Enthalpy and Calorimetry

Understand heat, enthalpy changes, calorimetry, Hess's law, and standard enthalpies of formation.

Enthalpy and Calorimetry

Energy and Heat

System: Part of universe we're studying Surroundings: Everything else

Energy transfer:

Heat (q): Energy transferred due to temperature difference
Work (w): Energy transferred by force through distance

First Law of Thermodynamics:

$\Delta E = q + w$

Sign conventions:

+q: Heat absorbed by system (endothermic)
-q: Heat released by system (exothermic)
+w: Work done on system
-w: Work done by system

Enthalpy (H)

Enthalpy: Heat content at constant pressure

$\Delta H = q_p$

At constant pressure (most reactions): Heat change = enthalpy change

Exothermic reaction:

ΔH < 0 (negative)
Releases heat to surroundings
Products lower energy than reactants
Feels warm

Endothermic reaction:

ΔH > 0 (positive)
Absorbs heat from surroundings
Products higher energy than reactants
Feels cold

Calorimetry

Calorimetry: Measuring heat changes

Heat capacity equation:

$q = mc\Delta T$

Where:

q = heat (J)
m = mass (g)
c = specific heat capacity (J/g·°C)
ΔT = T_{final} - T_{initial}

Common specific heats:

Water: 4.18 J/(g·°C)
Metals: 0.1-1 J/(g·°C)

Coffee Cup Calorimeter

For solution reactions:

Constant pressure (open to atmosphere)
Measures q_p = ΔH
Simple styrofoam cup

Assumption: All heat goes to solution (usually aqueous)

$q_{\text{solution}} = -q_{\text{reaction}}$

Bomb Calorimeter

For combustion:

Constant volume
Measures ΔE (not ΔH)
More precise, sealed container

$q = C_{\text{cal}}\Delta T$

C_{cal} = calorimeter constant (J/°C)

Hess's Law

Hess's Law: ΔH is path independent - only depends on initial and final states

Use: Calculate ΔH for reaction from known ΔH values

Rules:

Reverse reaction → change sign of ΔH
Multiply reaction by n → multiply ΔH by n
Add reactions → add ΔH values

Example approach:

Manipulate given equations to match target
Add them to get target equation
Add corresponding ΔH values

Standard Enthalpy of Formation (ΔH°_f)

ΔH°_f: Enthalpy change to form 1 mole from elements in standard states

Standard conditions:

25°C (298 K)
1 atm pressure
1 M concentration

Key rules:

ΔH°_f of element in standard state = 0
ΔH°_f values tabulated for compounds

Examples:

C(graphite): ΔH°_f = 0
O₂(g): ΔH°_f = 0
H₂O(l): ΔH°_f = -285.8 kJ/mol
CO₂(g): ΔH°_f = -393.5 kJ/mol

Calculating ΔH°_{rxn}

From ΔH°_f values:

$\Delta H°_{\text{rxn}} = \sum n\Delta H°_f(\text{products}) - \sum n\Delta H°_f(\text{reactants})$

Steps:

Sum (coefficient × ΔH°_f) for products
Sum (coefficient × ΔH°_f) for reactants
Subtract: products - reactants

Shortcut: Products minus reactants (with coefficients)

Bond Enthalpies

Bond enthalpy: Energy to break 1 mole of bonds (always positive)

Estimating ΔH_{rxn}:

$\Delta H_{\text{rxn}} = \sum \text{bonds broken} - \sum \text{bonds formed}$

Energy required to break bonds (positive) Energy released forming bonds (negative)

Note: Bond enthalpies are averages, less accurate than ΔH°_f

📚 Practice Problems

1Problem 1easy

❓ Question:

When 50.0 mL of 1.0 M HCl is mixed with 50.0 mL of 1.0 M NaOH in a coffee cup calorimeter, the temperature rises from 21.0°C to 27.5°C. Calculate ΔH for the reaction in kJ/mol. Assume solution density = 1.0 g/mL and c = 4.18 J/(g·°C).

💡 Show Solution

Given:

V_{HCl} = 50.0 mL, [HCl] = 1.0 M
V_{NaOH} = 50.0 mL, [NaOH] = 1.0 M
T_i = 21.0°C, T_f = 27.5°C
Density = 1.0 g/mL, c = 4.18 J/(g·°C)

Reaction: HCl + NaOH → NaCl + H₂O

Step 1: Calculate heat absorbed by solution

Total volume = 50.0 + 50.0 = 100.0 mL

Mass = 100.0 mL × 1.0 g/mL = 100.0 g

ΔT = 27.5 - 21.0 = 6.5°C

$q_{\text{solution}} = mc\Delta T$ $q_{\text{solution}} = (100.0)(4.18)(6.5)$ $q_{\text{solution}} = 2717 \text{ J} = 2.72 \text{ kJ}$

Step 2: Calculate heat of reaction

$q_{\text{reaction}} = -q_{\text{solution}} = -2.72 \text{ kJ}$

(Negative because reaction releases heat - exothermic)

Step 3: Calculate moles reacted

Moles HCl = (1.0 M)(0.050 L) = 0.050 mol Moles NaOH = (1.0 M)(0.050 L) = 0.050 mol

Limiting reactant: Both 0.050 mol (1:1 ratio) → 0.050 mol reacts

Step 4: Calculate ΔH per mole

$\Delta H = \frac{q_{\text{reaction}}}{\text{moles}} = \frac{-2.72 \text{ kJ}}{0.050 \text{ mol}}$

$\Delta H = -54.4 \text{ kJ/mol}$

Answer: ΔH = -54 kJ/mol (exothermic)

Note: Literature value is -57.1 kJ/mol - our answer is close!

2Problem 2medium

❓ Question:

A 50.0 g sample of aluminum is heated from 20.0°C to 95.0°C. (a) Calculate the heat absorbed by the aluminum (specific heat of Al = 0.900 J/g°C). (b) If this heat came from burning methane (CH₄), which releases 890 kJ/mol, how many grams of methane were burned?

💡 Show Solution

Solution:

(a) Heat absorbed by aluminum: q = mcΔT where m = mass, c = specific heat, ΔT = temperature change

q = (50.0 g)(0.900 J/g°C)(95.0 - 20.0)°C q = (50.0)(0.900)(75.0) q = 3,375 J = 3.38 kJ

(b) Mass of methane burned: Energy from CH₄: 890 kJ/mol

Moles of CH₄ = 3.38 kJ / 890 kJ/mol = 0.00380 mol

Molar mass of CH₄ = 12.01 + 4(1.008) = 16.04 g/mol Mass = 0.00380 mol × 16.04 g/mol = 0.0609 g

3Problem 3medium

❓ Question:

Given: (1) C(s) + O₂(g) → CO₂(g), ΔH° = -393.5 kJ, (2) H₂(g) + ½O₂(g) → H₂O(l), ΔH° = -285.8 kJ, (3) C₂H₅OH(l) + 3O₂(g) → 2CO₂(g) + 3H₂O(l), ΔH° = -1367 kJ. Use Hess's law to find ΔH° for: 2C(s) + 3H₂(g) + ½O₂(g) → C₂H₅OH(l)

💡 Show Solution

Target equation: 2C(s) + 3H₂(g) + ½O₂(g) → C₂H₅OH(l)

Given equations:

C(s) + O₂(g) → CO₂(g), ΔH° = -393.5 kJ
H₂(g) + ½O₂(g) → H₂O(l), ΔH° = -285.8 kJ
C₂H₅OH(l) + 3O₂(g) → 2CO₂(g) + 3H₂O(l), ΔH° = -1367 kJ

Strategy: Manipulate equations to match target

Target needs:

2 C(s) on left → multiply equation 1 by 2
3 H₂(g) on left → multiply equation 2 by 3
C₂H₅OH(l) on right → reverse equation 3

Equation 1 × 2: 2C(s) + 2O₂(g) → 2CO₂(g) ΔH° = 2(-393.5) = -787.0 kJ

Equation 2 × 3: 3H₂(g) + 3/2 O₂(g) → 3H₂O(l) ΔH° = 3(-285.8) = -857.4 kJ

Equation 3 reversed: 2CO₂(g) + 3H₂O(l) → C₂H₅OH(l) + 3O₂(g) ΔH° = -(-1367) = +1367 kJ

Add all three:

2C(s) + 2O₂(g) → 2CO₂(g) 3H₂(g) + 3/2 O₂(g) → 3H₂O(l) 2CO₂(g) + 3H₂O(l) → C₂H₅OH(l) + 3O₂(g)

Cancel species on both sides:

2CO₂(g): appears as product (eq 1) and reactant (eq 3) → cancel
3H₂O(l): appears as product (eq 2) and reactant (eq 3) → cancel
O₂(g): 2 + 3/2 on left, 3 on right → net ½ on left

Net equation: 2C(s) + 3H₂(g) + ½O₂(g) → C₂H₅OH(l) ✓

Add ΔH° values:

$\Delta H° = -787.0 + (-857.4) + 1367$ $\Delta H° = -787.0 - 857.4 + 1367$ $\Delta H° = -277.4 \text{ kJ}$

Answer: ΔH° = -277 kJ

This is ΔH°_f of ethanol!

4Problem 4hard

❓ Question:

Given the following thermochemical equations:

C(s) + O₂(g) → CO₂(g) ΔH° = -393.5 kJ
H₂(g) + ½O₂(g) → H₂O(l) ΔH° = -285.8 kJ
2C₂H₆(g) + 7O₂(g) → 4CO₂(g) + 6H₂O(l) ΔH° = -3119.6 kJ

Use Hess's Law to calculate ΔH°_f for C₂H₆(g).

💡 Show Solution

Solution:

We want: 2C(s) + 3H₂(g) → C₂H₆(g) ΔH°_f = ?

Strategy: Manipulate given equations to get target equation.

From equation 3: 4CO₂(g) + 6H₂O(l) → 2C₂H₆(g) + 7O₂(g) ΔH = +3119.6 kJ (reversed)

From equation 1 (×4): 4C(s) + 4O₂(g) → 4CO₂(g) ΔH = 4(-393.5) = -1574.0 kJ

From equation 2 (×6): 6H₂(g) + 3O₂(g) → 6H₂O(l) ΔH = 6(-285.8) = -1714.8 kJ

Adding all three: 4CO₂ + 6H₂O → 2C₂H₆ + 7O₂ +3119.6 kJ 4C + 4O₂ → 4CO₂ -1574.0 kJ 6H₂ + 3O₂ → 6H₂O -1714.8 kJ

4C + 6H₂ → 2C₂H₆ -169.2 kJ

For 1 mole of C₂H₆: 2C(s) + 3H₂(g) → C₂H₆(g) ΔH°_f = -169.2 kJ / 2 = -84.6 kJ/mol

5Problem 5hard

❓ Question:

Calculate ΔH°_{rxn} for: CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l) using ΔH°_f values: CH₄(g) = -74.8 kJ/mol, CO₂(g) = -393.5 kJ/mol, H₂O(l) = -285.8 kJ/mol, O₂(g) = 0.

💡 Show Solution

Reaction: CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l)

Given ΔH°_f values:

CH₄(g): -74.8 kJ/mol
O₂(g): 0 (element in standard state)
CO₂(g): -393.5 kJ/mol
H₂O(l): -285.8 kJ/mol

Formula:

$\Delta H°_{\text{rxn}} = \sum n\Delta H°_f(\text{products}) - \sum n\Delta H°_f(\text{reactants})$

Products:

1 mol CO₂: 1(-393.5) = -393.5 kJ 2 mol H₂O: 2(-285.8) = -571.6 kJ

Sum products: -393.5 + (-571.6) = -965.1 kJ

Reactants:

1 mol CH₄: 1(-74.8) = -74.8 kJ 2 mol O₂: 2(0) = 0 kJ

Sum reactants: -74.8 + 0 = -74.8 kJ

Calculate ΔH°_{rxn}:

$\Delta H°_{\text{rxn}} = -965.1 - (-74.8)$ $\Delta H°_{\text{rxn}} = -965.1 + 74.8$ $\Delta H°_{\text{rxn}} = -890.3 \text{ kJ}$

Answer: ΔH°_{rxn} = -890 kJ

Interpretation:

Highly exothermic (negative ΔH)
This is combustion of methane (natural gas)
Releases 890 kJ per mole CH₄ burned
Why natural gas is good fuel

Check:

Products more negative than reactants → exothermic ✓
Magnitude makes sense for combustion ✓

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