Thermochemistry — Hard Practice Quiz
A Chemistry cheat sheet for Thermochemistry — every key formula with its symbols defined — plus a hard-level practice quiz to test recall.
Formulas & key concepts
Kinetic Energy: Energy of motion of an object.
Where: m = mass, v = velocity
Change in Internal Energy: Difference between final and initial energy states.
First Law of Thermodynamics: Energy change equals heat plus work.
Where: q = heat, w = work
Work Done by Gas: Work done by an expanding gas at constant pressure.
Where: P = pressure, ΔV = volume change
Enthalpy Change: At constant pressure, enthalpy change equals heat transferred.
Where: q<sub>P</sub> = heat at constant pressure
Specific Heat: Heat capacity per gram of substance.
Where: q = heat, m = mass, ΔT = temperature change
Heat Calculation: Quantity of heat absorbed or released.
Where: C<sub>s</sub> = specific heat, m = mass, ΔT = temperature change
Standard Enthalpy of Reaction: Calculate reaction enthalpy from formation enthalpies.
Where: n, m = stoichiometric coefficients
| Substance | Specific Heat (J/g·K) |
|---|---|
| Water (H2O(l)) | 4.18 |
| Methane (CH4(g)) | 2.20 |
| Nitrogen (N2(g)) | 1.04 |
| Aluminum (Al(s)) | 0.90 |
| Carbon dioxide (CO2(g)) | 0.84 |
| Calcium carbonate (CaCO3(s)) | 0.82 |
| Iron (Fe(s)) | 0.45 |
| Mercury (Hg(l)) | 0.14 |
| Substance | ΔH°f (kJ/mol) |
|---|---|
| Water (H2O(l)) | -285.8 |
| Water vapor (H2O(g)) | -241.8 |
| Carbon dioxide (CO2(g)) | -393.5 |
| Methane (CH4(g)) | -74.80 |
| Ammonia (NH3(g)) | -46.19 |
| Sodium chloride (NaCl(s)) | -410.9 |
| Calcium carbonate (CaCO3(s)) | -1207.1 |
| Glucose (C6H12O6(s)) | -1273 |
| Ethanol (C2H5OH(l)) | -277.7 |
| Acetylene (C2H2(g)) | 226.7 |
| Benzene (C6H6(l)) | 49.0 |
Note: ΔH°f for elements in standard state = 0
Practice quiz
A gas in a piston-cylinder assembly absorbs $150 \text{ J}$ of heat and expands, doing $75 \text{ J}$ of work on the surroundings. If the gas initially had a kinetic energy of $200 \text{ J}$ and its mass remains constant, what is its final kinetic energy? Assume all internal energy change manifests as kinetic energy change for this simplified scenario.
- $125 \text{ J}$
- $200 \text{ J}$
- $275 \text{ J}$
- $350 \text{ J}$
Answer: $275 \text{ J}$
$50.0 \text{ g}$ of liquid water at $25.0^{\circ}\text{C}$ is heated to $75.0^{\circ}\text{C}$ at constant atmospheric pressure. Given the specific heat of water is $4.18 \text{ J/g} \cdot \text{K}$, what is the enthalpy change for this process?
- $5.23 \text{ kJ}$
- $10.45 \text{ kJ}$
- $20.90 \text{ kJ}$
- $2.09 \text{ kJ}$
Answer: $10.45 \text{ kJ}$
A system undergoes a process where its internal energy decreases by $100 \text{ J}$. If the system expands against a constant external pressure of $2.0 \text{ atm}$ and its volume changes from $1.0 \text{ L}$ to $3.0 \text{ L}$, how much heat was exchanged with the surroundings? (Note: $1 \text{ L} \cdot \text{atm} = 101.3 \text{ J}$)
- $-505.2 \text{ J}$
- $-305.2 \text{ J}$
- $305.2 \text{ J}$
- $505.2 \text{ J}$
Answer: $305.2 \text{ J}$
Consider the combustion of methane: $\text{CH}_4(\text{g}) + 2\text{O}_2(\text{g}) \rightarrow \text{CO}_2(\text{g}) + 2\text{H}_2\text{O}(\text{l})$. If $16.0 \text{ g}$ of methane is completely combusted, and all the heat released is absorbed by $1.00 \text{ kg}$ of water initially at $20.0^{\circ}\text{C}$, what will be the final temperature of the water? (Assume no heat loss to surroundings. Use molar mass of $\text{CH}_4 = 16.04 \text{ g/mol}$)
- $20.0^{\circ}\text{C}$
- $106.2^{\circ}\text{C}$
- $232.4^{\circ}\text{C}$
- $444.8^{\circ}\text{C}$
Answer: $232.4^{\circ}\text{C}$
For a chemical reaction occurring at constant temperature, under what conditions would the change in enthalpy ($\Delta H$) be approximately equal to the change in internal energy ($\Delta E$)?
- When the reaction involves only solids and liquids.
- When the reaction is carried out in an open container.
- When the system does a large amount of work on the surroundings.
- When the heat absorbed by the system is zero.
Answer: When the reaction involves only solids and liquids.
A substance of mass $m$ and specific heat $C_s$ absorbs heat $q$. If its initial temperature is $T_i$, derive an expression for its final temperature $T_f$.
- $T_f = T_i - \frac{q}{C_s \times m}$
- $T_f = \frac{q}{C_s \times m} - T_i$
- $T_f = T_i + \frac{q}{C_s \times m}$
- $T_f = q \times C_s \times m + T_i$
Answer: $T_f = T_i + \frac{q}{C_s \times m}$
An object has an initial kinetic energy $E_k$. If its mass is reduced to one-third ($\frac{1}{3}$) of its original value and its velocity is doubled, what is its new kinetic energy in terms of $E_k$?
- $\frac{1}{3}E_k$
- $\frac{2}{3}E_k$
- $\frac{4}{3}E_k$
- $\frac{8}{3}E_k$
Answer: $\frac{4}{3}E_k$
Calculate the change in internal energy ($\Delta E$) for the combustion of $1.00 \text{ mol}$ of acetylene ($\text{C}_2\text{H}_2(\text{g})$) at $25^{\circ}\text{C}$ and $1.00 \text{ atm}$ pressure, given the reaction: $2\text{C}_2\text{H}_2(\text{g}) + 5\text{O}_2(\text{g}) \rightarrow 4\text{CO}_2(\text{g}) + 2\text{H}_2\text{O}(\text{l})$. Assume ideal gas behavior and $R = 8.314 \text{ J/mol} \cdot \text{K}$.
- $-1303.2 \text{ kJ}$
- $-1299.5 \text{ kJ}$
- $-1295.8 \text{ kJ}$
- $-1292.1 \text{ kJ}$
Answer: $-1295.8 \text{ kJ}$
A $100.0 \text{ g}$ piece of iron (specific heat $0.45 \text{ J/g} \cdot \text{K}$) at $150.0^{\circ}\text{C}$ is dropped into $200.0 \text{ g}$ of water (specific heat $4.18 \text{ J/g} \cdot \text{K}$) at $20.0^{\circ}\text{C}$ in an insulated container. What is the final temperature of the system?
- $20.0^{\circ}\text{C}$
- $26.6^{\circ}\text{C}$
- $35.1^{\circ}\text{C}$
- $42.3^{\circ}\text{C}$
Answer: $26.6^{\circ}\text{C}$
A chemical reaction occurs in a closed system at constant pressure. The system releases $50 \text{ kJ}$ of heat to the surroundings and does $10 \text{ kJ}$ of work on the surroundings. Which of the following statements is true regarding the change in internal energy ($\Delta E$) and enthalpy ($\Delta H$) for the system?
- $\Delta E = -60 \text{ kJ}$ and $\Delta H = -50 \text{ kJ}$
- $\Delta E = -40 \text{ kJ}$ and $\Delta H = -50 \text{ kJ}$
- $\Delta E = -60 \text{ kJ}$ and $\Delta H = -60 \text{ kJ}$
- $\Delta E = -50 \text{ kJ}$ and $\Delta H = -40 \text{ kJ}$
Answer: $\Delta E = -60 \text{ kJ}$ and $\Delta H = -50 \text{ kJ}$
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