Chapter 06: Chemical Energetics
Short Questions & Flashcards Study Portal
Short Questions
Enthalpy Change
Q.1
and Endothermic Reactions.
Answer
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
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Entropy
Free Energy Change
Q.3
(S) and Gibbs Free Energy (G):
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Q.3
its two applications.
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Q.4
Distinguish and standard enthalpy of reaction standard enthalpy of Formation.
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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
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.
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
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.
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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
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
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
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.
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Q.15
What is an open, closed, and isolated system?
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Q.16
Differentiate exothermic and endothermic reactions.
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Q.17
What is enthalpy (AH)?
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Q.18
State Thermodynamics.
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Q.19
What is internal energy (AU)?
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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®).
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Q.22
Define heat of combustion.
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Q.23
What is enthalpy of neutralization?
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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.
Q.27 State an example where AH is positive.
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Q.28
What is calorimetry?
Answer
It is the science of measuring heat changes during chemical reactions using a calorimeter.
Q.29
Define specific heat capacity.
Answer
Q.30
Give formula calculation.
Answer
Q.31
How can calorimetry be used to measure enthalpy?
Answer
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
Q.34
What is the standard state of an element?
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Hess's Law
Q.35
What is Hess's Law?
Answer
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
Q.39
What is the relationship between bond energy and enthalpy of reaction?
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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
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
Entropy
Q.48
What is entropy (AS)?
Answer
Free Energy Change
Q.49
Define free energy (4G). whether a reaction is
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