Heat, calorimetry, PV work, the first law of thermodynamics and enthalpy

Thermochemistry studies the heat exchanged in reactions. For the exam two question types dominate: calorimetry calculations (going from a temperature change to the heat of reaction) and stoichiometry with ΔH (grams are given, how many kJ come out). In between there is a conceptual part: the first law and the sign rules. A student who mixes up the signs finds the right number and still loses points.

Golden rules:

1. The sign is from the system’s point of view. If the system takes in energy, +; if it gives out, −.
2. Heat is not lost in a calorimeter. What one gives, the other takes: $\sum q = 0$.
3. The heat measured at constant pressure is ΔH, at constant volume it is ΔU.
4. ΔH is read together with the equation. If the coefficients change, ΔH changes too.

Heat, specific heat and heat capacity

Heat ($q$) is the energy transferred from a hot body to a cold one because of a temperature difference. Temperature, on the other hand, is a measure of the average kinetic energy of the particles. They are not the same thing: a bathtub of lukewarm water stores more heat than a cup of boiling water.

The heat needed to warm a substance:

$q = m\,c\,\Delta T \qquad \Delta T = T_{\text{final}} - T_{\text{initial}}$

• $c$: specific heat, the heat that raises the temperature of 1 g of substance by 1 °C, J/(g·°C).
• $C = mc$: heat capacity, the heat that raises the temperature of the whole body by 1 °C, J/°C.
• Molar heat capacity: J/(mol·°C).

For $\Delta T$ the difference in °C and K is the same; there is no need to convert one to the other.

The specific heat of water is very high; copper receiving the same heat warms up ten times more than water.

Example 1. To raise 75.0 g of copper from 20.0 °C to 95.0 °C:

$q = 75.0\ \text{g} \times 0.385\ \tfrac{\text{J}}{\text{g·°C}} \times 75.0\ ^\circ\text{C} = 2.17\times10^3\ \text{J}$

The sign rule

Calorimetry

A calorimeter is an insulated container in which the heat of a reaction is measured without escaping to the surroundings. The logic: the heat given by the reaction is taken by the water and the container.

$q_{\text{rxn}} = -\left(q_{\text{water}} + q_{\text{calorimeter}}\right)$

The coffee-cup calorimeter

Two foam cups, a lid and a thermometer. Because it is open to the atmosphere, it works at constant pressure. It is used for reactions in solution (neutralization, dissolution). The solution is generally assumed to behave like water (density 1.00 g/mL, c ≈ 4.18).

Example 2 (calorimeter constant). 50.0 g of water in the calorimeter is at 20.0 °C. 50.0 g of water at 60.0 °C is added to it, and the final temperature is 39.2 °C. Because the calorimeter absorbed some heat, the average temperature of 40.0 °C was not reached.

hot water gave: 50.0 × 4.184 × (39.2 − 60.0) = −4351 J
cold water took: 50.0 × 4.184 × (39.2 − 20.0) = +4017 J
calorimeter took: 4351 − 4017 = 334 J   (ΔT = 19.2 °C)
C_cal = 334 / 19.2 = 17.4 J/°C

The bomb calorimeter

The sample is burned in excess oxygen in a thick steel container (the bomb); the bomb is immersed in water. Because the container is closed and rigid, it works at constant volume, and the measured heat is $\Delta U$. Usually the heat capacity of the whole setup is given as a single number:

$q_{\text{rxn}} = -C_{\text{cal}}\,\Delta T$

The energy values of foods and the heats of combustion of fuels are measured this way.

The heat capacity of the setup is first measured with a standard whose heat of combustion is known (usually benzoic acid, $-26.42$ kJ/g), and then the unknown sample is burned.

Example 3 (bomb calibration and measurement).

calibration: 1.000 g benzoic acid → 26.42 kJ;  ΔT = 5.07 °C
  C_cal = 26.42 kJ / 5.07 °C = 5.21 kJ/°C
measurement: 0.750 g unknown fuel;  ΔT = 3.88 °C
  q_cal = 5.211 × 3.88 = 20.22 kJ
  heat of combustion = −20.22 / 0.750 = −27.0 kJ/g

PITFALL · kJ/g versus kJ/mol. The calorimeter first gives the total heat of the burned sample. If it is asked per mole, divide by the number of moles burned; if per gram, divide by the grams. Put the sign at the very end: combustion is exothermic, so the answer is negative.

PV work

If a reaction produces gas, the gas expands against a constant external pressure and pushes the surroundings. This work done by the system is called pressure-volume work:

$w = -P_{\text{ext}}\,\Delta V$

• Expansion ($\Delta V > 0$): $w < 0$, the system does work and loses energy.
• Compression ($\Delta V < 0$): $w > 0$, the surroundings do work on the system.
• Convert L·atm to joules: $1\ \text{L·atm} = 101.325\ \text{J}$.

Example 4. A gas is compressed from 4.00 L to 1.50 L at a constant external pressure of 1.00 atm:

$w = -1.00 \times (1.50 - 4.00) = +2.50\ \text{L·atm} = +253\ \text{J}$

The first law of thermodynamics

Energy cannot be created or destroyed. The internal energy ($U$) of a system changes only by exchange of heat or work:

$\Delta U = q + w$

Example 5. A system loses 450 J of heat, and the surroundings do 200 J of work on the system: $\Delta U = -450 + 200 = -250$ J.

State functions

These are quantities whose value depends only on the present state of the system; it does not matter by which path it got there. $U$, $H$, $P$, $V$, $T$ are state functions. $q$ and $w$ are not: the same $\Delta U$ can be obtained with different combinations of $q$/$w$. It is like climbing a mountain: the difference in altitude (a state function) is independent of the path, but the distance you walked (q, w) is not.

Enthalpy

Most reactions take place in an open container at constant atmospheric pressure. For the heat measured under this condition, a new state function is defined, enthalpy:

$H = U + PV \qquad \Delta H = q_P \quad (\text{constant } P)$

At constant volume, $\Delta U = q_V$. The difference between the two is the expansion work of gases:

$\Delta H = \Delta U + \Delta n_{\text{gas}}\,RT$

$\Delta n_{\text{gas}}$ = moles of gas in the products − moles of gas in the reactants. If there are only solids and liquids, or the moles of gas are equal, $\Delta H \approx \Delta U$.

Thermochemical equations

The $\Delta H$ value belongs to the written equation:

$\mathrm{2H_2(g) + O_2(g) \rightarrow 2H_2O(l)} \qquad \Delta H = -571.6\ \text{kJ}$

The rules:

1. If you multiply the equation by k, $\Delta H$ is also multiplied by k (an extensive property).
2. If you reverse the equation, the sign of $\Delta H$ changes.
3. Physical states matter: if $\mathrm{H_2O(g)}$ forms, $\Delta H$ is less negative, because the energy of the heat of vaporization stays with the product.

Stoichiometry with ΔH

Use $\Delta H$ like a conversion factor: the equation above says “571.6 kJ per 2 mol of $\mathrm{H_2}$”.

Example 6. The heat released when 5.00 g of $\mathrm{H_2}$ burns:

$5.00\ \text{g} \times \frac{1\ \text{mol H}_2}{2.016\ \text{g}} \times \frac{571.6\ \text{kJ}}{2\ \text{mol H}_2} = 709\ \text{kJ}$

TIP · divide by the coefficient. The most common mistake is to take 571.6 kJ as “per 1 mol of $\mathrm{H_2}$”. There are 2 mol of $\mathrm{H_2}$ in the equation; per mole it is 285.8 kJ.

Common mistakes