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Level 2 — RecallThermodynamics (Chemical)

Thermodynamics (Chemical)

40 marksprintable — key stays hidden on paper

Level: 2 (Recall & standard problems) Time: 30 minutes Total Marks: 40

Use R=8.314 J mol−1K−1R = 8.314\ \text{J mol}^{-1}\text{K}^{-1} where required. Assume ideal gas behaviour.

Q1. Define the following, giving one example of each: (a) an isolated system, (b) a closed system. State clearly what can and cannot be exchanged with the surroundings in each case. (4 marks)

Q2. Classify each of the following as a state function or a path function: internal energy UU, work ww, enthalpy HH, heat qq. Briefly explain the difference between the two categories. (4 marks)

Q3. State the first law of thermodynamics using the chemist's sign convention (ΔU=q+w\Delta U = q + w). A gas absorbs 150 J150\ \text{J} of heat and has 80 J80\ \text{J} of work done on it. Calculate ΔU\Delta U. (4 marks)

Q4. Two moles of an ideal gas expand isothermally and reversibly at 300 K300\ \text{K} from a volume of 5.0 L5.0\ \text{L} to 15.0 L15.0\ \text{L}. Calculate the work done by the gas. (4 marks)

Q5. (a) Derive the relationship Cp−Cv=nRC_p - C_v = nR for one mole of an ideal gas. (b) Given Cv,m=20.8 J mol−1K−1C_{v,m} = 20.8\ \text{J mol}^{-1}\text{K}^{-1} for an ideal gas, calculate Cp,mC_{p,m}. (5 marks)

Q6. Using standard enthalpies of formation: ΔHf∘(CO2,g)=−393.5 kJ mol−1,ΔHf∘(H2O,l)=−285.8 kJ mol−1,ΔHf∘(CH4,g)=−74.8 kJ mol−1\Delta H^\circ_f(\text{CO}_2, g) = -393.5\ \text{kJ mol}^{-1}, \quad \Delta H^\circ_f(\text{H}_2\text{O}, l) = -285.8\ \text{kJ mol}^{-1}, \quad \Delta H^\circ_f(\text{CH}_4, g) = -74.8\ \text{kJ mol}^{-1} Calculate the standard enthalpy of combustion of methane: CH4(g)+2O2(g)→CO2(g)+2H2O(l)\text{CH}_4(g) + 2\text{O}_2(g) \rightarrow \text{CO}_2(g) + 2\text{H}_2\text{O}(l) (4 marks)

Q7. State Hess's law and explain why it is a direct consequence of enthalpy being a state function. (3 marks)

Q8. Estimate ΔHrxn\Delta H_{\text{rxn}} for H2(g)+Cl2(g)→2HCl(g)\text{H}_2(g) + \text{Cl}_2(g) \rightarrow 2\text{HCl}(g) using the bond enthalpies: B(H–H)=436 kJ mol−1,B(Cl–Cl)=242 kJ mol−1,B(H–Cl)=431 kJ mol−1B(\text{H–H}) = 436\ \text{kJ mol}^{-1}, \quad B(\text{Cl–Cl}) = 242\ \text{kJ mol}^{-1}, \quad B(\text{H–Cl}) = 431\ \text{kJ mol}^{-1} (4 marks)

Q9. For a reaction at 298 K298\ \text{K}: ΔH=−92.2 kJ mol−1\Delta H = -92.2\ \text{kJ mol}^{-1} and ΔS=−198.8 J mol−1K−1\Delta S = -198.8\ \text{J mol}^{-1}\text{K}^{-1}. (a) Calculate ΔG\Delta G. (b) State whether the reaction is spontaneous at this temperature. (4 marks)

Q10. Write the relationship between standard Gibbs free energy and the equilibrium constant. For a reaction at 298 K298\ \text{K}, ΔG∘=−22.8 kJ mol−1\Delta G^\circ = -22.8\ \text{kJ mol}^{-1}. Calculate KK. (4 marks)

Answer keyMark scheme & solutions

Q1. (4 marks)

  • Isolated system: neither matter nor energy can be exchanged with surroundings. Example: an ideal (perfectly insulated, sealed) thermos flask / bomb calorimeter. (2) [1 for definition, 1 for example]
  • Closed system: energy (heat/work) can be exchanged but not matter. Example: a sealed flask of gas, or a stoppered reaction vessel. (2) [1 for definition, 1 for example]

Why: classification is based on what crosses the boundary.

Q2. (4 marks)

  • State functions: UU, HH — depend only on state (initial/final), not path. (1)
  • Path functions: ww, qq — depend on the route taken. (1)
  • Explanation: a state function's change is fixed by end states (ΔX=Xf−Xi\Delta X = X_f - X_i); path functions have different values for different paths between the same states. (2)

Q3. (4 marks)

  • Statement: energy of the universe is conserved; ΔU=q+w\Delta U = q + w, where qq is heat added to system, ww is work done on the system. (2)
  • q=+150 Jq = +150\ \text{J} (absorbed), w=+80 Jw = +80\ \text{J} (done on gas). (1)
  • ΔU=150+80=+230 J\Delta U = 150 + 80 = +230\ \text{J}. (1)

Q4. (4 marks)

  • Reversible isothermal work done on gas: w=−nRTln⁡(V2/V1)w = -nRT\ln(V_2/V_1). (1)
  • w=−(2)(8.314)(300)ln⁡(15/5)=−(4988.4)ln⁡3=−(4988.4)(1.0986)w = -(2)(8.314)(300)\ln(15/5) = -(4988.4)\ln 3 = -(4988.4)(1.0986). (1)
  • w=−5480 J=−5.48 kJw = -5480\ \text{J} = -5.48\ \text{kJ} (work done on gas). (1)
  • Work done by the gas =+5.48 kJ= +5.48\ \text{kJ}. (1)

Q5. (5 marks) (a) At constant VV: qV=ΔU=Cv dTq_V = \Delta U = C_v\,dT. At constant PP: qP=ΔH=Cp dTq_P = \Delta H = C_p\,dT. (1) Since H=U+PV=U+nRTH = U + PV = U + nRT (ideal gas), dH=dU+nR dTdH = dU + nR\,dT. (1) Dividing by dTdT: Cp=Cv+nRC_p = C_v + nR, hence Cp−Cv=nRC_p - C_v = nR. (1) (b) Cp,m=Cv,m+R=20.8+8.314=29.1 J mol−1K−1C_{p,m} = C_{v,m} + R = 20.8 + 8.314 = 29.1\ \text{J mol}^{-1}\text{K}^{-1}. (2)

Q6. (4 marks) ΔH∘=∑ΔHf∘(products)−∑ΔHf∘(reactants)\Delta H^\circ = \sum \Delta H^\circ_f(\text{products}) - \sum \Delta H^\circ_f(\text{reactants}) (1) =[(−393.5)+2(−285.8)]−[(−74.8)+0]= [(-393.5) + 2(-285.8)] - [(-74.8) + 0] (1) =(−393.5−571.6)−(−74.8)=−965.1+74.8= (-393.5 - 571.6) - (-74.8) = -965.1 + 74.8 (1) =−890.3 kJ mol−1= -890.3\ \text{kJ mol}^{-1} (1)

Q7. (3 marks)

  • Hess's law: the total enthalpy change for a reaction is the same regardless of the number of steps or route taken, provided initial and final states are the same. (2)
  • Because HH is a state function, ΔH\Delta H depends only on initial and final states, not the path; hence intermediate routes sum to the same overall ΔH\Delta H. (1)

Q8. (4 marks) ΔH=∑B(bonds broken)−∑B(bonds formed)\Delta H = \sum B(\text{bonds broken}) - \sum B(\text{bonds formed}) (1) Broken: B(H–H)+B(Cl–Cl)=436+242=678 kJB(\text{H–H}) + B(\text{Cl–Cl}) = 436 + 242 = 678\ \text{kJ}. (1) Formed: 2×B(H–Cl)=2×431=862 kJ2 \times B(\text{H–Cl}) = 2 \times 431 = 862\ \text{kJ}. (1) ΔH=678−862=−184 kJ mol−1\Delta H = 678 - 862 = -184\ \text{kJ mol}^{-1} (1)

Q9. (4 marks) (a) ΔG=ΔH−TΔS\Delta G = \Delta H - T\Delta S; convert ΔS=−0.1988 kJ mol−1K−1\Delta S = -0.1988\ \text{kJ mol}^{-1}\text{K}^{-1}. (1) ΔG=−92.2−(298)(−0.1988)=−92.2+59.24=−32.96 kJ mol−1\Delta G = -92.2 - (298)(-0.1988) = -92.2 + 59.24 = -32.96\ \text{kJ mol}^{-1}. (2) (b) ΔG<0⇒\Delta G < 0 \Rightarrow reaction is spontaneous at 298 K. (1)

Q10. (4 marks)

  • Relationship: ΔG∘=−RTln⁡K\Delta G^\circ = -RT\ln K. (1)
  • ln⁡K=−ΔG∘RT=−−22800(8.314)(298)=228002477.6=9.203\ln K = -\dfrac{\Delta G^\circ}{RT} = -\dfrac{-22800}{(8.314)(298)} = \dfrac{22800}{2477.6} = 9.203. (1)
  • K=e9.203≈9.9×103K = e^{9.203} \approx 9.9 \times 10^{3}. (2)
[
  {"claim":"Q4 reversible isothermal work on gas ≈ -5480 J","code":"w=-2*8.314*300*log(3); result = abs(float(w)-(-5480))<5"},
  {"claim":"Q6 combustion enthalpy of methane = -890.3 kJ/mol","code":"dH=(-393.5+2*(-285.8))-(-74.8); result = abs(float(dH)-(-890.3))<0.05"},
  {"claim":"Q8 bond enthalpy estimate = -184 kJ/mol","code":"dH=(436+242)-2*431; result = dH==-184"},
  {"claim":"Q9 Gibbs energy ≈ -32.96 kJ/mol","code":"dG=-92.2-298*(-0.1988); result = abs(float(dG)-(-32.96))<0.05"},
  {"claim":"Q10 equilibrium constant K ≈ 9900","code":"K=exp(22800/(8.314*298)); result = abs(float(K)-9900)/9900<0.05"}
]