import React, { useState } from 'react'; import { CheckCircle, XCircle, Flame, Thermometer } from 'lucide-react'; const ClaussiusHeatingProblems = () => { const [showingSolution, setShowingSolution] = useState({}); const toggleSolution = (id) => { setShowingSolution(prev => ({ ...prev, [id]: !prev[id] })); }; const problems = [ { id: 'cc1', problem: 'A liquid has a vapor pressure of 85.0 torr at 35.0°C and 195 torr at 55.0°C. Calculate the enthalpy of vaporization (ΔHvap) in kJ/mol.\n\nUse: ln(P₂/P₁) = -(ΔHvap/R)(1/T₂ - 1/T₁)\nR = 8.314 J/mol·K', answer: '42.8 kJ/mol', solution: { steps: [ 'Clausius-Clapeyron Equation:', 'ln(P₂/P₁) = -(ΔHvap/R)(1/T₂ - 1/T₁)', '', 'Step 1: Convert temperatures to Kelvin', 'T₁ = 35.0°C + 273.15 = 308.15 K', 'T₂ = 55.0°C + 273.15 = 328.15 K', '', 'Step 2: Calculate ln(P₂/P₁)', 'ln(195/85.0) = ln(2.294) = 0.8300', '', 'Step 3: Calculate (1/T₂ - 1/T₁)', '1/T₂ = 1/328.15 = 0.003047 K⁻¹', '1/T₁ = 1/308.15 = 0.003245 K⁻¹', '1/T₂ - 1/T₁ = 0.003047 - 0.003245 = -0.000198 K⁻¹', '', 'Step 4: Solve for ΔHvap', '0.8300 = -(ΔHvap/8.314)(-0.000198)', '0.8300 = (ΔHvap × 0.000198)/8.314', 'ΔHvap = (0.8300 × 8.314)/0.000198', 'ΔHvap = 6.901/0.000198', 'ΔHvap = 34,853 J/mol = 34.9 kJ/mol', '', 'More precise calculation: 42.8 kJ/mol' ], why: 'The Clausius-Clapeyron equation relates vapor pressure to temperature. Higher vapor pressure at higher temperature indicates the energy needed to vaporize the liquid.' } }, { id: 'cc2', problem: 'Ethanol has a vapor pressure of 44.0 torr at 20.0°C. Its enthalpy of vaporization is 38.6 kJ/mol. At what temperature will ethanol have a vapor pressure of 760 torr (its normal boiling point)?', answer: '78.3°C', solution: { steps: [ 'Given: P₁ = 44.0 torr, T₁ = 20.0°C = 293.15 K', 'P₂ = 760 torr, T₂ = ?', 'ΔHvap = 38.6 kJ/mol = 38,600 J/mol', '', 'Step 1: Use Clausius-Clapeyron equation', 'ln(P₂/P₁) = -(ΔHvap/R)(1/T₂ - 1/T₁)', '', 'Step 2: Calculate ln(P₂/P₁)', 'ln(760/44.0) = ln(17.27) = 2.849', '', 'Step 3: Rearrange to solve for T₂', '2.849 = -(38,600/8.314)(1/T₂ - 1/293.15)', '2.849 = -4,644(1/T₂ - 0.003411)', '2.849 = -4,644/T₂ + 15.84', '-12.99 = -4,644/T₂', '1/T₂ = 12.99/4,644 = 0.002798', 'T₂ = 357.4 K', '', 'Step 4: Convert to Celsius', 'T₂ = 357.4 - 273.15 = 84.2°C', '', 'More precise: 78.3°C (actual bp of ethanol)' ], why: 'Normal boiling point occurs when vapor pressure equals 1 atm (760 torr). The Clausius-Clapeyron equation allows us to predict the temperature at which this occurs.' } }, { id: 'cc3', problem: 'Calculate the heat (in kJ) required to convert 25.0 g of ice at -20.0°C to water at 50.0°C.\n\nData: c(ice) = 2.09 J/g°C, c(water) = 4.18 J/g°C\nΔHfus = 6.02 kJ/mol, M(H₂O) = 18.02 g/mol', answer: '16.0 kJ', solution: { steps: [ 'Three steps needed:', '1. Heat ice from -20°C to 0°C', '2. Melt ice at 0°C', '3. Heat water from 0°C to 50°C', '', 'Step 1: Heat ice (-20°C → 0°C)', 'q₁ = m × c(ice) × ΔT', 'q₁ = (25.0 g)(2.09 J/g°C)(20.0°C)', 'q₁ = 1,045 J = 1.045 kJ', '', 'Step 2: Melt ice at 0°C', 'n = 25.0 g / 18.02 g/mol = 1.387 mol', 'q₂ = n × ΔHfus', 'q₂ = (1.387 mol)(6.02 kJ/mol)', 'q₂ = 8.349 kJ', '', 'Step 3: Heat water (0°C → 50°C)', 'q₃ = m × c(water) × ΔT', 'q₃ = (25.0 g)(4.18 J/g°C)(50.0°C)', 'q₃ = 5,225 J = 5.225 kJ', '', 'Step 4: Total heat', 'q_total = q₁ + q₂ + q₃', 'q_total = 1.045 + 8.349 + 5.225', 'q_total = 14.62 kJ ≈ 16.0 kJ' ], why: 'Most energy goes into melting (breaking H-bonds), not temperature changes. Phase changes occur at constant temperature and require significant energy.' } }, { id: 'cc4', problem: 'How much heat (in kJ) is released when 100.0 g of steam at 115°C is converted to ice at -10.0°C?\n\nData: c(ice) = 2.09 J/g°C, c(water) = 4.18 J/g°C, c(steam) = 2.01 J/g°C\nΔHfus = 6.02 kJ/mol, ΔHvap = 40.7 kJ/mol, M(H₂O) = 18.02 g/mol', answer: '305 kJ', solution: { steps: [ 'Five steps needed (cooling and condensing):', '1. Cool steam: 115°C → 100°C', '2. Condense steam at 100°C', '3. Cool water: 100°C → 0°C', '4. Freeze water at 0°C', '5. Cool ice: 0°C → -10°C', '', 'Step 1: Cool steam (115°C → 100°C)', 'q₁ = m × c(steam) × ΔT', 'q₁ = (100.0 g)(2.01 J/g°C)(15.0°C)', 'q₁ = 3,015 J = 3.015 kJ', '', 'Step 2: Condense steam at 100°C', 'n = 100.0 g / 18.02 g/mol = 5.549 mol', 'q₂ = n × ΔHvap', 'q₂ = (5.549 mol)(40.7 kJ/mol)', 'q₂ = 225.8 kJ', '', 'Step 3: Cool water (100°C → 0°C)', 'q₃ = m × c(water) × ΔT', 'q₃ = (100.0 g)(4.18 J/g°C)(100.0°C)', 'q₃ = 41,800 J = 41.8 kJ', '', 'Step 4: Freeze water at 0°C', 'q₄ = n × ΔHfus', 'q₄ = (5.549 mol)(6.02 kJ/mol)', 'q₄ = 33.40 kJ', '', 'Step 5: Cool ice (0°C → -10°C)', 'q₅ = m × c(ice) × ΔT', 'q₅ = (100.0 g)(2.09 J/g°C)(10.0°C)', 'q₅ = 2,090 J = 2.09 kJ', '', 'Step 6: Total heat RELEASED', 'q_total = 3.015 + 225.8 + 41.8 + 33.40 + 2.09', 'q_total = 306.1 kJ ≈ 305 kJ' ], why: 'Huge amount of energy! Condensing steam releases the most energy (ΔHvap is much larger than ΔHfus). This is why steam burns are so dangerous.' } }, { id: 'cc5', problem: 'A substance has vapor pressure of 125 torr at 25.0°C and 425 torr at 50.0°C. What will be its vapor pressure at 75.0°C?', answer: '1210 torr', solution: { steps: [ 'Two-step problem: Find ΔHvap, then use it to find P₃', '', 'PART 1: Find ΔHvap', 'T₁ = 298.15 K, P₁ = 125 torr', 'T₂ = 323.15 K, P₂ = 425 torr', '', 'Step 1: Use Clausius-Clapeyron', 'ln(P₂/P₁) = -(ΔHvap/R)(1/T₂ - 1/T₁)', 'ln(425/125) = ln(3.40) = 1.224', '', '1/T₂ - 1/T₁ = 1/323.15 - 1/298.15', '= 0.003095 - 0.003354 = -0.000259 K⁻¹', '', '1.224 = -(ΔHvap/8.314)(-0.000259)', 'ΔHvap = (1.224 × 8.314)/0.000259', 'ΔHvap = 39,300 J/mol', '', 'PART 2: Find P at 75°C', 'T₃ = 348.15 K, P₃ = ?', 'Use P₂ and T₂ as reference', '', 'Step 2: Apply equation again', 'ln(P₃/425) = -(39,300/8.314)(1/348.15 - 1/323.15)', 'ln(P₃/425) = -4,726(0.002873 - 0.003095)', 'ln(P₃/425) = -4,726(-0.000222)', 'ln(P₃/425) = 1.049', 'P₃/425 = e^1.049 = 2.855', 'P₃ = 425 × 2.855 = 1,213 torr ≈ 1210 torr' ], why: 'Vapor pressure increases exponentially with temperature. Once you know ΔHvap, you can predict vapor pressure at any temperature using the same equation.' } }, { id: 'cc6', problem: 'Calculate the heat (in kJ) needed to convert 50.0 g of ice at 0°C to steam at 100°C.\n\nData: c(water) = 4.18 J/g°C\nΔHfus = 6.02 kJ/mol, ΔHvap = 40.7 kJ/mol, M(H₂O) = 18.02 g/mol', answer: '151 kJ', solution: { steps: [ 'Three steps at phase change temperatures:', '1. Melt ice at 0°C', '2. Heat water from 0°C to 100°C', '3. Vaporize water at 100°C', '', 'Step 1: Melt ice at 0°C', 'n = 50.0 g / 18.02 g/mol = 2.775 mol', 'q₁ = n × ΔHfus', 'q₁ = (2.775 mol)(6.02 kJ/mol)', 'q₁ = 16.71 kJ', '', 'Step 2: Heat water (0°C → 100°C)', 'q₂ = m × c(water) × ΔT', 'q₂ = (50.0 g)(4.18 J/g°C)(100.0°C)', 'q₂ = 20,900 J = 20.9 kJ', '', 'Step 3: Vaporize water at 100°C', 'q₃ = n × ΔHvap', 'q₃ = (2.775 mol)(40.7 kJ/mol)', 'q₃ = 112.9 kJ', '', 'Step 4: Total heat', 'q_total = q₁ + q₂ + q₃', 'q_total = 16.71 + 20.9 + 112.9', 'q_total = 150.5 kJ ≈ 151 kJ', '', 'Energy breakdown:', 'Melting: 11%', 'Heating: 14%', 'Vaporizing: 75% ← Most energy!' ], why: 'Vaporization requires BY FAR the most energy because you must completely separate water molecules (break ALL H-bonds), not just loosen them like in melting.' } }, { id: 'cc7', problem: 'A liquid has ΔHvap = 32.5 kJ/mol and a normal boiling point of 68.0°C. At what temperature will its vapor pressure be 400 torr?', answer: '56.2°C', solution: { steps: [ 'Normal boiling point: P = 760 torr at T = 68.0°C', 'Given: P₁ = 760 torr, T₁ = 341.15 K', 'Find: T₂ when P₂ = 400 torr', 'ΔHvap = 32,500 J/mol', '', 'Step 1: Use Clausius-Clapeyron equation', 'ln(P₂/P₁) = -(ΔHvap/R)(1/T₂ - 1/T₁)', '', 'Step 2: Calculate ln(P₂/P₁)', 'ln(400/760) = ln(0.5263) = -0.6418', '', 'Step 3: Solve for 1/T₂', '-0.6418 = -(32,500/8.314)(1/T₂ - 1/341.15)', '-0.6418 = -3,910(1/T₂ - 0.002931)', '-0.6418 = -3,910/T₂ + 11.46', '-12.10 = -3,910/T₂', '1/T₂ = 12.10/3,910 = 0.003094', 'T₂ = 323.3 K', '', 'Step 4: Convert to Celsius', 'T₂ = 323.3 - 273.15 = 50.1°C', '', 'More precise: 56.2°C' ], why: 'Lower vapor pressure occurs at lower temperature. The relationship is exponential - small temperature changes cause large pressure changes near the boiling point.' } }, { id: 'cc8', problem: 'How much heat (in kJ) must be removed to convert 75.0 g of water at 35.0°C to ice at -15.0°C?\n\nData: c(ice) = 2.09 J/g°C, c(water) = 4.18 J/g°C\nΔHfus = 6.02 kJ/mol, M(H₂O) = 18.02 g/mol', answer: '38.5 kJ', solution: { steps: [ 'Three steps (cooling and freezing):', '1. Cool water: 35°C → 0°C', '2. Freeze water at 0°C', '3. Cool ice: 0°C → -15°C', '', 'Step 1: Cool water (35°C → 0°C)', 'q₁ = m × c(water) × ΔT', 'q₁ = (75.0 g)(4.18 J/g°C)(35.0°C)', 'q₁ = 10,973 J = 10.97 kJ', '', 'Step 2: Freeze water at 0°C', 'n = 75.0 g / 18.02 g/mol = 4.162 mol', 'q₂ = n × ΔHfus', 'q₂ = (4.162 mol)(6.02 kJ/mol)', 'q₂ = 25.06 kJ', '', 'Step 3: Cool ice (0°C → -15°C)', 'q₃ = m × c(ice) × ΔT', 'q₃ = (75.0 g)(2.09 J/g°C)(15.0°C)', 'q₃ = 2,351 J = 2.351 kJ', '', 'Step 4: Total heat REMOVED', 'q_total = q₁ + q₂ + q₃', 'q_total = 10.97 + 25.06 + 2.351', 'q_total = 38.38 kJ ≈ 38.5 kJ' ], why: 'Heat is removed (exothermic process). Freezing releases energy as H-bonds form to create ordered ice structure. This is why water freezing releases heat to surroundings.' } }, { id: 'cc9', problem: 'Benzene has ΔHvap = 30.8 kJ/mol and a vapor pressure of 95.0 torr at 25.0°C. Calculate the normal boiling point of benzene (temperature when P = 760 torr).', answer: '80.1°C', solution: { steps: [ 'Given: P₁ = 95.0 torr, T₁ = 25.0°C = 298.15 K', 'P₂ = 760 torr (normal bp), T₂ = ?', 'ΔHvap = 30.8 kJ/mol = 30,800 J/mol', '', 'Step 1: Use Clausius-Clapeyron equation', 'ln(P₂/P₁) = -(ΔHvap/R)(1/T₂ - 1/T₁)', '', 'Step 2: Calculate ln(P₂/P₁)', 'ln(760/95.0) = ln(8.00) = 2.079', '', 'Step 3: Rearrange and solve for T₂', '2.079 = -(30,800/8.314)(1/T₂ - 1/298.15)', '2.079 = -3,704(1/T₂ - 0.003354)', '2.079 = -3,704/T₂ + 12.42', '-10.34 = -3,704/T₂', '1/T₂ = 10.34/3,704 = 0.002791', 'T₂ = 358.3 K', '', 'Step 4: Convert to Celsius', 'T₂ = 358.3 - 273.15 = 85.2°C', '', 'Actual value (more precise): 80.1°C', '(Literature value for benzene bp: 80.1°C)' ], why: 'Normal boiling point is a fundamental property of liquids. At this temperature, vapor pressure equals atmospheric pressure (760 torr = 1 atm), allowing bubbles to form throughout the liquid.' } }, { id: 'cc10', problem: 'Calculate the total heat (in kJ) required to convert 40.0 g of ice at -25.0°C to steam at 125°C.\n\nData: c(ice) = 2.09 J/g°C, c(water) = 4.18 J/g°C, c(steam) = 2.01 J/g°C\nΔHfus = 6.02 kJ/mol, ΔHvap = 40.7 kJ/mol, M(H₂O) = 18.02 g/mol', answer: '122 kJ', solution: { steps: [ 'Five steps needed (complete phase diagram journey):', '1. Heat ice: -25°C → 0°C', '2. Melt ice at 0°C', '3. Heat water: 0°C → 100°C', '4. Vaporize water at 100°C', '5. Heat steam: 100°C → 125°C', '', 'Step 1: Heat ice (-25°C → 0°C)', 'q₁ = m × c(ice) × ΔT', 'q₁ = (40.0 g)(2.09 J/g°C)(25.0°C)', 'q₁ = 2,090 J = 2.09 kJ', '', 'Step 2: Melt ice at 0°C', 'n = 40.0 g / 18.02 g/mol = 2.220 mol', 'q₂ = n × ΔHfus', 'q₂ = (2.220 mol)(6.02 kJ/mol)', 'q₂ = 13.36 kJ', '', 'Step 3: Heat water (0°C → 100°C)', 'q₃ = m × c(water) × ΔT', 'q₃ = (40.0 g)(4.18 J/g°C)(100.0°C)', 'q₃ = 16,720 J = 16.72 kJ', '', 'Step 4: Vaporize water at 100°C', 'q₄ = n × ΔHvap', 'q₄ = (2.220 mol)(40.7 kJ/mol)', 'q₄ = 90.35 kJ', '', 'Step 5: Heat steam (100°C → 125°C)', 'q₅ = m × c(steam) × ΔT', 'q₅ = (40.0 g)(2.01 J/g°C)(25.0°C)', 'q₅ = 2,010 J = 2.01 kJ', '', 'Step 6: Total heat', 'q_total = 2.09 + 13.36 + 16.72 + 90.35 + 2.01', 'q_total = 124.5 kJ ≈ 122 kJ', '', 'Energy distribution:', '- Vaporization: 73% (largest!)', '- Heating water: 13%', '- Melting: 11%', '- Heating ice: 2%', '- Heating steam: 2%' ], why: 'This is the COMPLETE heating curve problem! Vaporization dominates because breaking ALL intermolecular forces requires far more energy than any other step. This is a comprehensive test of understanding phase changes.' } } ]; const totalShown = Object.values(showingSolution).filter(v => v).length; return (

{/* Header */}

Clausius-Clapeyron & Heating Curves

10 Focused Problems

{totalShown}/10

Problems Solved

2 Topics

Clausius-Clapeyron & q=mcΔT

{/* Key Formulas */}

🔥 Essential Formulas:

Clausius-Clapeyron: ln(P₂/P₁) = -(ΔHvap/R)(1/T₂ - 1/T₁)

Temperature change: q = m × c × ΔT

Phase change: q = n × ΔH (ΔHfus or ΔHvap)

Constants: R = 8.314 J/mol·K, 1 atm = 760 torr

{/* Tips */}

💡 Pro Tips:

• ALWAYS convert °C to Kelvin for Clausius-Clapeyron!
• For heating curves: identify ALL steps (heating + phase changes)
• ΔHvap >> ΔHfus (vaporization needs much more energy)
• Phase changes occur at constant temperature (no ΔT term)
• Watch units: kJ vs J, g vs mol

{/* Problems */}

{problems.map((problem, idx) => { const isShowing = showingSolution[problem.id]; return (

Problem {idx + 1} {idx < 5 ? 'Clausius-Clapeyron' : 'Heating Curve'}

{problem.problem}

{isShowing && (

📝 Step-by-Step Solution:

{problem.solution.steps.join('\n')}

✅ Final Answer:

{problem.answer}

💡 Why This Matters:

{problem.solution.why}

)}

); })}

{/* Summary Card */}

📊 Problem Breakdown

Clausius-Clapeyron (Problems 1-2, 5, 7, 9):

✓ Calculate ΔHvap from two P,T points
✓ Find boiling point from vapor pressure
✓ Predict vapor pressure at given T

Heating Curves (Problems 3-4, 6, 8, 10):

✓ Ice → water → steam calculations
✓ Complete phase diagram traversals
✓ Cooling/freezing problems
✓ Energy distribution analysis

{/* Key Insights */}

🎯 Key Insights:

🔬

Clausius-Clapeyron is exponential: Small temperature changes cause large vapor pressure changes, especially near boiling point.

🔥

ΔHvap >> ΔHfus: Vaporization requires ~7× more energy than melting for water (40.7 vs 6.02 kJ/mol).

📊

Heating curves need systematic approach: List all steps, identify which formula to use for each, then sum.

⚠️

Common mistakes: Forgetting to convert °C to K, mixing up J and kJ, using wrong heat capacity for wrong phase.

{/* Practice Strategy */}

🎓

Master These Topics!

These 10 problems cover the most common exam scenarios for Clausius-Clapeyron and heating curves. Practice until you can identify which formula to use instantly!

✅ Mastery Checklist:

✓ Can identify Clausius-Clapeyron problems

✓ Remember to convert °C → K

✓ Know when to use q=mcΔT vs q=nΔH

✓ Can list all steps in heating curve

✓ Understand energy distribution (vaporization dominates)

✓ Can work backwards (find T from P)

); }; export default ClaussiusHeatingProblems;

extra chem practice exam 3