Unit 6: Chemical Energetics — Short Questions

Ans.
• Exothermic Reaction: Heat is released; ΔH is negative.
Example: Combustion of methane.
• Endothermic Reaction: Heat is absorbed; ΔH is positive.
Example: Decomposition of calcium carbonate.

Ans. Enthalpy (H) is the total heat content of a system at constant pressure. It is the sum of the internal energy and the product of pressure and volume.

Ans.
• 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)

Ans.
• Standard Enthalpy of Reaction (ΔH°r): Enthalpy change when a chemical reaction occurs under standard conditions.
• Standard Enthalpy of Formation (ΔH°f): Enthalpy change when 1 mole of a compound is formed from its elements in standard states.

Ans.
(i) Standard Enthalpy of Solution
ΔH when 1 mole of a substance dissolves in a solvent.
Example: NaCl(s) → Na+(aq) + Cl-(aq)

(ii) Standard Enthalpy of Hydration
ΔH when 1 mole of gaseous ions hydrated.
Example: Na+(g) → Na+(aq)

(iii) Standard Enthalpy of Atomization
ΔH when 1 mole of gaseous atoms is formed from an element in standard state.
Example: ½H2(g) → H(g)

(iv) Standard Enthalpy of Combustion
ΔH when 1 mole of a substance completely burns in oxygen.
Example:
CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)

Ans. 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.

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

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

Ans. Bonds broken 4 C – H bonds in CH4
2 O = O double bonds
• Bonds formed:
2 C = O bonds in CO2
4 O – H bonds in H2O

Ans. For a reaction to be spontaneous, ΔG must be negative
Always spontaneous when:
ΔH is negative and ΔS is positive

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

Ans.
• Charge density = Charge /Volume (or radius')
• Higher charge density (small size or high charge) means stronger ion-dipole interaction and more exothermic hydration enthalpy.

Ans. 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 + 2O2 → CO2 + 2H2O + heat

Ans.
• 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.

Ans.
• 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.

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

Ans. Enthalpy is the heat content of a system at constant pressure. It is a state function.
Unit: kJ/mol
ΔH = H(products) – H(reactants)

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

Ans. It is the total energy stored in a system (kinetic + potential).
ΔU = q + w

Ans. It is the amount of heat absorbed or released during a chemical reaction at constant pressure.
Example
ΔH for:
H2 + ½O2 → H2O = -285.8 kJ/mol
(exothermic)

Ans. Enthalpy change when 1 mole of compound is formed from its elements in their standard states.
Example
C (graphite) + O2 → CO2
ΔHf° = -393.5 kJ/mol

Ans. Heat change when 1 mole of a substance burns completely in oxygen.
Example
CH4 + 2O2 → CO2 + 2H2O
ΔH = -890 kJ/mol

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

Ans. Because complete ionization occurs and same reaction (H+ + OH- → H2O) happens.

Ans. Reaction that absorbs heat
Example: Photosynthesis
6CO2 + 6H2O + sunlight → C6H12O6 + O2

Ans. Because molecules absorb heat to overcome intermolecular forces and escape.

Ans. N2 + O2 → 2NO
ΔH = +180.5 kJ/mol

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

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

Ans.
q = mcΔT
Where
• m = mass (g)
• c = specific heat
• ΔT = change in temperature

Ans. By measuring the temperature change (ΔT) in a known mass of water:
q = mcΔT

Ans. It's an exothermic process due to strong hydration energy of Na+ and OH- ions.

Ans. A balanced chemical equation that includes the heat change (ΔH).
Example
H2 + ½O2 → H2O(l); ΔH = -285.8 kJ/mol

Ans. Its most stable physical form at 298 K and 1 atm pressure.
Example
Standard state of oxygen = O2(g)

Ans. The total enthalpy change is the same regardless of the path taken.
Example
C + O2 → CO2
= (C + ½O2 → CO) + (CO + ½O2 → CO2)

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

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

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

Ans. ΔH = Σ(Bonds broken) – Σ(Bonds formed)
More energy released in bond formation = exothermic

Ans. Energy released when one mole of an ionic compound forms from gaseous ions.
Example
Na+(g) + Cl-(g) → NaCl(s) ΔH°latt = -787kJmol-1

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

Ans. Energy released when 1 mole of gaseous ions dissolve in water.

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

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

Ans. A thermochemical cycle used to calculate lattice energy of ionic compounds.

Ans. Due to large lattice energy released during crystal formation.

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

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

Ans. It tells whether a reaction is spontaneous.
ΔG = ΔH – TΔS
• ΔG < 0: spontaneous
• ΔG > 0: non-spontaneous

Ans. See Q.7 from theory.

Ans. See Q.13 from theory.

Ans. Given • Mass of NaOH = 0.400g
• Mass of water = 100.0g
• Initial temperature = 25.00°C
• Final temperature = 26.03°C
• ΔT = 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)
q = m c ΔT
q = 100.0g × 4.18J/g°C × 1.03°C
q = 430.54J

Answer (i) qwater = 430.54J

(ii): Calculate ΔH for the solution process (in kJ/mol)
1 Molar mass of NaOH = 23 +16 + 1 = 40 g/mol
2. Moles of NaOH
n = 0.400g / 40g / mol = 0.010mol

Now use ΔH = q / n = 430.54J / 0.010mol = 43054J / mol = 43.05kJ / mol

Answer (ii) ΔH(solution) = -43.05kJ/mol

Ans. Target reaction NH3(g) + HCl(g) → NH4Cl(aq)

Given 1 NH3(g) → NH3(aq) ΔH = -35.16kJ / mol
2. HCl(g) → HCl(aq) ΔH = -72.41kJ / mol
3. NH3(aq) + HCl(aq) → NH4Cl(aq) ΔH = -51.48kJ / mol

Using Hess's Law

Add all three steps to get the overall reaction:
NH3(g) + HCl(g) → NH4Cl(aq)
ΔH = (-35.16) + (-72.41) + (-51.48)
ΔH = -159.05kJ / mol

Answer ΔH = -159.05kJ / mol

Ans. Given information 1. Heat of combustion of ethyl alcohol:
C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l) ΔH = -1367kJ / mol

ΔH°f of CO2(g) = -393.7kJ/mol
ΔH°f of H2O(l) = -285.8kJ/mol

Step 1 Use Hess's Law Equation
According to Hess's Law:
ΔH°combustion = [∑ΔH°f (products)] - [∑ΔH°f (reactants)]

Rewriting to find the ΔHf° of C2H5OH(l):
ΔH°f (C2H5OH) = [∑ΔH°f (products)] – ΔH°combustion

Step 2 Products
• 2 mol CO2: 2 × (-393.7) = -787.4kJ
• 3 mol H2O: 3 × (-285.8) = -857.4kJ
Total ΔH°f of products =

• 787.4 + (-857.4) = -1644.8kJ

Now calculate ΔH°f of ethanol:
ΔH°f (C2H5OH) = -1644.8 – (-1367) = -1644.8 + 1367 = -277.8kJ / mol

Answer ΔH°f (C2H5OH) = -277.8kJ / mol

Ans. Given Reaction ΔH(kJ/mol)
K(s) + 1/2 Br2(l) → KBr(s) -392
K(s) → K(g) +90
K(g) → K+(g) + e- +420
1/2 Br2(l) → Br(g) +112
Br(g) + e- → Br-(g) -324

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+(g) + Br-(g) → KBr(s) Lattice energy = ΔH_latt

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) ΔH = +90kJ / mol

2. Ionization of K(g) to K+(g):
K(g) → K+(g) + e- ΔH = +420kJ / mol

3. Dissociation of Br2(l) to Br(g):
½ Br2(l) → Br(g) ΔH = +112kJ / mol

4. Electron affinity of Br(g) to Br-(g):
Br(g) + e- → Br-(g) ΔH = -324kJ / mol

5. Formation of KBr(s) from gaseous ions (Lattice energy):
K+(g) + Br-(g) → KBr(s) ΔH_latt = ?

Step 3 Write the enthalpy cycle equation
The total enthalpy change for the formation of KBr(s) from K(s) and ½ Br2(l) is:
ΔHf = Sublimation + Ionization + Dissociation + Electron affinity + Lattice energy
-392 = (90 + 420 + 112 – 324) + ΔH_latt

Calculate the sum inside the parentheses:
-392 = 298 + ΔH_latt
ΔH_latt = -392 – 298 = -690kJ/mol

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

Ans. Given Reaction
2Ca(s) + O2(g) → 2CaO(s)
• ΔH°reaction = -1270.2kJmol-1
• 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:
ΔS°surroundings = - ΔH°reaction / T

Here • ΔH°reaction should be in Joules (not kJ) for consistency with entropy units.
• Temperature in Kelvin.

Step 2 Convert enthalpy change to Joules
ΔH°reaction = -1270.2kJ / mol = -1270.2 × 10³ = -1,270,200 J / mol

Step 3 Calculate ΔS surrounding°
ΔS°surrounding = -1,270,200 / 298 = +4262.75JK^-1 mol^-1

Answer ΔS°surrounding = +4262.75JK^-1 mol^-1

Ans.
Reaction
CaSO4(s) → Ca2+(aq) + SO42-(aq)

Given data Enthalpy of formation (ΔH°f) (in kJ/mol)
ΔH°f(CaSO4(s)) = -1432.7
ΔH°f(Ca2+(aq)) = -543.0
ΔH°f(SO42-(aq)) = -907.5

Standard Entropies So (in J/K mol):
S°(CaSO4(s)) = 106.7
S°(Ca2+(aq)) = -55.2
S°(SO42-(aq)) = 17.2

Step 1 Calculate ΔH° of the reaction
ΔH° = ∑ΔH°f(products) – ∑ΔH°f(reactants)
= [(-543.0) + (-907.5)] – (-1432.7)
= (-1450.5) + 1432.7 = -17.8kJ / mol Answer

Step 2 Calculate ΔS° of the reaction
ΔS° = ∑S°(products) – ∑S°(reactants)
= [(-55.2) + (17.2)] – (106.7) = -38.0 – 106.7 = -144.7J / K Answer

Convert to kJ for use in ΔG° calculation:
ΔS° = -0.1447kJ/K

Step 3 Calculate ΔG° of the reaction
ΔG° = ΔH° - TΔS°
= -17.8 – (298 × (-0.1447)) = -17.8 + 43.1 = +25.3kJ / mol Answer

Step 4 Discuss spontaneity
ΔG° = +25.3 kJ/mol, which is positive.
Therefore, the reaction is non-spontaneous under standard conditions.