Exercises
1: A thermodynamic state function is a quantity (choose correct answer)
Answer
(ii) whose value is independent of path.
A state function depends only on the current state of the system, not on how that state was reached. Examples: internal energy (U), enthalpy (H), entropy (S), Gibbs energy (G).
2: For the process to occur under adiabatic conditions, the correct condition is
Answer
(iii) q = 0
Adiabatic process means no heat exchange between system and surroundings. Therefore q = 0. By first law: ΔU = q + w = 0 + w = w.
3: The enthalpies of all elements in their standard states are
Answer
(ii) zero
By convention, the standard enthalpy of formation of all elements in their standard state (most stable allotrope at 298 K and 1 bar) is taken as zero.
4: ΔU of combustion of methane is –X kJ mol⁻¹. The value of ΔH is
Answer
(iii) < ΔU
Combustion of methane: CH₄(g) + 2O₂(g)
⇒ CO₂(g) + 2H₂O(l)
Δng = moles of gaseous products − moles of gaseous reactants = 1 − 3 = −2
ΔH = ΔU + ΔngRT = −X + (−2)(RT) = −X − 2RT
Since −X − 2RT < −X, ΔH < ΔU.
Q 5.5: Enthalpy of formation of CH₄(g)
Given: ΔH°comb(CH₄) = −890.3 kJ mol⁻¹, ΔH°comb(C) = −393.5 kJ mol⁻¹, ΔH°comb(H₂) = −285.8 kJ mol⁻¹
Answer: (i) −74.8 kJ mol⁻¹
Using Hess's law: ΔH°f(CH₄) = (1) + 2×(2) − (3) = −393.5 + 2(−285.8) − (−890.3) = −74.8 kJ mol⁻¹
6: A reaction A + B → C + D + q has positive entropy change. The reaction will be:
Answer
(v) possible at any temperature.
ΔH < 0 (exothermic) and ΔS > 0, so ΔG = ΔH − TΔS is always negative
⇒ spontaneous at all temperatures.
Q 5.7: 701 J absorbed, 394 J work done by system. ΔU = ?
ΔU = q + w = 701 + (−394) = 307 J
Q 5.8: ΔH for combustion of NH₂CN(s) at 298 K, ΔU = −742.7 kJ mol⁻¹
NH₂CN(s) + (3/2)O₂(g) → N₂(g) + CO₂(g) + H₂O(l)
Δng = (1+1) − 3/2 = 1/2
ΔH = ΔU + ΔngRT = −742.7 + (0.5)(8.314 × 10⁻³)(298) = −742.7 + 1.239 = −741.5 kJ mol⁻¹
9: Heat to raise 60.0 g Al from 35°C to 55°C; Cm(Al) = 24 J mol⁻¹ K⁻¹
Answer
n = 60.0/27 = 2.222 mol; ΔT = 20 K
q = 2.222 × 24 × 20 = 1066.7 J ≈ 1.067 kJ
Q 5.10: Enthalpy change: freeze 1.0 mol water at 10°C to ice at −10°C
Step 1 – cool water 10°C → 0°C: ΔH₁ = 1.0 × 75.3 × (−10) = −753 J
Step 2 – freeze at 0°C: ΔH₂ = −6030 J
Step 3 – cool ice 0°C → −10°C: ΔH₃ = 1.0 × 36.8 × (−10) = −368 J
Total ΔH = −753 + (−6030) + (−368) = −7151 J = −7.151 kJ mol⁻¹
Q 5.11: Heat released forming 35.2 g CO₂; ΔH°c(C) = −393.5 kJ mol⁻¹
n(CO₂) = 35.2/44 = 0.8 mol; Heat = 0.8 × 393.5 = 314.8 kJ
Q 5.12: ΔrH for N₂O₄(g) + 3CO(g) → N₂O(g) + 3CO₂(g)
ΔrH = [81 + 3(−393)] − [9.7 + 3(−110)] = −1098 − (−320.3) = −777.7 kJ mol⁻¹
Q 5.13: Standard enthalpy of formation of NH₃(g)
N₂(g) + 3H₂(g) → 2NH₃(g); ΔH = −92.4 kJ mol⁻¹
ΔH°f(NH₃) = −92.4/2 = −46.2 kJ mol⁻¹
Q 5.14: Standard enthalpy of formation of CH₃OH(l)
Target: C(s) + 2H₂(g) + ½O₂(g) → CH₃OH(l)
= (2) + 2×(3) − (1) = −393 + 2(−286) − (−726) = −239 kJ mol⁻¹
Q 5.15: Bond enthalpy of C–Cl in CCl₄(g)
ΔH for CCl₄(g) → C(g) + 4Cl(g):
= −30.5 + 135.5 + 715 + 2(242) = 1304 kJ mol⁻¹
Bond enthalpy C–Cl = 1304/4 = 326 kJ mol⁻¹
Q 5.16: For isolated system ΔU = 0, what is ΔS?
ΔS > 0. For any spontaneous process in an isolated system, entropy increases (Second Law).
17: At what temperature does 2A + B → C become spontaneous?
Answer
ΔH = 400 kJ mol⁻¹, ΔS = 0.2 kJ K⁻¹ mol⁻¹
T = ΔH/ΔS = 400/0.2 = 2000 K (spontaneous above this temperature)
Q 5.18: Signs of ΔH and ΔS for 2Cl(g) → Cl₂(g)
ΔH < 0 (bond formation, exothermic); ΔS < 0 (2 mol gas → 1 mol gas, less randomness)
19: ΔG for 2A(g) + B(g) → 2D(g); ΔU = −10.5 kJ, ΔS = −44.1 J K⁻¹
Answer
Δng = 2 − 3 = −1; ΔH = −10.5 + (−1)(8.314×10⁻³)(298) = −12.978 kJ
ΔG = −12978 − 298(−44.1) = −12978 + 13141.8 = +163.8 J mol⁻¹
ΔG > 0 → reaction is non-spontaneous
Q 5.20: ΔG° when K = 10, T = 300 K
ΔG° = −RT ln K = −(8.314)(300)(2.3026) = −5744 J mol⁻¹ ≈ −5.74 kJ mol⁻¹
Q 5.21: Thermodynamic stability of NO(g)
ΔH°f(NO) = +90 kJ mol⁻¹ (positive → thermodynamically unstable w.r.t. elements). NO also converts exothermically to NO₂ (ΔH = −74 kJ mol⁻¹), making it thermodynamically unstable. However it is kinetically stable under normal conditions.
Q 5.22: Entropy change in surroundings when 1.00 mol H₂O(l) forms; ΔH°f = −286 kJ mol⁻¹
ΔSsurr = +286000/298 = +959.7 J K⁻¹ mol⁻¹
Important Extra Questions
Q A: Difference between enthalpy and internal energy; when is ΔH = ΔU?
Internal energy U = total energy of system. Enthalpy H = U + pV (includes pV work).
ΔH = ΔU + ΔngRT for reactions with gases. ΔH = ΔU when Δng = 0 (e.g., C + O₂ → CO₂).
Q B: State and explain Hess's Law with an example.
Hess's Law: Total enthalpy change is independent of the path taken (state function property).
Example: C + O₂ → CO₂ directly (−393.5 kJ) = C → CO (−110.5 kJ) + CO → CO₂ (−283.0 kJ) = −393.5 kJ. ✓
Q C: Gibbs energy and criteria for spontaneity.
ΔG = ΔH − TΔS. Spontaneous if ΔG < 0; at equilibrium ΔG = 0; non-spontaneous if ΔG > 0.
When ΔH − and ΔS +: always spontaneous. When ΔH + and ΔS −: never spontaneous.
Q D: Bond enthalpy and its use in calculating reaction enthalpy.
Bond enthalpy = energy to break 1 mol of bond in gas phase. ΔrH = Σ(bonds broken) − Σ(bonds formed).
Example: H₂ + Cl₂ → 2HCl: ΔH = (436 + 242) − 2(431) = −184 kJ mol⁻¹.
Q E: Cp vs Cv; derive Cp − Cv = R.
Cv = heat at constant volume; Cp = heat at constant pressure (includes expansion work).
Derivation: H = U + nRT → dH = dU + nR dT → Cp = Cv + R (per mole, ideal gas).