrounding 298 . surrounding = +4262.75JK mol Answer: AS® surrounding =+4262.75JK 'mol-1

Q.10

For a reaction to be spontaneous, what is the required sign of the Gibbs free energy change (AG)? under what conditions of enthalpy change (AH) and entropy change (AH) will a reaction always be spontaneous?

Answer

For a reaction to be spontaneous, AG must be negative Always spontaneous when: AH is negative and AS is positive

Q.10

For the reaction: CaSO 4(5) -Cat Calculate AH°, AS® and AG° at 25°C using the following data; and discuss its spontaneity. Enthalpy of formation: AH i =- 1432.7 kJ, AH (ca) = -543.0 kJ, AH, ku Standard entropy: S°(as04(s)) = 106.7J/K, Sicam.)

Answer

Reaction CaSOA(s) Given data: Enthalpy of formation (AH, ) (in kJ/mol) =-1432.7 ДН® =-543.0 AH® (501) = =-907.5 for the reaction at 298K AH® reaction =1270.2kJmoll → 2CaO(s) (aq) + SO 4 (ag) (013) = - 907.5 = + 17.2 J/K] = - 55.2 J/K, Sis,) *(a) + S03 (ag) Standard Entropies So (in J/K mol): S° (CaS04(s)) = 106.7 Step 1: Calculate AH° of the reaction f (reactants) AH° = ZAH° "(products) - ZAH° =|(-543.0) + (-907.5)|-(-1432.7) = (-1450.5) +1432.7 = -17.8kJ / mol Answer Step 2: Calculate AS of the reaction (reactants) (produces) - 2 S° =[(-55.2) + (17.2)] - (106.7) = -38.0-106.7 = -144.7J/ K| Answer Convert to kJ for use in AG° calculation: AS° = -0.1447kJ/K Step 3: Calculate AG° of the reaction AG° = AH° - TAS® =-17.8 - (298 X (-0.1447)) =-17.8 + 43.1 = +25.3kJ/ mol Answer Step 4: Discuss spontaneity AG° = +25.3 kJ/mol, which is positive. Therefore, the reaction is non-spontaneous under standard. conditions.

Q.11

The enthalpy of solution can be either positive or negative. Explain what a positive AHsol and a negative AHsol indicate about the energy changes during the dissolution process.

Answer

Positive AHsol: Energy absorbed; solution process is endothermic. Negative AHsol: Energy released; solution process is exothermic.

Q.12

Consider two ions with similar charges but different sizes, or similar sizes but different charges. Explain how the concept of density can be used to predict which ion will have a more exothermic enthalpy of hydration and why?

Answer

SLO BASED SHORT QUESTION ANSWERS

Enthalpy Change

Q.13

What is chemical energetics?

Answer

Chemical energetics deals with the study of energy changes during chemical reactions. It tells us whether a reaction is exothermic (releases heat) or endothermic (absorbs heat). Example: Combustion of methane is exothermic: CH4 + 202 → CO2 + 2H2O + heat

Q.14

Define system and surroundings.

Answer

•System: Part of the universe under study (e.g., reactants and products). • Surroundings: Everything outside the system. Example: In a beaker where a reaction is happening, the contents are the system; the beaker and air are the surroundings.

Q.15

What is an open, closed, and isolated system?

Answer

•Open system: Can exchange both matter and energy. • Closed system: Only energy exchange is possible. • Isolated system: Neither energy nor matter can be exchanged. Example • Open: boiling water in an open pan: • Closed: sealed bottle of hot water • Isolated: thermos flask. • Charge density = Charge /Volume or radius?) • Higher charge density (small size or high means stronger ion-dipole charge) interaction and more exothermic hydration enthalpy. between

Q.16

Differentiate exothermic and endothermic reactions.

Answer

•Exothermic: Release heat, AH is negative. • Endothermic: Absorb heat, AH is positive. Examples • Exo: C + 02 → CO2 + heat • Endo: N2 + O2 → 2NO - requires heat

Q.17

What is enthalpy (AH)?

Answer

Enthalpy is the heat content of a system . It is a state function. at constant pressure. Unit: kJ/mol AH = (products) - H(reactants) the First Law of

Q.18

State Thermodynamics.

Answer

Energy cannot be created or destroyed, only transformed. Mathematically: AU = q + ₩ where AU = change in internal energy, q= heat, w = work done.

Q.19

What is internal energy (AU)?

Answer

It is the total energy stored in a system (kinetic + potential). AU = 9 + w Standard

Enthalpy Change

Q.20

Define heat of reaction.

Answer

It is the amount of heat absorbed or released during a chemical reaction at constant pressure. AH for: Example H2 + ½02 → H2O=-285.8 kJ/mol (exothermic)

Q.21
Define standard enthalpy of formation (AHf®).

Answer

Enthalpy change when 1 mole of compound is formed from its elements in their standard states. Example C (graphite) + Oz → CO2 AHf = -393.5 kJ/mol

Q.22

Define heat of combustion.

Answer

Heat change when 1 mole of a substance burns completely in oxygen. Example CH4 + 202 → CO2 + 2H20 AH = -890 kJ/mol

Q.23

What is enthalpy of neutralization?

Answer

Heat released when 1 mole of acid reacts with base to form water. Example Hcl + NaOH → NaCl + H2O AH = -57.1 kJ/mol

Q.24

Why is enthalpy of neutralization constant for strong acid-base pairs?

Answer

Because complete ionization occurs and same reaction (Ht + OH → HO) happens.

Q.25

Define endothermic reaction with example.

Answer

Reaction that absorbs heat. Example: Photosynthesis 6CО2 + 6H2О + sunlight → C.Hi206 + 02 is an 026. Why evaporation endothermic process? Ans. Because molecules absorb heat to overcome intermolecular forces and escape.

Illustration (added) - Surface Tension Forces Bulk (Uniform Pull) Surface (Net Downward Pull)

Q.27
State an example where AH is positive.

Answer

N2 + 02 → 2NO ДН = +180.5 kJ/mol Energetics Change and Calorie Content

Q.28

What is calorimetry?

Answer

It is the science of measuring heat changes during chemical reactions using a calorimeter.

Illustration (added) - Bomb Calorimeter Setup Bomb Thermometer Water Jacket surrounding Bomb

Q.29

Define specific heat capacity.

Answer

The amount of heat required to raise the temperature of 1 gram of a substance by 1°C. Unit: J/g. °C Water's value = 4.18 J/g. °C for heat (g)

Q.30

Give formula calculation.

Answer

9 = mcAT Where • m = mass (g) • c = specific heat • AT = change in temperature

Q.31

How can calorimetry be used to measure enthalpy?

Answer

By measuring the temperature change (AT) in a known mass of water: q = mcAT

Q.32

Why does dissolving NaOH increase temperature?

Answer

It's an exothermic process due to strong hydration energy of Nat and OHions.

Q.33

What is a thermochemical equation?

Answer

A balanced chemical equation that includes the heat change (AH). Example H2 + 202 → H2O(1); AH =-285.8 kJ/mol

Q.34

What is the standard state of an element?

Answer

Its most stable physical form at 298K and 1 atm pressure. Example Standard state of oxygen = 02(g)

Hess's Law

Q.35

What is Hess's Law?

Answer

The total enthalpy change is the same regardless of the path taken. Example C + 02 → CO2 =(C+½0г →СО).+(СО+½0г → C02)

Q.36

Why is Hess's Law useful?

Answer

It helps to calculate enthalpy changès that are difficult to measure directly using known enthalpies.

Q.37

What are state functions?

Answer

Properties that depend only on the initial and final states, not the path taken. Examples: Enthalpy, internal energy. Bond Energy and

Enthalpy Change

Q.38

What is bond enthalpy?

Answer

Energy required to break one mole of bonds in gaseous molecules. Example: Bond enthalpy of H-H = 436 kJ/mol

Q.39

What is the relationship between bond energy and enthalpy of reaction?

Answer

AH = E(Bonds broken) - (Bonds formed) More energy released in bond formation = exothermic attice Energy

Q.40

What is lattice energy?

Answer

Energy released when one mole of an ionic compound forms from gaseous ions. Example Na* (g) + C1(g) → NaCks) AH att=-787kJmol Energetics of Solution

Q.41

What factors affect lattice energy?

Answer

•Ionic charge (higher charge = more energy) • Ionic radius (smaller radius = more energy)

Q.42

Define hydration enthalpy.

Answer

Energy released when 1 mole of gaseous ions dissolve in water. DESCRIPTIVE QUESTIONS (EXERCISE) State and explain Hess's law. Give

Q.43

Define enthalpy of solution.

Answer

Heat change when 1 mole of a substance dissolves in a solvent.

Q.44

Why is dissolution of NH&NO3 endothermic?

Answer

Because the lattice energy absorbed is more than the hydration energy released

Born-Haber's Cycle

Q.45

What is the Born-Haber cycle?

Answer

A thermochemical cycle used to calculate lattice energy of ionic compounds. 046. Why is formation of NaCl exothermic? Ans. Due to large lattice energy released during crystal formation.

Q.47

Define spontaneous reaction.

Answer

A reaction that occurs on its own without external energy input. Example Rusting of iron

Illustration (added) - Mechanism of Iron Corrosion (Rusting) Water Droplet Anode: Fe → Fe²⁺ + 2e⁻ Cathode: O₂ + 4H⁺ + 4e⁻

Entropy

Q.48

What is entropy (AS)?

Answer

Measure of randomness or disorder. Higher AS = more disordered. Example: S→ L → G → increasing entropy

Free Energy Change

Q.49

Define free energy (4G). whether a reaction is

Answer

It tells spontaneous. AG = AH - TAS • AG < O: spontaneous • AG > 0: non-spontaneous

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