Thermochemistry
Unit 6 of AP Chemistry, worth 7–9% of the exam. 13 questions below, each with the working. Every answer was checked by a second pass before it was published.
Endothermic and exothermic processes, energy diagrams, heat transfer and thermal equilibrium, heat capacity and calorimetry, energy of phase changes, enthalpy of reaction, bond enthalpies, Hess's law.
How this unit is tested
Most Unit 6 problems fall into one of four buckets: classify a process as exothermic or endothermic (or read an energy diagram), calculate heat transferred using calorimetry, calculate energy for a phase change, or calculate ΔH for a reaction using formation enthalpies, bond enthalpies, or Hess's law. Identify which bucket you're in before you start plugging into formulas.
For any calorimetry problem, first decide what is losing heat and what is gaining heat. Write the conservation statement $q_{lost} = -q_{gained}$, then substitute $q = mc\Delta T$ for each side, using $\Delta T = T_{final} - T_{initial}$ so the signs come out correctly on their own — do not force a sign by hand.
For enthalpy of reaction, check what data you're given. Standard enthalpies of formation call for $\Delta H^\circ_{rxn} = \sum \Delta H_f^\circ(\text{products}) - \sum \Delta H_f^\circ(\text{reactants})$. Bond enthalpies (only usable for gas-phase reactions where you can count bonds) call for $\Delta H = \sum(\text{bonds broken}) - \sum(\text{bonds formed})$, treating breaking bonds as energy input and forming bonds as energy release. A set of reference reactions with known ΔH values calls for Hess's law: manipulate the given equations — reversing and/or scaling them — until they sum to the target equation, applying the identical operations to their ΔH values.
For energy diagrams, locate reactants, products, and the peak (transition state). ΔH is the vertical gap between reactants and products (products lower means exothermic, negative ΔH). Activation energy is the vertical gap from reactants up to the peak for the forward reaction, and from products up to the peak for the reverse reaction.
Finally, always attach the correct sign and units to your final answer, and read the question carefully to see whether it wants a signed ΔH or the plain magnitude of heat released or absorbed.
What you have to know
First Law of Thermodynamics
The change in internal energy of a system equals heat added plus work done on it: $\Delta E = q + w$.
Specific heat equation
Heat transferred is $q = mc\Delta T$, where m is mass, c is specific heat capacity, and $\Delta T = T_{final}-T_{initial}$.
Enthalpy of reaction from formation enthalpies
$\Delta H^\circ_{rxn} = \sum \Delta H_f^\circ(\text{products}) - \sum \Delta H_f^\circ(\text{reactants})$, each term multiplied by its stoichiometric coefficient.
Enthalpy of reaction from bond enthalpies
$\Delta H_{rxn} = \sum(\text{bond enthalpies of bonds broken}) - \sum(\text{bond enthalpies of bonds formed})$.
Hess's Law
If a reaction is the sum of two or more steps, $\Delta H$ for the overall reaction equals the sum of the ΔH values for those steps. Reversing a reaction reverses the sign of its ΔH; multiplying a reaction's coefficients by a factor multiplies its ΔH by that same factor.
Calorimetry energy conservation
In an isolated calorimeter system, heat lost by one component equals heat gained by the other: $q_{lost} = -q_{gained}$.
13 practice questions
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In an energy diagram for an exothermic reaction, how does the energy of the products compare to the energy of the reactants?
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Answer. The products are lower in energy than the reactants.
Exothermic reactions release energy from the system to the surroundings, so the system ends with less energy than it started with, giving a negative ΔH and a downward-sloping diagram. -
How much heat is released when 25.0 g of water cools from 80.0°C to 20.0°C? (c_water = 4.18 J/g·°C)
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Answer. 6.27 kJ is released.
Use $q = mc\Delta T = 25.0(4.18)(20.0-80.0) = -6270\text{ J} = -6.27\text{ kJ}$. The negative sign shows the water loses heat, so 6.27 kJ is released to the surroundings. -
A 45.0 g piece of metal at 95.0°C is dropped into a coffee-cup calorimeter containing 100.0 g of water at 22.0°C. The mixture reaches a final temperature of 25.0°C. Calculate the specific heat capacity of the metal.
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Answer. About 0.398 J/(g·°C).
Heat gained by water: $q_{water}=100.0(4.18)(25.0-22.0)=1254\text{ J}$. By conservation, $q_{metal}=-1254\text{ J}=45.0\,c\,(25.0-95.0)$, so $c=1254/(45.0\times70.0)=0.398\text{ J/(g·°C)}$. -
A hot metal block at 80°C and a cool metal block at 20°C are placed in contact and insulated from everything else. What determines their final common temperature?
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Answer. The final temperature depends on the masses and specific heat capacities of both blocks, not simply their average.
Heat flows from the hotter block to the cooler one until $q_{lost}=-q_{gained}$ is satisfied; solving that equation, which involves each block's mass and specific heat, gives the true final temperature, which equals the numerical average only if both blocks have identical mc products. -
How much energy is required to vaporize 36.0 g of water at 100°C, given ΔH_vap = 40.7 kJ/mol?
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Answer. 81.4 kJ.
Convert mass to moles: $36.0\text{ g}/18.0\text{ g/mol}=2.00\text{ mol}$. Multiply by the molar heat of vaporization: $2.00\times40.7=81.4\text{ kJ}$. -
On a heating curve, why does the temperature stay constant during a phase change even though heat is continuously being added?
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Answer. The added energy is going into overcoming intermolecular attractions (increasing potential energy), not into increasing average kinetic energy.
Temperature is a measure of average kinetic energy of particles. During melting or boiling, energy is used to separate particles against intermolecular forces rather than to speed them up, so temperature plateaus until the phase change is complete. -
Use standard enthalpies of formation to calculate ΔH°rxn for $CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(l)$, given ΔHf°: CH4(g) = −74.8 kJ/mol, CO2(g) = −393.5 kJ/mol, H2O(l) = −285.8 kJ/mol, O2(g) = 0.
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Answer. ΔH°rxn = −890.3 kJ/mol.
$\Delta H = [(-393.5)+2(-285.8)] - [(-74.8)+2(0)] = -965.1-(-74.8) = -890.3\text{ kJ}$. The large negative value confirms methane combustion is strongly exothermic. -
Calculate ΔH for $H_2(g)+Cl_2(g)\rightarrow 2HCl(g)$ using bond enthalpies: H–H = 436 kJ/mol, Cl–Cl = 243 kJ/mol, H–Cl = 431 kJ/mol.
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Answer. ΔH = −183 kJ/mol.
$\Delta H = \text{(bonds broken)} - \text{(bonds formed)} = (436+243) - 2(431) = 679-862 = -183\text{ kJ}$. More energy is released forming two H–Cl bonds than is needed to break one H–H and one Cl–Cl bond. -
Given $N_2(g)+O_2(g)\rightarrow 2NO(g)$, ΔH₁ = +180.5 kJ, and $2NO(g)+O_2(g)\rightarrow 2NO_2(g)$, ΔH₂ = −114.2 kJ, find ΔH for $N_2(g)+2O_2(g)\rightarrow 2NO_2(g)$.
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Answer. ΔH = +66.3 kJ.
The two given equations, added directly, sum to the target equation (2NO cancels). By Hess's law, simply add the ΔH values: $180.5+(-114.2)=+66.3\text{ kJ}$. -
For an endothermic reaction, the activation energy of the forward reaction is 125 kJ/mol and the activation energy of the reverse reaction is 80 kJ/mol. What is ΔH for the forward reaction?
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Answer. ΔH = +45 kJ/mol.
On an energy diagram, ΔH equals the forward activation energy minus the reverse activation energy: $125-80=45\text{ kJ/mol}$. The positive value confirms the reaction is endothermic, consistent with the higher forward barrier. -
Why is it valid to add together the ΔH values of several reference reactions to find the ΔH of a target reaction, as in Hess's Law?
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Answer. Because enthalpy is a state function, so ΔH depends only on the initial and final states of a reaction, not on the pathway taken to get there.
Since the total ΔH for any given overall change is fixed regardless of route, breaking a reaction into hypothetical steps that sum to the target reaction and adding their individual ΔH values gives the same correct total ΔH. -
Combustion of a 1.500 g sample of a fuel in a bomb calorimeter with a heat capacity of 9.20 kJ/°C raises the calorimeter's temperature by 3.25°C. If the fuel's molar mass is 180 g/mol, calculate ΔH of combustion per mole.
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Answer. About −3590 kJ/mol.
Heat absorbed by the calorimeter: $q=C\Delta T=9.20(3.25)=29.9\text{ kJ}$, released by the reaction, so $q_{rxn}=-29.9\text{ kJ}$. Moles of fuel: $1.500/180=0.00833\text{ mol}$. $\Delta H=-29.9/0.00833\approx-3590\text{ kJ/mol}$. -
A calorimeter contains 250. g of water. What is the heat capacity of this water sample?
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Answer. 1.05 × 10³ J/°C (1.045 kJ/°C).
Heat capacity is mass times specific heat: $C=mc=250.\times4.18=1045\text{ J/°C}$. Unlike specific heat, heat capacity depends on the total amount of substance present.
What people get wrong
- Writing ΔH as positive for an exothermic process because energy is 'released.' Instead, remember the sign describes the system: exothermic reactions lose enthalpy, so ΔH is negative, even though the surroundings gain heat.
- In calorimetry, forgetting the sign flip between the two components (treating q_metal and q_water as equal instead of opposite). Instead, always start from $q_{lost} = -q_{gained}$ before substituting numbers.
- In bond-enthalpy calculations, computing (bonds formed) − (bonds broken) instead of the reverse. Instead, remember breaking bonds always costs energy (positive) and forming bonds always releases energy (subtracted), so it's always broken minus formed.
- When reversing or scaling an equation for Hess's law, changing the equation but forgetting to apply the identical operation to its ΔH. Instead, treat the equation and its ΔH as a single unit — whatever you do to one, do to the other.
- Assuming that when two objects at different temperatures reach thermal equilibrium, the final temperature is simply the average of the two starting temperatures. Instead, the final temperature depends on the masses and specific heats of both objects, so it must be calculated from $q_{lost}=-q_{gained}$.
- On an energy diagram, confusing the height of the activation-energy hump with ΔH. Instead, ΔH is the net vertical difference between reactants and products; activation energy is the height of the peak above the reactants (or products, for the reverse reaction).
Drill this unit until it sticks
These questions come back on a schedule built from what you get wrong, alongside the rest of AP Chemistry. Free, and no account needed to start.