Exothermic, Endothermic Reactions & Hess's Law
Enthalpy and Enthalpy Change
Enthalpy (H) is the total heat content of a system. It accounts for both the internal energy of the system and the work done by the system when it expands or contracts against a constant external pressure. Since it depends only on the current state of the system and not on how that state was reached, enthalpy is a state function.
$$H = E + PV$$
Enthalpy equals internal energy plus the product of pressure and volume
$H$=Enthalpy of the system(J or kJ)
$E$=Internal energy of the system(J or kJ)
$P$=Pressure of the system(Pa or atm)
$V$=Volume of the system(m³ or L)
Absolute enthalpy cannot be measured: only the change in enthalpy ($\Delta H$) between two states is experimentally accessible
At constant pressure: $\Delta H = q_p$, meaning the enthalpy change equals the heat exchanged at constant pressure, making $\Delta H$ the most practical way to track heat in chemical reactions
For solids and liquids: volume changes are negligible ($\Delta V \approx 0$), so $\Delta H \approx \Delta E$
The change in enthalpy relates to the change in internal energy through the pressure-volume work term. Starting from the definition of enthalpy and expanding the change, we arrive at the key relationship between $\Delta H$ and $\Delta E$.
$$\Delta H = \Delta E + P\Delta V$$
Enthalpy change equals the change in internal energy plus the pressure-volume work done by the system
$\Delta H$=Change in enthalpy at constant pressure(kJ mol⁻¹)
$\Delta E$=Change in internal energy(kJ mol⁻¹)
$P$=Constant external pressure(atm)
$\Delta V$=Change in volume of the system(L)
$\Delta V = 0$ (solids/liquids)
→$\Delta H = \Delta E$
$\Delta n$ moles of gas are produced
→$P\Delta V = \Delta n RT$
From first law: $\Delta E = q + w$ where $w = -P\Delta V$, so $\Delta E = q - P\Delta V$
Substituting into $\Delta H$: $\Delta H = (q - P\Delta V) + P\Delta V = q = q_p$
Gas-phase shortcut: $P\Delta V = \Delta nRT$ where $\Delta n = n_{\text{products}} - n_{\text{reactants}}$ (moles of gas only)
Exothermic and Endothermic Reactions
In an exothermic reaction, the enthalpy of the products is less than the enthalpy of the reactants. Heat is released to the surroundings, and the $\Delta H$ value is negative. The energy profile shows reactants at a higher energy level and products at a lower energy level.
$$\Delta H = H_{\text{products}} - H_{\text{reactants}} < 0$$
For exothermic reactions, products have lower enthalpy than reactants, so $\Delta H$ is negative
$\Delta H$=Enthalpy change of the reaction(kJ mol⁻¹)
$H_{\text{products}}$=Total enthalpy of products(kJ mol⁻¹)
$H_{\text{reactants}}$=Total enthalpy of reactants(kJ mol⁻¹)
Heat flows out: the reaction vessel and surroundings feel warmer
Energy diagram: reactants start high, products end low — the drop represents energy released as heat
Examples: combustion of fuels, neutralization of strong acids and bases, formation of water from $H_2$ and $O_2$
Neutralization example: $H^+_{(aq)} + OH^-_{(aq)} \rightarrow H_2O_{(l)}$ with $\Delta H^\circ_n = -57.4$ kJ mol⁻¹
Exothermic Reaction Examples
•
$C_{(s)} + O_{2(g)} \rightarrow CO_{2(g)}$ — $\Delta H^\circ_f = -393.7$ kJ mol⁻¹
•
$2H_{2(g)} + O_{2(g)} \rightarrow 2H_2O_{(l)}$ — $\Delta H^\circ = -285.8$ kJ mol⁻¹
•
$NaOH_{(aq)} + HCl_{(aq)} \rightarrow NaCl_{(aq)} + H_2O_{(l)}$ — $\Delta H^\circ_n = -57.4$ kJ mol⁻¹
•
$Mg_{(s)} + \frac{1}{2}O_{2(g)} \rightarrow MgO_{(s)}$ — $\Delta H^\circ_f = -692$ kJ mol⁻¹
In an endothermic reaction, the enthalpy of the products is greater than the enthalpy of the reactants. Heat is absorbed from the surroundings, and the $\Delta H$ value is positive. The energy profile shows reactants at a lower energy level and products at a higher energy level.
$$\Delta H = H_{\text{products}} - H_{\text{reactants}} > 0$$
For endothermic reactions, products have higher enthalpy than reactants, so $\Delta H$ is positive
$\Delta H$=Enthalpy change of the reaction(kJ mol⁻¹)
$H_{\text{products}}$=Total enthalpy of products(kJ mol⁻¹)
$H_{\text{reactants}}$=Total enthalpy of reactants(kJ mol⁻¹)
Heat flows in: the reaction vessel and surroundings feel cooler
Energy diagram: reactants start low, products end high — the rise represents energy absorbed from surroundings
Examples: thermal decomposition, dissolution of $NH_4Cl$ in water, photosynthesis
Dissolution example: $NH_4Cl_{(s)} \rightarrow NH^+_{4(aq)} + Cl^-_{(aq)}$ with $\Delta H^\circ_{\text{sol}} = +16.2$ kJ mol⁻¹
Endothermic Reaction Examples
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Dissolution of $NH_4Cl$: $\Delta H^\circ_{\text{sol}} = +16.2$ kJ mol⁻¹ (solution cools)
•
Decomposition of $CaCO_3$: $CaCO_{3(s)} \rightarrow CaO_{(s)} + CO_{2(g)}$ — requires continuous heating
•
Photosynthesis: $6CO_2 + 6H_2O \rightarrow C_6H_{12}O_6 + 6O_2$ — absorbs solar energy
Several specific types of enthalpy change are important for classifying and comparing reactions. Each type describes a particular process under standard conditions (25°C, 1 atm), and the sign convention ($+$ or $-$) tells you whether the process is endothermic or exothermic.
Standard enthalpy of formation ($\Delta H^\circ_f$): heat absorbed or evolved when one mole of a compound forms from its elements in their standard states — e.g., $\Delta H^\circ_f$ of $CO_{2(g)} = -393.7$ kJ mol⁻¹
Standard enthalpy of combustion ($\Delta H^\circ_c$): heat evolved when one mole of a substance burns completely in excess oxygen — e.g., $\Delta H^\circ_c$ of ethanol = −1368 kJ mol⁻¹
Standard enthalpy of neutralization ($\Delta H^\circ_n$): heat evolved when one mole of $H^+$ reacts with one mole of $OH^-$ — always ≈ −57.4 kJ mol⁻¹ for strong acid–strong base pairs
Standard enthalpy of solution ($\Delta H^\circ_{\text{sol}}$): heat absorbed or evolved when one mole of solute dissolves in a large excess of solvent — can be positive (endothermic, e.g., $NH_4Cl$: +16.2 kJ mol⁻¹) or negative (exothermic, e.g., $Na_2CO_3$: −25.0 kJ mol⁻¹)
Sign Convention Summary
•
$\Delta H < 0$: Exothermic — heat released, surroundings gain energy
•
$\Delta H > 0$: Endothermic — heat absorbed, surroundings lose energy
Measuring Enthalpy Changes
A glass calorimeter is an insulated container fitted with a thermometer and stirrer, used to measure enthalpy changes of reactions in solution. Reactants in stoichiometric amounts are mixed inside, and the temperature change is recorded before and after the reaction. The heat exchanged is calculated from the mass, specific heat, and temperature change of the reaction mixture.
$$q = m \times s \times \Delta T$$
Heat absorbed or released by the reaction mixture equals its mass times its specific heat times the temperature change
$q$=Heat exchanged (negative for exothermic)(J or kJ)
$m$=Mass of the reaction mixture (solution)(g)
$s$=Specific heat capacity of the solution (~4.18 J g⁻¹ K⁻¹ for water)(J g⁻¹ K⁻¹)
$\Delta T$=Temperature change ($T_{\text{final}} - T_{\text{initial}}$)(K or °C)
Density assumption: for dilute aqueous solutions, density ≈ 1 g cm⁻³, so volume in cm³ ≈ mass in grams
Moles from molarity: moles = $\frac{M \times V \text{ (cm}^3\text{)}}{1000}$ where M is molarity
Per-mole enthalpy: divide total heat $q$ by the number of moles of the limiting reagent to get $\Delta H$ in kJ mol⁻¹
A bomb calorimeter is used for accurate determination of enthalpy of combustion. The sample is placed in a sealed steel vessel (the bomb) pressurized with oxygen to about 20 atm, then immersed in an insulated water jacket. The sample is ignited electrically, and the temperature rise of the water is measured. Since the bomb has a fixed volume, the heat measured equals the internal energy change, though for most purposes this approximates $\Delta H$.
$$q = c \times \Delta T$$
Heat released by combustion equals the heat capacity of the calorimeter system times the temperature rise
$q$=Heat evolved by combustion (negative)(kJ)
$c$=Heat capacity of the entire calorimeter system (bomb + water + hardware)(kJ K⁻¹)
$\Delta T$=Temperature rise recorded(K)
Heat capacity: given directly for the calorimeter as a whole — no need to find separate mass and specific heat
Convert to per mole: divide $q$ by the number of moles of the sample burnt (mass / molar mass)
Always negative: combustion always releases heat, so $\Delta H^\circ_c$ is always negative
Hess's Law of Constant Heat Summation
Hess's law states that if a chemical change takes place by several different routes, the overall enthalpy change is the same regardless of the route taken, provided the initial and final conditions are identical. This is a direct consequence of the first law of thermodynamics (conservation of energy) because enthalpy is a state function. Mathematically, the sum of all enthalpy changes around a complete cycle equals zero.
$$\Sigma \Delta H_{\text{(cycle)}} = 0$$
Around any closed thermodynamic cycle, the total enthalpy change is zero
$\Sigma \Delta H_{\text{(cycle)}}$=Sum of all enthalpy changes in the cycle (going one way is positive, returning is negative)(kJ mol⁻¹)
$A \rightarrow D$ directly
→$\Delta H = \Delta H_1 + \Delta H_2 + \Delta H_3$ via intermediates B and C
Why it matters: many $\Delta H$ values cannot be measured directly — e.g., $\Delta H^\circ_f$ of CO cannot be measured because burning carbon always produces a mixture of CO and $CO_2$
Practical approach: combine known enthalpy values for other reactions algebraically (adding, subtracting, reversing) to find the unknown enthalpy
Reversing rule: if a reaction is reversed, the sign of $\Delta H$ changes; if coefficients are multiplied by a factor, $\Delta H$ is multiplied by the same factor
Hess's law can be verified experimentally. Consider the formation of sodium carbonate: it can occur in a single step or in two steps via sodium hydrogen carbonate. The total enthalpy change is identical in both routes, confirming that enthalpy is path-independent.
Single step: $2NaOH_{(aq)} + CO_{2(g)} \rightarrow Na_2CO_{3(aq)} + H_2O_{(l)}$ with $\Delta H = -89.08$ kJ
Step 1: $NaOH_{(aq)} + CO_{2(g)} \rightarrow NaHCO_{3(aq)}$ with $\Delta H_1 = -48.06$ kJ
Step 2: $NaHCO_{3(aq)} + NaOH_{(aq)} \rightarrow Na_2CO_{3(aq)} + H_2O_{(l)}$ with $\Delta H_2 = -41.02$ kJ
Verification: $\Delta H_1 + \Delta H_2 = -48.06 + (-41.02) = -89.08$ kJ $= \Delta H$
Applying Hess's Law — Energy Cycles
A classic application of Hess's law is finding the enthalpy of formation of carbon monoxide. Since burning carbon in limited oxygen always produces a mixture of CO and $CO_2$, $\Delta H^\circ_f$ of CO cannot be measured directly. Instead, we use the known combustion enthalpies of graphite and CO in an energy cycle.
$$\Delta H = \Delta H_1 + \Delta H_2$$
The enthalpy change for the direct route (graphite to $CO_2$) equals the sum of the two-step route (graphite to CO, then CO to $CO_2$)
$\Delta H$=Enthalpy of combustion of graphite to $CO_2$(kJ mol⁻¹)
$\Delta H_1$=Enthalpy of formation of CO from graphite (unknown)(kJ mol⁻¹)
$\Delta H_2$=Enthalpy of combustion of CO to $CO_2$(kJ mol⁻¹)
Direct route: $C_{(graphite)} + O_{2(g)} \rightarrow CO_{2(g)}$ with $\Delta H = -393.7$ kJ mol⁻¹
Indirect step 1: $C_{(graphite)} + \frac{1}{2}O_{2(g)} \rightarrow CO_{(g)}$ with $\Delta H_1 = ?$
Indirect step 2: $CO_{(g)} + \frac{1}{2}O_{2(g)} \rightarrow CO_{2(g)}$ with $\Delta H_2 = -283$ kJ mol⁻¹
Solving: $\Delta H_1 = \Delta H - \Delta H_2 = -393.7 - (-283) = -110.7$ kJ mol⁻¹
Energy Cycle for CO Formation
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$C_{(s)} + O_{2(g)}$ — starting point
•
Route A (direct): $\rightarrow CO_{2(g)}$ — $\Delta H = -393.7$ kJ mol⁻¹
•
Route B (via CO): $\rightarrow CO_{(g)} + \frac{1}{2}O_{2(g)}$ — $\Delta H_1$; then $\rightarrow CO_{2(g)}$ — $\Delta H_2 = -283$ kJ mol⁻¹
The Born-Haber cycle is a specific application of Hess's law used to calculate lattice energy — the enthalpy change when one mole of an ionic solid forms from its gaseous ions under standard conditions. Lattice energy cannot be measured directly, so the Born-Haber cycle combines measurable quantities to find it indirectly.
$$\Delta H^\circ_f = \Delta H^\circ_x + \Delta H^\circ_{\text{Latt}}$$
The standard enthalpy of formation equals the total energy to form gaseous ions plus the lattice energy
$\Delta H^\circ_f$=Standard enthalpy of formation of the ionic compound(kJ mol⁻¹)
$\Delta H^\circ_x$=Sum of all steps to convert elements to gaseous ions (atomization + ionization + electron affinity)(kJ mol⁻¹)
$\Delta H^\circ_{\text{Latt}}$=Lattice energy of the ionic compound(kJ mol⁻¹)
Step 1 — Atomization of metal: $Na_{(s)} \rightarrow Na_{(g)}$ — $\Delta H_{\text{at}} = +108$ kJ mol⁻¹
Step 2 — Ionization of metal: $Na_{(g)} \rightarrow Na^+_{(g)} + e^-$ — $\Delta H_i = +496$ kJ mol⁻¹
Step 3 — Atomization of non-metal: $\frac{1}{2}Cl_{2(g)} \rightarrow Cl_{(g)}$ — $\Delta H_{\text{at}} = +121$ kJ mol⁻¹
Step 4 — Electron affinity: $Cl_{(g)} + e^- \rightarrow Cl^-_{(g)}$ — $\Delta H_e = -349$ kJ mol⁻¹
$\Delta H^\circ_x$ for NaCl: $108 + 496 + 121 + (-349) = +376$ kJ mol⁻¹
Lattice energy: $\Delta H^\circ_{\text{Latt}} = \Delta H^\circ_f - \Delta H^\circ_x = -411 - 376 = -787$ kJ mol⁻¹
Born-Haber Cycle Steps for NaCl
1
Start: $Na_{(s)} + \frac{1}{2}Cl_{2(g)}$
2
Atomize Na: $\rightarrow Na_{(g)} + \frac{1}{2}Cl_{2(g)}$ — $+108$ kJ mol⁻¹
3
Ionize Na: $\rightarrow Na^+_{(g)} + e^- + \frac{1}{2}Cl_{2(g)}$ — $+496$ kJ mol⁻¹
4
Atomize Cl: $\rightarrow Na^+_{(g)} + e^- + Cl_{(g)}$ — $+121$ kJ mol⁻¹
5
Electron affinity: $\rightarrow Na^+_{(g)} + Cl^-_{(g)}$ — $-349$ kJ mol⁻¹
6
Lattice formation: $\rightarrow NaCl_{(s)}$ — $\Delta H^\circ_{\text{Latt}}$ (unknown)
7
Direct: $Na_{(s)} + \frac{1}{2}Cl_{2(g)} \rightarrow NaCl_{(s)}$ — $\Delta H^\circ_f = -411$ kJ mol⁻¹
Application Strategy
Worked Examples
Problem Statement
When 2.00 moles of $H_2$ and 1.00 mole of $O_2$ at 100°C and 1 atm react to produce 2.00 moles of gaseous water, 484.5 kJ of energy are evolved. Calculate (a) $\Delta H$ and (b) $\Delta E$ for the production of one mole of $H_2O_{(g)}$. ($R = 8.314$ J K⁻¹ mol⁻¹)
Enthalpy change at constant pressureInternal energy from enthalpyPressure-volume work for gases
Strategic Toolkit
$$\Delta H = \Delta E + P\Delta V$$
Step 1
Find $\Delta H$ per mole
Since the reaction occurs at constant pressure, the enthalpy change equals the heat evolved. Divide the total heat by the number of moles of water produced.
$$\Delta H = \frac{-484.5\text{ kJ}}{2 \text{ mol}} = -242.2 \text{ kJ mol}^{-1}$$
Step 2
Calculate $\Delta n$ for the reaction
Count moles of gaseous products minus moles of gaseous reactants to find the change in gas moles.
$$\Delta n = 2 - (2+1) = -1 \text{ mol}$$
Step 3
Find $P\Delta V$ using the ideal gas relation
Use $P\Delta V = \Delta nRT$ with $T = 373$ K and $R = 8.314$ J K⁻¹ mol⁻¹. This gives the value for 2 moles of water.
$$P\Delta V = (-1)(8.314)(373) = -3100 \text{ J} = -3.10 \text{ kJ}$$
Step 4
Per-mole $P\Delta V$ and $\Delta E$
Divide $P\Delta V$ by 2 to get the per-mole value, then use $\Delta E = \Delta H - P\Delta V$.
$$P\Delta V_{\text{per mol}} = \frac{-3.10}{2} = -1.55 \text{ kJ mol}^{-1}$$$$\Delta E = -242.2 - (-1.55) = -240.6 \text{ kJ mol}^{-1}$$
Problem Statement
Neutralization of 100 cm³ of 0.5 M NaOH at 25°C with 100 cm³ of 0.5 M HCl at 25°C raised the temperature to 28.5°C. Find the enthalpy of neutralization. (Specific heat of water = 4.2 J g⁻¹ K⁻¹)
Glass calorimeterEnthalpy of neutralizationHeat calculation from temperature change
Strategic Toolkit
$$q = m \times s \times \Delta T$$
Step 1
Find the total mass of solution
Total volume = 200 cm³. Assuming density of water ≈ 1 g cm⁻³, mass ≈ 200 g.
$$m = 100 + 100 = 200 \text{ g}$$
Step 2
Calculate the temperature rise
Subtract the initial temperature from the final temperature.
$$\Delta T = 28.5 - 25.0 = 3.5 \text{ K}$$
Step 3
Calculate total heat evolved
Multiply mass, specific heat, and temperature change. The reaction is exothermic, so assign a negative sign.
$$q = 200 \times 4.2 \times 3.5 = 2940 \text{ J} = 2.94 \text{ kJ}$$$$q = -2.94 \text{ kJ}$$
Step 4
Find moles of limiting reagent
100 cm³ of 0.5 M solution contains 0.05 moles of solute.
$$n = \frac{0.5 \times 100}{1000} = 0.05 \text{ mol}$$
Step 5
Calculate $\Delta H^\circ_n$ per mole
Divide the total heat by the number of moles to get the per-mole enthalpy of neutralization.
$$\Delta H^\circ_n = \frac{-2.94}{0.05} = -58.8 \text{ kJ mol}^{-1}$$
Problem Statement
10.16 g of graphite is burnt in a bomb calorimeter and the temperature rise is 3.87 K. Calculate the enthalpy of combustion of graphite if the heat capacity of the calorimeter is 86.02 kJ K⁻¹.
Bomb calorimeterEnthalpy of combustionHeat capacity method
Strategic Toolkit
$$q = c \times \Delta T$$
Step 1
Calculate total heat evolved
Multiply the heat capacity of the calorimeter by the temperature rise.
$$q = 86.02 \times 3.87 = 332.89 \text{ kJ}$$
Step 2
Convert sample mass to moles
Divide the mass of graphite by its molar mass (12 g mol⁻¹).
$$n = \frac{10.16}{12} = 0.847 \text{ mol}$$
Step 3
Find enthalpy of combustion per mole
Divide the total heat by the number of moles. Combustion releases heat, so $\Delta H^\circ_c$ is negative.
$$\Delta H^\circ_c = \frac{-332.89}{0.847} = -393 \text{ kJ mol}^{-1}$$
Problem Statement
Given: combustion of graphite gives $\Delta H = -393.7$ kJ mol⁻¹, and combustion of CO gives $\Delta H_2 = -283$ kJ mol⁻¹. Use Hess's law to find the enthalpy of formation of CO from graphite.
Hess's law energy cyclePath-independent enthalpyAlgebraic manipulation of $\Delta H$
Strategic Toolkit
$$\Delta H = \Delta H_1 + \Delta H_2$$
Step 1
Write the energy cycle equations
Direct route: graphite + $O_2 \rightarrow CO_2$ with $\Delta H = -393.7$ kJ mol⁻¹. Indirect route: graphite + $\frac{1}{2}O_2 \rightarrow CO$ ($\Delta H_1$), then CO + $\frac{1}{2}O_2 \rightarrow CO_2$ ($\Delta H_2 = -283$ kJ mol⁻¹).
$$\Delta H = \Delta H_1 + \Delta H_2$$
Step 2
Solve for the unknown
Rearrange and substitute known values.
$$\Delta H_1 = \Delta H - \Delta H_2 = -393.7 - (-283) = -110.7 \text{ kJ mol}^{-1}$$
Problem Statement
Calculate the lattice energy of NaCl given: $\Delta H^\circ_f(\text{NaCl}) = -411$ kJ mol⁻¹, atomization enthalpy of Na = +108 kJ mol⁻¹, first ionization energy of Na = +496 kJ mol⁻¹, atomization enthalpy of Cl = +121 kJ mol⁻¹, and electron affinity of Cl = −349 kJ mol⁻¹.
Born-Haber cycleLattice energy calculationSummation of enthalpy steps
Strategic Toolkit
$$\Delta H^\circ_{\text{Latt}} = \Delta H^\circ_f - \Delta H^\circ_x$$
Step 1
Calculate $\Delta H^\circ_x$ (total energy to form gaseous ions)
Add all the steps from elements to gaseous ions: atomization and ionization of Na, atomization and electron affinity of Cl.
$$\Delta H^\circ_x = 108 + 496 + 121 + (-349) = +376 \text{ kJ mol}^{-1}$$
Step 2
Apply the Born-Haber equation
Use $\Delta H^\circ_f = \Delta H^\circ_x + \Delta H^\circ_{\text{Latt}}$ and rearrange to find lattice energy.
$$\Delta H^\circ_{\text{Latt}} = -411 - 376 = -787 \text{ kJ mol}^{-1}$$
Problem Statement
Calculate the enthalpy of formation of aqueous $NH_4Cl$ from $NH_3$ gas and HCl gas. Given: (i) $NH_{3(g)} + aq \rightarrow NH_{3(aq)}$, $\Delta H = -35.16$ kJ mol⁻¹; (ii) $HCl_{(g)} + aq \rightarrow HCl_{(aq)}$, $\Delta H = -72.41$ kJ mol⁻¹; (iii) $NH_{3(aq)} + HCl_{(aq)} \rightarrow NH_4Cl_{(aq)}$, $\Delta H = -51.48$ kJ mol⁻¹.
Hess's law summationCombining sequential enthalpy steps
Strategic Toolkit
$$\Delta H_{\text{total}} = \Delta H_1 + \Delta H_2 + \Delta H_3$$
Step 1
Identify the route and add all steps
The overall reaction $NH_{3(g)} + HCl_{(g)} + aq \rightarrow NH_4Cl_{(aq)}$ proceeds via three steps: dissolution of $NH_3$, dissolution of HCl, and their neutralization in solution.
$$\Delta H = -35.16 + (-72.41) + (-51.48) = -159.05 \text{ kJ mol}^{-1}$$
Problem Statement
Calculate the heat of formation of ethyl alcohol ($C_2H_5OH$) from: (i) heat of combustion of ethanol = −1367 kJ mol⁻¹; (ii) $\Delta H^\circ_f$ of $CO_2$ = −393.7 kJ mol⁻¹; (iii) $\Delta H^\circ_f$ of $H_2O$ = −285.8 kJ mol⁻¹.
Hess's law with combustion dataReversing reactions and changing signs
Strategic Toolkit
$$\Delta H^\circ_f = \Delta H_{\text{combustion}} + 2\Delta H^\circ_f(CO_2) + 3\Delta H^\circ_f(H_2O)$$
Step 1
Write the formation and combustion equations
Formation: $2C_{(s)} + 3H_{2(g)} + \frac{1}{2}O_{2(g)} \rightarrow C_2H_5OH_{(l)}$ — target with unknown $\Delta H^\circ_f$. Combustion: $C_2H_5OH_{(l)} + 3O_{2(g)} \rightarrow 2CO_{2(g)} + 3H_2O_{(l)}$ with $\Delta H = -1367$ kJ mol⁻¹.
Step 2
Reverse the combustion and add formation steps for products
Reverse the combustion equation (sign of $\Delta H$ flips to +1367). Add formation of 2 moles of $CO_2$ and 3 moles of $H_2O$, subtracting their formation enthalpies.
$$\Delta H^\circ_f = +1367 + 2(-393.7) + 3(-285.8)$$$$\Delta H^\circ_f = 1367 - 787.4 - 857.4 = -277.8 \text{ kJ mol}^{-1}$$
Memorize formulas
Enthalpy Definition
Total heat content; H = E + PV. Units: kJ. $\Delta H$ is measurable but H is not.
Formula
$$H = E + PV$$
Enthalpy Change (constant pressure)
Relates $\Delta H$ to $\Delta E$ via pressure-volume work. At constant pressure, $\Delta H = q_p$. Units: kJ mol⁻¹.
Formula
$$\Delta H = \Delta E + P\Delta V$$
PV Work in Terms of Gas Moles
Converts $P\Delta V$ to a form using moles of gas. $\Delta n$ = moles of gaseous products − moles of gaseous reactants. $R$ = 8.314 J K⁻¹ mol⁻¹.
Formula
$$P\Delta V = \Delta nRT$$
Exothermic / Endothermic Criterion
$\Delta H < 0$: exothermic (heat released). $\Delta H > 0$: endothermic (heat absorbed).
Formula
$$\Delta H = H_{\text{products}} - H_{\text{reactants}}$$
Heat from Mass and Temperature Change
Glass calorimeter formula. $m$ in grams, $s$ = 4.18 J g⁻¹ K⁻¹ for water, $\Delta T$ in K. Gives total heat in J or kJ.
Formula
$$q = ms\Delta T$$
Heat from Heat Capacity
Bomb calorimeter formula. $c$ = heat capacity of the entire system in kJ K⁻¹, $\Delta T$ in K.
Formula
$$q = c\Delta T$$
Hess's Law — Cycle Summation
Sum of all $\Delta H$ around a complete cycle is zero. Equivalently, direct route $\Delta H$ equals sum of all indirect step $\Delta H$ values.
Formula
$$\Sigma \Delta H_{\text{cycle}} = 0$$
Born-Haber — Lattice Energy
Rearrange to find lattice energy: $\Delta H^\circ_{\text{Latt}} = \Delta H^\circ_f - \Delta H^\circ_x$. All values in kJ mol⁻¹.
Formula
$$\Delta H^\circ_f = \Delta H^\circ_x + \Delta H^\circ_{\text{Latt}}$$
Review
1
The enthalpy $H$ of a thermodynamic system is defined as:
Ans
$H = E + PV$
Review
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Enthalpy Definition
By definition, enthalpy combines internal energy with the product of pressure and volume
$$H = E + PV$$
2
Which of the following correctly describes an exothermic reaction?
Ans
$\Delta H < 0$ and heat is released to the surroundings
Review
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Enthalpy Change Criterion
The sign of $\Delta H$ determines whether the reaction is exothermic or endothermic
$$\Delta H = H_{\text{products}} - H_{\text{reactants}}$$
Exothermic Condition
When products have lower enthalpy than reactants, $\Delta H$ is negative and heat is released
$$\Delta H < 0 \rightarrow \text{exothermic}$$
3
Which of the following processes is endothermic?
Ans
$CaCO_3(s) \rightarrow CaO(s) + CO_2(g)$
Review
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Identify the Reaction Type
Thermal decomposition reactions typically require heat absorption to break chemical bonds
$$CaCO_3(s) \xrightarrow{\text{heat}} CaO(s) + CO_2(g)$$
Classify by Enthalpy Change
Decomposition of $CaCO_3$ has $\Delta H = +178 \text{ kJ/mol}$, confirming it is endothermic
$$\Delta H > 0 \rightarrow \text{endothermic}$$
4
For the reaction $\frac{1}{2}N_2(g) + \frac{3}{2}H_2(g) \rightarrow NH_3(g)$, $\Delta H^\circ = -46$ kJ/mol. Which statement is true?
Ans
The reaction releases 46 kJ per mole of $NH_3$ formed
Review
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Interpret the Sign of $\Delta H$
$\Delta H = -46$ kJ/mol means the enthalpy of products is less than reactants
$$H_{\text{products}} < H_{\text{reactants}} \rightarrow \Delta H < 0$$
Determine Heat Flow
A negative $\Delta H$ means heat is released from the system to the surroundings
$$\text{Heat released} = 46 \text{ kJ per mole of } NH_3$$
5
For the reaction $2C(s) + O_2(g) \rightarrow 2CO(g)$, what is the value of $\Delta n$ (change in moles of gas)?
Ans
$+1$
Review
Tap to reveal
Count Gas Moles on Each Side
Only gaseous species contribute to $\Delta n$; solids are excluded
$$n_{\text{gas, reactants}} = 1 \quad (O_2)$$
Calculate $\Delta n$
Products have 2 moles of gas ($2CO$), reactants have 1 mole ($O_2$)
$$\Delta n = 2 - 1 = +1$$
6
For $H_2(g) + \frac{1}{2}O_2(g) \rightarrow H_2O(g)$, $\Delta H = -241.8$ kJ/mol at 298 K. Given $\Delta n = -0.5$ and $R = 8.314$ J/mol·K, what is $\Delta E$?
Ans
$-240.6$ kJ/mol
Review
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Calculate PV Work
Use $P\Delta V = \Delta nRT$ with the given values
$$P\Delta V = (-0.5)(8.314)(298) = -1239 \text{ J} = -1.24 \text{ kJ}$$
Find $\Delta E$ from $\Delta H$
Rearrange $\Delta H = \Delta E + P\Delta V$ to solve for $\Delta E$
$$\Delta E = -241.8 - (-1.24) = -240.6 \text{ kJ/mol}$$
7
100 g of water is heated from $25$°C to $35$°C. If the specific heat capacity of water is $4.18$ J/g·°C, how much heat energy is absorbed?
Ans
$4180$ J
Review
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Identify Given Values
Extract mass, specific heat, and temperature change from the problem
$$m = 100 \text{ g}, \quad s = 4.18 \text{ J/g·°C}, \quad \Delta T = 35 - 25 = 10 \text{°C}$$
Apply Heat Formula
Substitute into $q = ms\Delta T$
$$q = (100)(4.18)(10) = 4180 \text{ J}$$
8
Which apparatus is used to measure the enthalpy of combustion at constant volume?
Ans
Bomb calorimeter
Review
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Identify Constant-Volume Apparatus
A bomb calorimeter has a strong steel container that maintains constant volume during combustion
$$\text{Bomb calorimeter} \rightarrow \text{constant volume} \rightarrow \text{measures } \Delta E$$
Relate to Constant-Pressure Alternative
Glass calorimeter operates at constant pressure (atmospheric), measuring $\Delta H$ directly
$$\text{Glass calorimeter} \rightarrow \text{constant pressure} \rightarrow \text{measures } \Delta H$$
9
According to Hess's law, the total enthalpy change for a chemical reaction:
Ans
Is the same regardless of the pathway taken
Review
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State Hess's Law
Enthalpy is a state function, so the total change is path-independent
$$\Sigma \Delta H_{\text{cycle}} = 0$$
Implication
Whether the reaction proceeds directly or through intermediates, the net $\Delta H$ is identical
$$\Delta H_{\text{direct}} = \Delta H_{\text{step 1}} + \Delta H_{\text{step 2}} + \cdots$$
10
Given: $C(s) + O_2(g) \rightarrow CO_2(g)$; $\Delta H_1 = -393.5$ kJ/mol and $CO(g) + \frac{1}{2}O_2(g) \rightarrow CO_2(g)$; $\Delta H_2 = -283.0$ kJ/mol. What is the enthalpy of formation of $CO(g)$?
Ans
$-110.5$ kJ/mol
Review
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Reverse the Second Equation
Reverse $CO(g) + \frac{1}{2}O_2(g) \rightarrow CO_2(g)$ to get $CO_2(g) \rightarrow CO(g) + \frac{1}{2}O_2(g)$ and change the sign of $\Delta H$
$$\Delta H_{\text{reversed}} = +283.0 \text{ kJ/mol}$$
Apply Hess's Law
Add the reversed equation to the first equation and cancel common terms ($CO_2$ and $\frac{1}{2}O_2$)
$$C(s) + \frac{1}{2}O_2(g) \rightarrow CO(g); \quad \Delta H = -393.5 + 283.0 = -110.5 \text{ kJ/mol}$$
11
In a Born-Haber cycle for NaCl, the enthalpy of formation $\Delta H^\circ_f$ is related to the other enthalpy changes by:
Ans
The sum of all enthalpy changes around the complete cycle
Review
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Born-Haber Cycle Equation
The formation enthalpy equals the sum of all intermediate steps in the cycle
$$\Delta H^\circ_f = \Delta H^\circ_{\text{sub}} + IE + \tfrac{1}{2}BDE + EA + U$$
Lattice Energy Relation
Since $\Sigma \Delta H_{\text{cycle}} = 0$, the lattice energy can be isolated by summing all other terms
$$U = \Delta H^\circ_f - (\Delta H^\circ_{\text{sub}} + IE + \tfrac{1}{2}BDE + EA)$$
12
Using the Born-Haber cycle for NaCl: $\Delta H^\circ_f = -411$ kJ/mol, $\Delta H^\circ_{\text{sub}}(Na) = +108$ kJ/mol, $IE(Na) = +496$ kJ/mol, $\frac{1}{2}BDE(Cl_2) = +122$ kJ/mol, $EA(Cl) = -349$ kJ/mol. What is the lattice energy of NaCl?
Ans
$-788$ kJ/mol
Review
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Sum Known Enthalpy Changes
Add all enthalpy changes except lattice energy
$$108 + 496 + 122 + (-349) = 377 \text{ kJ/mol}$$
Solve for Lattice Energy
Set $\Delta H^\circ_f = \Sigma(\text{all steps})$ and isolate $U$
$$U = -411 - 377 = -788 \text{ kJ/mol}$$
13
Which of the following is a state function?
Ans
Enthalpy ($H$)
Review
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Definition of State Function
A state function's value depends only on the current state, independent of how that state was reached
$$\text{State function: } H, E, P, V, T$$
Path Functions vs State Functions
Heat and work depend on the specific process path taken, making them path functions
$$\text{Path function: } q, w$$
14
Hess's law applied to any complete thermodynamic cycle requires that:
Ans
$\Sigma \Delta H_{\text{cycle}} = 0$
Review
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State Function Property
A state function has zero net change over a complete cycle
$$\Sigma \Delta H_{\text{cycle}} = 0$$
Application
This allows unknown enthalpy changes to be found by rearranging the cycle equation
$$\Delta H_{\text{unknown}} = -\Sigma \Delta H_{\text{known}}$$
15
Given: $S(s) + O_2(g) \rightarrow SO_2(g)$; $\Delta H_1 = -297$ kJ/mol and $SO_2(g) + \frac{1}{2}O_2(g) \rightarrow SO_3(g)$; $\Delta H_2 = -98$ kJ/mol. What is the enthalpy change for $S(s) + \frac{3}{2}O_2(g) \rightarrow SO_3(g)$?
Ans
$-395$ kJ/mol
Review
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Add Both Equations
Combine the two given reactions to obtain the target equation
$$S(s) + O_2(g) + SO_2(g) + \tfrac{1}{2}O_2(g) \rightarrow SO_2(g) + SO_3(g)$$
Cancel Common Species and Sum Enthalpies
Remove $SO_2(g)$ from both sides and add the $\Delta H$ values
$$S(s) + \tfrac{3}{2}O_2(g) \rightarrow SO_3(g); \quad \Delta H = -297 + (-98) = -395 \text{ kJ/mol}$$
Question 1 of 15
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1
The enthalpy $H$ of a thermodynamic system is defined as: