Master Thermochemistry

Your comprehensive guide to understanding energy, heat, enthalpy, and the thermodynamics of chemical reactions.

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Comprehensive Study Guide

1. Energy, Heat, & Temperature

  • Energy: The capacity to do work or produce heat. Measured in Joules (J).
  • Temperature ($T$): A measure of the average kinetic energy of particles. It is not energy itself!
  • Heat ($q$): The transfer of energy between objects due to a temperature difference. Heat flows from hot to cold.
  • System vs. Surroundings: The system is the chemical reaction; surroundings are everything else (beaker, air, your hand).

2. Exothermic vs. Endothermic

  • Exothermic Reactions: Energy is released. The system loses heat ($\Delta H \lt 0$). The surroundings get hot. (e.g., Combustion, freezing water).
  • Endothermic Reactions: Energy is absorbed. The system gains heat ($\Delta H \gt 0$). The surroundings get cold. (e.g., Melting ice, photosynthesis).

3. Calorimetry & Specific Heat

Used to calculate heat transfer when temperature changes:

$$q = m \cdot c \cdot \Delta T$$
  • $q$ = Heat (Joules)
  • $m$ = Mass (grams)
  • $c$ = Specific heat capacity ($\text{J/g}^\circ\text{C}$). Water is $4.184 \text{ J/g}^\circ\text{C}$.
  • $\Delta T$ = Change in temp ($T_{final} - T_{initial}$)

4. Phase Changes (Latent Heat)

During a phase change, temperature remains constant. Energy is used to break intermolecular forces, not increase speed.

$$q = m \cdot \Delta H \quad \text{or} \quad q = n \cdot \Delta H$$
  • Heat of Fusion ($\Delta H_{fus}$): Solid to Liquid (Melting)
  • Heat of Vaporization ($\Delta H_{vap}$): Liquid to Gas (Boiling)

5. Calculating Enthalpy of Reaction ($\Delta H_{rxn}$)

Method A: Standard Formations

Using a table of standard enthalpies of formation ($\Delta H_f^\circ$). Elements in standard states = 0.

$\Delta H_{rxn} = \sum(\text{Products}) - \sum(\text{Reactants})$

Method B: Hess's Law

Adding multiple reactions together to get a target reaction. If you flip a reaction, flip the sign of $\Delta H$. If you multiply a reaction, multiply $\Delta H$.

Method C: Bond Enthalpies

Energy required to break bonds minus energy released when forming bonds.

$\Delta H_{rxn} = \sum(\text{Bonds Broken}) - \sum(\text{Bonds Formed})$

Recommended Videos

Visual explanations of core concepts.

Enthalpy & Thermochemistry

Calorimetry & Specific Heat

Step-by-Step Walkthroughs

Click "Show Walkthrough" on any question to see exactly how to approach and solve the problem step-by-step.

1. A $25.0 \text{ g}$ piece of iron is heated from $20.0^\circ\text{C}$ to $85.0^\circ\text{C}$. How much heat was absorbed? ($c_{iron} = 0.450 \text{ J/g}^\circ\text{C}$) Show Walkthrough +

Step-by-Step Solution:

  1. Identify the given information:
    Mass ($m$) = $25.0 \text{ g}$
    Initial Temp ($T_i$) = $20.0^\circ\text{C}$
    Final Temp ($T_f$) = $85.0^\circ\text{C}$
    Specific Heat ($c$) = $0.450 \text{ J/g}^\circ\text{C}$
  2. Calculate the temperature change ($\Delta T$):
    $\Delta T = T_f - T_i = 85.0^\circ\text{C} - 20.0^\circ\text{C} = 65.0^\circ\text{C}$
  3. Choose the correct formula: Because temperature is changing, use $q = mc\Delta T$.
  4. Plug in the numbers and calculate:
    $q = (25.0 \text{ g})(0.450 \text{ J/g}^\circ\text{C})(65.0^\circ\text{C})$
    $q = 731.25 \text{ J}$
  5. Check sig figs: Given values have 3 sig figs, so the final answer is $731 \text{ J}$.
2. Using Standard Enthalpies of Formation to find $\Delta H_{rxn}$ for the combustion of methane: $\text{CH}_4(g) + 2\text{O}_2(g) \rightarrow \text{CO}_2(g) + 2\text{H}_2\text{O}(l)$ Show Walkthrough +

Given $\Delta H_f^\circ$ values (kJ/mol): $\text{CH}_4(g) = -74.8$, $\text{CO}_2(g) = -393.5$, $\text{H}_2\text{O}(l) = -285.8$

Step-by-Step Solution:

  1. Write the formula:
    $\Delta H_{rxn} = \sum n\Delta H_f^\circ(\text{Products}) - \sum m\Delta H_f^\circ(\text{Reactants})$
  2. Identify the $\Delta H_f^\circ$ for elements:
    $\text{O}_2(g)$ is an element in its standard state, so its $\Delta H_f^\circ = 0 \text{ kJ/mol}$.
  3. Calculate sum for Products:
    $[1 \cdot \text{CO}_2 + 2 \cdot \text{H}_2\text{O}]$
    $[1(-393.5) + 2(-285.8)] = -393.5 - 571.6 = -965.1 \text{ kJ}$
  4. Calculate sum for Reactants:
    $[1 \cdot \text{CH}_4 + 2 \cdot \text{O}_2]$
    $[1(-74.8) + 2(0)] = -74.8 \text{ kJ}$
  5. Subtract Reactants from Products:
    $\Delta H_{rxn} = (-965.1) - (-74.8)$
    $\Delta H_{rxn} = -965.1 + 74.8 = \mathbf{-890.3 \text{ kJ}}$ (Exothermic)
3. Hess's Law: Find $\Delta H$ for $A + 2B \rightarrow C$ given the following two steps:
(1) $A \rightarrow 2D \quad \Delta H = 40 \text{ kJ}$
(2) $C \rightarrow 2D + 2B \quad \Delta H = -15 \text{ kJ}$ Show Walkthrough +

Step-by-Step Solution:

Goal Equation: $A + 2B \rightarrow C$

  1. Analyze Step 1: $A \rightarrow 2D$.
    We need $A$ on the reactant side in our goal equation. Step 1 has $A$ on the reactant side. So, we keep Step 1 exactly as is.
    Current $\Delta H_1 = 40 \text{ kJ}$
  2. Analyze Step 2: $C \rightarrow 2D + 2B$.
    We need $C$ on the product side, and $2B$ on the reactant side. Step 2 has them reversed. Therefore, we must reverse Step 2.
    Reversed Step 2: $2D + 2B \rightarrow C$.
    New $\Delta H_2$ (flip the sign) $= +15 \text{ kJ}$
  3. Add the equations together:
    $A \rightarrow 2D$
    $2D + 2B \rightarrow C$
    -----------------------
    Notice that $2D$ is produced in the first reaction and consumed in the second, so it cancels out.
    Result: $A + 2B \rightarrow C$ (This matches our goal!)
  4. Add the $\Delta H$ values:
    $\Delta H_{total} = \Delta H_1 + \text{New } \Delta H_2$
    $\Delta H_{total} = 40 \text{ kJ} + 15 \text{ kJ} = \mathbf{55 \text{ kJ}}$

Additional Practice Questions

Test your knowledge. Write down your answers, then check the answer key below.

  1. A cold pack is placed on an injured leg. Is the chemical reaction inside the cold pack endothermic or exothermic? What is the sign of $\Delta H$?
  2. How much heat is required to warm $100.0 \text{ g}$ of water from $25.0^\circ\text{C}$ to $100.0^\circ\text{C}$? ($c_{water} = 4.184 \text{ J/g}^\circ\text{C}$)
  3. Why does the temperature of a glass of ice water remain at $0^\circ\text{C}$ until all the ice has melted?
  4. Calculate the enthalpy change for $2\text{H}_2\text{O}(l) \rightarrow 2\text{H}_2(g) + \text{O}_2(g)$ if the formation of 2 moles of liquid water releases $572 \text{ kJ}$ of energy.
  5. Identify the substance that has a standard enthalpy of formation ($\Delta H_f^\circ$) equal to exactly zero: $\text{H}_2\text{O}(g)$, $\text{Na}(s)$, or $\text{CO}_2(g)$?
  6. A $50 \text{ g}$ block of metal at $100^\circ\text{C}$ is dropped into $50 \text{ g}$ of water at $20^\circ\text{C}$. Will the final temperature of the mixture be exactly $60^\circ\text{C}$? Why or why not?
  7. Use bond enthalpies to estimate $\Delta H$ for: $\text{H}_2 + \text{F}_2 \rightarrow 2\text{HF}$. (Bond energies in kJ/mol: $\text{H-H} = 436$, $\text{F-F} = 155$, $\text{H-F} = 567$).
  8. If a reaction releases $450 \text{ J}$ of heat to its surroundings, what is the value of $q$ for the system?
  9. Is the freezing of water into ice an exothermic or endothermic process? Explain based on energy flow.
  10. Substance A has a specific heat of $0.5 \text{ J/g}^\circ\text{C}$ and Substance B has a specific heat of $2.0 \text{ J/g}^\circ\text{C}$. If equal masses of both absorb the same amount of heat, which one will experience the larger temperature change?
Show Answer Key
  1. Endothermic. The reaction absorbs heat from its surroundings (your leg), making it feel cold. The sign of $\Delta H$ is positive ($+$).
  2. $31,380 \text{ J}$ (or $31.4 \text{ kJ}$). $q = (100.0 \text{ g})(4.184 \text{ J/g}^\circ\text{C})(100.0 - 25.0)^\circ\text{C} = 31380 \text{ J}$.
  3. Latent Heat of Fusion. Energy entering the system is being used to break the intermolecular forces holding the ice lattice together, rather than increasing the kinetic energy (temperature) of the molecules.
  4. $+572 \text{ kJ}$. The reaction given is the exact reverse of the formation of water. Reversing the reaction changes the sign of $\Delta H$.
  5. $\text{Na}(s)$. Sodium is a pure element in its standard stable state at room temperature, so its formation enthalpy is zero.
  6. No, it will be much lower than $60^\circ\text{C}$. Water has a much higher specific heat capacity than metals, meaning it resists temperature changes more effectively. The water's temp will rise less than the metal's temp will fall.
  7. $-543 \text{ kJ}$. $\text{Bonds Broken} = (436 + 155) = 591$. $\text{Bonds Formed} = 2(567) = 1134$. $\Delta H = 591 - 1134 = -543 \text{ kJ}$.
  8. $q = -450 \text{ J}$. Because the system is losing/releasing energy, the sign must be negative.
  9. Exothermic. Liquid water must lose heat (release energy to the surroundings) to form the stronger intermolecular bonds of solid ice.
  10. Substance A. A lower specific heat means it takes less energy to change its temperature. Therefore, it will heat up much faster and have a larger $\Delta T$.