Difficulty: Intermediate
Prerequisites: Part 1 of these study notes (energy, heat transfer, internal energy, sign conventions for q and w). You will also need to be comfortable with unit conversions between grams, moles, joules, kilojoules and L atm.
This section moves from the theory of energy transfer to its measurement (calorimetry) and to the quantity chemists care about most in practice: enthalpy change (Delta H). Enthalpy is the go-to measure for "how much heat does this reaction produce or consume?" and it connects directly to Hess's Law and to standard enthalpies of formation, which let you calculate the heat of any reaction from tabulated data. If you can handle Delta H, you are ready for chemical equilibrium and thermodynamic spontaneity later in the course.
Calorimetry measures heat by tracking temperature changes: q = m s Delta T. At constant pressure, the heat of a reaction equals the enthalpy change (Delta H). Because enthalpy is a state function, you can calculate Delta H for any reaction by combining known reactions (Hess's Law) or by using standard enthalpies of formation from a reference table.
Calorimetry
The science of measuring heat. It works by observing the temperature change (Delta T) when a substance absorbs or gives off heat.
In simple terms, you measure how much the temperature moves and work backwards to figure out how much energy was involved.
Heat capacity (C)
The amount of energy (as heat) required to raise the temperature of a substance by 1 degree C or 1 K. Heat capacity does not specify the amount of substance, so it applies to whatever quantity you have.
Think of it as: the bigger the heat capacity, the more energy you need to warm it up, and the smaller the temperature change for a given input of heat.
Specific heat capacity (s)
The energy (as heat) required to raise the temperature of 1 gram of a substance by 1 degree C (or 1 K). This is heat capacity per gram. Units: J/(g degree C) or J/(g K).
In simple terms, specific heat tells you how resistant one gram of a substance is to temperature change. Water has a high specific heat (4.18 J/g degree C), so it takes a lot of energy to heat up. Metals have low specific heats, so they heat up quickly.
Molar heat capacity
The energy required to raise the temperature of 1 mole of a substance by 1 degree C. Calculated by multiplying the specific heat by the molar mass. Units: J/(mol degree C).
Coffee-cup calorimeter
A simple constant-pressure calorimeter made from an insulated cup (often a polystyrene cup). Used in general chemistry labs to measure the heat of reactions in solution.
Think of it as: a thermos with a thermometer. You mix two solutions, watch the temperature change, and calculate q.
Enthalpy (H)
The energy of a system measured as heat at constant pressure. You cannot measure H directly, but the change in enthalpy (Delta H) is what matters in practice.
Enthalpy change (Delta H)
The heat absorbed or released by a reaction at constant pressure. At constant pressure, Delta H = q. Sometimes called the "heat of reaction."
If Delta H < 0, the reaction is exothermic. If Delta H > 0, the reaction is endothermic.
State function
A property whose value depends only on the current state of the system, not on how it got there. Enthalpy is a state function: the enthalpy change for going from A to B is the same regardless of the route taken.
Think of it as altitude: the height difference between the base and the summit of a mountain is the same whether you took the steep path or the gentle one.
Hess's Law
In going from a set of reactants to a set of products, the change in enthalpy is the same regardless of how many steps the process takes. This lets you add known reactions together to find the Delta H of a reaction you cannot measure directly.
Standard enthalpy of formation (Delta H_f degree)
The enthalpy change that accompanies the formation of 1 mole of a compound from its elements in their standard states (25 degree C, 1 atm). By definition, Delta H_f degree for any element in its standard state is zero.
In simple terms, it is the energy cost (or payoff) of building one mole of a compound from scratch.
Calorimetry is based on observing the temperature change (Delta T) when a substance absorbs or releases heat.
Different substances respond very differently to the same amount of heat:
Substances with a high specific heat (e.g. water, 4.18 J/g degree C) require a lot of heat to raise their temperature. They make good coolants.
Substances with a low specific heat (e.g. metals like Cu at 0.39 J/g degree C, Fe at 0.45 J/g degree C) heat up quickly with very little energy input. This is why a metal pan gets hot much faster than the water inside it.
Heat capacity (C): energy to raise the temperature of an entire sample by 1 degree. Depends on how much substance you have.
Specific heat capacity (s): energy to raise the temperature of 1 gram by 1 degree. An intrinsic property of the material, independent of sample size.
Note: 4.18 J/(g degree C) = 4.18 J/(g K). The size of a degree Celsius equals the size of a kelvin.
Substance | Specific heat, s (J/g degree C) |
|---|---|
H2O (l) | 4.18 |
Al (s) | 0.89 |
Fe (s) | 0.45 |
Cu (s) | 0.39 |
Ag (s) | 0.24 |
Heat gained or lost:
q = m s Delta T
where m = mass (g), s = specific heat (J/g degree C), Delta T = T(final) - T(initial).
When Delta T > 0 (temperature rises), q > 0: the substance gained heat.
When Delta T < 0 (temperature drops), q < 0: the substance lost heat.
Heat gained = heat lost (conservation of energy in a calorimeter):
When two substances at different temperatures are mixed, the heat lost by the hotter substance equals the heat gained by the cooler one:
|q(cold)| = -|q(hot)|
A 30.0 g sample of H2O at 280 K is mixed with 50.0 g of H2O at 330 K. Calculate the final temperature, assuming no heat loss to the surroundings.
Heat lost by hot water = heat gained by cold water
-[(50.0 g)(4.18 J/g K)(T_f - 330 K)] = (30.0 g)(4.18 J/g K)(T_f - 280 K)
Solving: -209(T_f) + 68 970 = 125.4(T_f) - 35 112
334.4(T_f) = 104 082
T_f = 311.3 K
Molar heat capacity = specific heat times molar mass. Units: J/(mol degree C).
Example: the specific heat of Ag is 0.24 J/(g degree C). Molar mass of Ag = 107.9 g/mol. Molar heat capacity = 0.24 x 107.9 = 26 J/(mol degree C).
It takes 1.25 kJ of energy to heat a sample of pure silver from 12.0 degree C to 15.2 degree C. Calculate the mass of the silver sample.
q = 1.25 kJ = 1250 J
s = 0.24 J/(g degree C), Delta T = 15.2 - 12.0 = 3.2 degree C
1250 = m (0.24)(3.2)
m = 1250 / 0.768 = 1.6 x 10^3 g Ag
Enthalpy (H) is the energy of a system as heat. What matters in chemistry is the change: Delta H.
At constant pressure, the enthalpy change equals the heat flow:
Delta H = q_p
This is why "heat of reaction" and "enthalpy change" mean the same thing.
Sign of Delta H:
Delta H < 0: exothermic (energy released to surroundings).
Delta H > 0: endothermic (energy absorbed from surroundings).
Delta H is sometimes written as Delta H degree (rxn) when the reaction runs at standard conditions (25 degree C, 1 atm).
Worked example: 890 kJ of energy is released as heat when 1 mol of methane (CH4) is burned. Calculate Delta H for the process in which 5.8 g of CH4 is burned at 1 atm.
For 1 mol CH4: q = -890 kJ (released, so negative). Therefore Delta H = -890 kJ/mol.
5.8 g CH4 / 16.04 g/mol = 0.362 mol.
Delta H = 0.362 mol x (-890 kJ/mol) = -322 kJ (approximately).
Enthalpy change (Delta H) is a state function. Its value depends only on the initial and final states, not on the path taken.
This means: it does not matter whether a reaction goes from reactants to products in one step or in five steps. The total Delta H is the same.
Because Delta H is a state function, you can calculate it by adding up the enthalpy changes of a series of intermediate steps. This is Hess's Law.
Two important rules for manipulating reactions:
If a reaction is reversed, the sign of Delta H is reversed.
If a reaction is multiplied by a coefficient, Delta H is also multiplied by that coefficient.
Worked example: Calculate Delta H for the synthesis of diborane (B2H6) from its elements:
2 B(s) + 3 H2(g) -> B2H6(g)
Given reactions:
(a) 2 B(s) + 3/2 O2(g) -> B2O3(s), Delta H = -1273 kJ
(b) B2H6(g) + 3 O2(g) -> B2O3(s) + 3 H2O(g), Delta H = -2035 kJ
(c) H2(g) + 1/2 O2(g) -> H2O(l), Delta H = -286 kJ
(d) H2O(l) -> H2O(g), Delta H = +44 kJ
Strategy: keep (a) as is (it has B(s) on the left). Reverse (b) to put B2H6 on the right. Multiply (c) by 3 and (d) by 3.
Use (a): Delta H = -1273 kJ
Reverse (b): Delta H = +2035 kJ
Add: 2 B(s) + 3 H2O(g) -> B2H6(g) + 3/2 O2(g), Delta H = +762 kJ
Use 3 x (c): Delta H = 3(-286) = -858 kJ
Add: 2 B(s) + 3 H2(g) + 3 H2O(l) -> B2H6(g) + 3 H2O(l), Delta H = -96 kJ
Use 3 x (d): Delta H = 3(+44) = +132 kJ
Final: 2 B(s) + 3 H2(g) -> B2H6(g), Delta H = +36 kJ
The standard enthalpy of formation is the enthalpy change when 1 mole of a compound is formed from its elements in their standard states.
Standard states: 25 degree C (298 K) and 1 atm. Elements in their most stable form at these conditions.
By definition, Delta H_f degree for any element in its standard state = 0.
Examples of formation reactions:
NaCl: Na(s) + 1/2 Cl2(g) -> NaCl(s)
H2O: H2(g) + 1/2 O2(g) -> H2O(l)
C6H12O6: 6 C(s) + 6 H2(g) + 3 O2(g) -> C6H12O6(s)
PbSO4: Pb(s) + S(s) + 2 O2(g) -> PbSO4(s)
Notice: reactants are always elements in their standard states, and products are exactly 1 mole of the compound.
Calculating Delta H of a reaction from formation values:
Delta H degree (rxn) = sum of Delta H_f degree (products) - sum of Delta H_f degree (reactants)
Remember to multiply each Delta H_f degree by its stoichiometric coefficient in the balanced equation.
Worked example: The thermite reaction
2 Al(s) + Fe2O3(s) -> Al2O3(s) + 2 Fe(s)
Delta H_f degree for Al2O3(s) = -1676 kJ/mol
Delta H_f degree for Fe2O3(s) = -826 kJ/mol
Delta H_f degree for Al(s) = 0 (element in standard state)
Delta H_f degree for Fe(s) = 0 (element in standard state)
Delta H degree = [-1676 + 2(0)] - [-826 + 2(0)] = -1676 + 826 = -850 kJ/mol
The thermite reaction is strongly exothermic.
Worked example: Oxidation of ammonia
4 NH3(g) + 5 O2(g) -> 4 NO(g) + 6 H2O(g)
Delta H_f degree for NO(g) = +91.3 kJ/mol
Delta H_f degree for H2O(g) = -241.8 kJ/mol
Delta H_f degree for NH3(g) = -45.9 kJ/mol
Delta H_f degree for O2(g) = 0
Delta H degree (rxn) = [4(+91.3) + 6(-241.8)] - [4(-45.9) + 5(0)]
= [+365.2 + (-1450.8)] - [-183.6 + 0]
= [-1085.6] - [-183.6]
= -902.0 kJ
Formula | Meaning | Units |
|---|---|---|
q = m s Delta T | Heat gained or lost by a substance | J or kJ |
Delta T = T(final) - T(initial) | Temperature change | degree C or K |
q(cold) | = - | |
Molar heat capacity = s x molar mass | Heat capacity per mole | J/(mol degree C) |
Delta H = q_p | Enthalpy change at constant pressure | kJ or kJ/mol |
Delta H degree (rxn) = sum Delta H_f degree (products) - sum Delta H_f degree (reactants) | Standard enthalpy of reaction from formation values | kJ/mol |
Key reminder: Delta H_f degree = 0 for any element in its standard state.
The reason your car radiator has fins and is filled with a water-based coolant: water's high specific heat (4.18 J/g degree C) lets it absorb a large amount of engine heat with a relatively small temperature rise. The fins increase surface area for heat transfer to the air.
Hess's Law and standard enthalpies of formation are how chemical engineers calculate whether a proposed industrial reaction will produce or consume energy, before ever running it in a lab. Designing a fertiliser plant or an explosives process starts with these calculations.
"High specific heat means the substance heats up quickly." The opposite is true. A high specific heat means it takes more energy to raise the temperature. Water is hard to heat up and hard to cool down, which is exactly why it is such a good coolant.
"Delta H_f degree for water is zero because it is so common." No. Delta H_f degree is zero only for elements in their standard states. Water is a compound, so it has a nonzero Delta H_f degree (-285.8 kJ/mol for liquid water).
"You can use any reaction to calculate Delta H from Hess's Law." You can use any set of reactions, provided they add up algebraically to give the target reaction. You will not necessarily use every reaction you are given, and you might need to reverse or multiply some of them.
"Reversing a reaction changes Delta H from negative to positive, so the reaction switches from exothermic to endothermic." This is correct, and it catches students off guard. If the forward reaction is exothermic (Delta H < 0), the reverse reaction is endothermic (Delta H > 0) by exactly the same magnitude.
Coffee-cup calorimeter problems are extremely common on exams. You will be given masses, temperatures, and a specific heat (usually water's) and asked to calculate q, then Delta H per mole of a product or reactant. Watch for the step where you need to find the total mass of the solution and convert units.
Hess's Law problems are a favourite. The exam will give you 3 or 4 reactions with their Delta H values and ask you to combine them to find the Delta H of a target reaction. Practise identifying which reactions to reverse and which to multiply.
Standard enthalpies of formation: know that Delta H_f degree = 0 for elements in their standard states. A common error is forgetting to multiply Delta H_f degree by the stoichiometric coefficient.
Know the difference between Delta H and Delta E. At constant pressure, Delta H = q. At constant volume, Delta E = q. Most reactions in open containers are at constant pressure.
Fill in the blank: The equation for heat gained or lost is q = ______ x ______ x ______.
m x s x Delta T.
True or false: The specific heat of a substance tells you how much energy is needed to raise the temperature of 1 mole by 1 degree.
False. That is molar heat capacity. Specific heat is per gram.
True or false: If Delta H for a reaction is negative, the reaction is endothermic.
False. Negative Delta H means exothermic.
Fill in the blank: According to Hess's Law, if you reverse a reaction, the sign of Delta H is ______.
Reversed (changes from + to - or from - to +).
True or false: Delta H_f degree for O2(g) is zero.
True. O2 is an element in its standard state.
Q: A 200.0 g sample of water at 80.0 degree C is mixed with 150.0 g of water at 20.0 degree C. What is the final temperature? (Assume no heat loss.)
A: Heat lost by hot water = heat gained by cold water. -(200.0)(4.18)(T_f - 80.0) = (150.0)(4.18)(T_f - 20.0). The 4.18 cancels. -200(T_f - 80) = 150(T_f - 20). -200 T_f + 16 000 = 150 T_f - 3000. 350 T_f = 19 000. T_f = 54.3 degree C.
Q: In a coffee-cup calorimeter, 50.0 mL of 1.0 M HCl is mixed with 50.0 mL of 1.0 M NaOH, both at 25.0 degree C. The temperature rises to 31.9 degree C. Calculate Delta H per mole of water formed. (Assume density = 1 g/mL, s = 4.18 J/g degree C.)
A: Total mass = 100.0 g. Delta T = 31.9 - 25.0 = 6.9 degree C. q = (100.0)(4.18)(6.9) = 2884 J = 2.88 kJ. Moles of water formed = 0.050 L x 1.0 mol/L = 0.050 mol. The reaction is exothermic (temperature rose), so Delta H = -2.88 kJ / 0.050 mol = -57.6 kJ/mol.
Q: Using standard enthalpies of formation, calculate Delta H degree (rxn) for: 2 C2H6(g) + 7 O2(g) -> 4 CO2(g) + 6 H2O(l). Given: Delta H_f degree for C2H6(g) = -84.7 kJ/mol, CO2(g) = -393.5 kJ/mol, H2O(l) = -285.8 kJ/mol.
A: Delta H degree = [4(-393.5) + 6(-285.8)] - [2(-84.7) + 7(0)] = [-1574.0 + (-1714.8)] - [-169.4] = -3288.8 - (-169.4) = -3119.4 kJ.
Q: If the reaction A -> B has Delta H = -150 kJ, what is Delta H for 2B -> 2A?
A: Reverse the reaction (sign flips): B -> A, Delta H = +150 kJ. Multiply by 2: 2B -> 2A, Delta H = +300 kJ.
Q: Write the formation reaction for glucose, C6H12O6(s).
A: 6 C(s, graphite) + 6 H2(g) + 3 O2(g) -> C6H12O6(s). Reactants are elements in their standard states; exactly 1 mol of product is formed.
Enthalpy reappears immediately in the discussion of entropy (Delta S) and Gibbs free energy (Delta G = Delta H - T Delta S), which determines whether a reaction is spontaneous.
Calorimetry techniques in this chapter are the same ones used in bomb calorimetry (constant volume) later, where Delta E rather than Delta H is measured directly.
The sign conventions and energy-conservation principles from Part 1 underpin every calculation in this section. If those feel shaky, revisit Part 1 before tackling these problems.
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