Easy MCAT Thermochemistry Practice Questions

Mastering thermochemistry is essential for success on the Chemical and Physical Foundations of Biological Systems section of the MCAT. This guide provides Easy MCAT Thermochemistry Practice Questions designed to build your foundational knowledge of energy, heat, and enthalpy. By focusing on these core principles using retrieval practice, you can ensure that you are prepared for the more complex applications you will encounter on exam day.

Concept Explanation

Thermochemistry is the study of the energy and heat associated with chemical reactions and physical transformations. At its core, it focuses on the First Law of Thermodynamics, which states that energy cannot be created or destroyed, only transferred or transformed. In the context of the MCAT, you must understand how energy moves between a system (the reaction itself) and the surroundings (everything else). Key variables include enthalpy ($\Delta H$), which measures heat flow at constant pressure; entropy ($\Delta S$), which measures disorder; and Gibbs Free Energy ($\Delta G$), which determines reaction spontaneity.

According to LibreTexts Chemistry, enthalpy changes are classified as exothermic (releasing heat, $\Delta H < 0$) or endothermic (absorbing heat, $\Delta H > 0$). Specific heat capacity ($c$) is another vital concept, defined by the equation $q = mc\Delta T$, where $q$ is heat, $m$ is mass, and $\Delta T$ is the change in temperature. Understanding these relationships is a key part of mastering STEM subjects like general chemistry. You should also be familiar with Hess’s Law, which allows you to calculate the total enthalpy change of a reaction by summing the enthalpy changes of its individual steps.

Solved Examples

Review these step-by-step solutions to understand how to approach thermochemistry calculations and conceptual problems.

1. Specific Heat Calculation: How much heat is required to raise the temperature of 50 g of water from $25^\circ \text{C}$ to $45^\circ \text{C}$? (Specific heat of water $c = 4.18 \text{ J/g}\cdot^\circ \text{C}$).
   1. Identify the variables: $m = 50 \text{ g}$, $c = 4.18 \text{ J/g}\cdot^\circ \text{C}$, $\Delta T = 45 - 25 = 20^\circ \text{C}$.
   2. Use the formula: $q = mc\Delta T$.
   3. Substitute the values: $q = (50 \text{ g})(4.18 \text{ J/g}\cdot^\circ \text{C})(20^\circ \text{C})$.
   4. Calculate the final result: $q = 4,180 \text{ J}$ or $4.18 \text{ kJ}$.
2. Hess's Law Application: Given the following reactions, find the $\Delta H$ for $A \rightarrow C$.
   • $A \rightarrow B \quad \Delta H = +100 \text{ kJ/mol}$
   • $B \rightarrow C \quad \Delta H = -150 \text{ kJ/mol}$
   1. Recognize that enthalpy is a state function, meaning the path does not matter.
   2. Add the two reactions together: $(A \rightarrow B) + (B \rightarrow C) = A \rightarrow C$.
   3. Add the corresponding enthalpy changes: $+100 \text{ kJ/mol} + (-150 \text{ kJ/mol})$.
   4. The total enthalpy change is $-50 \text{ kJ/mol}$.
3. Gibbs Free Energy and Spontaneity: A reaction has $\Delta H = -200 \text{ kJ/mol}$ and $\Delta S = -0.5 \text{ kJ/mol}\cdot \text{K}$. At what temperature does the reaction become non-spontaneous?
   1. Set the Gibbs Free Energy equation to zero to find the equilibrium point: $\Delta G = \Delta H - T\Delta S = 0$.
   2. Rearrange for temperature: $T = \frac{\Delta H}{\Delta S}$.
   3. Substitute values: $T = \frac{-200 \text{ kJ/mol}}{-0.5 \text{ kJ/mol}\cdot \text{K}}$.
   4. Calculate: $T = 400 \text{ K}$. The reaction is spontaneous below this temperature.

Practice Questions

Test your knowledge with these Easy MCAT Thermochemistry Practice Questions. Using active recall while answering these will help solidify the concepts in your long-term memory.

1. Which of the following describes a process where the system absorbs heat from the surroundings?

2. A 100 g sample of an unknown metal absorbs 1000 J of heat, resulting in a temperature increase of $20^\circ \text{C}$. What is the specific heat capacity of the metal?

3. If a reaction is exothermic and results in an increase in entropy, under what temperature conditions is it spontaneous?

4. Define a "state function" and identify whether heat ($q$) and work ($w$) are state functions.

5. Calculate the standard enthalpy of reaction ($\Delta H^\circ_{rxn}$) for the combustion of methane given: $\Delta H^\circ_f \text{ [CH}_4 \text{]} = -75 \text{ kJ/mol}$, $\Delta H^\circ_f \text{ [CO}_2 \text{]} = -394 \text{ kJ/mol}$, and $\Delta H^\circ_f \text{ [H}_2 \text{O(l)]} = -286 \text{ kJ/mol}$.

6. What is the value of $\Delta G$ at the melting point of ice at 1 atm?

7. A gas expands against a constant external pressure of 2 atm from a volume of 5 L to 10 L. Calculate the work done by the gas in L·atm.

8. According to the Second Law of Thermodynamics, what must be true about the total entropy of the universe for any spontaneous process?

9. If a reaction has a positive $\Delta H$ and a positive $\Delta S$, will it be spontaneous at high or low temperatures?

10. Which phase change is associated with the largest positive change in entropy: sublimation, deposition, or condensation?

Answers & Explanations

1. Endothermic. By definition, endothermic processes absorb heat ($\Delta H > 0$), while exothermic processes release heat ($\Delta H < 0$).
2. $0.5 \text{ J/g}\cdot^\circ \text{C}$. Use $q = mc\Delta T$. Rearranging gives $c = \frac{q}{m\Delta T}$. Plugging in the numbers: $c = \frac{1000 \text{ J}}{(100 \text{ g})(20^\circ \text{C})} = \frac{1000}{2000} = 0.5$.
3. Spontaneous at all temperatures. Using $\Delta G = \Delta H - T\Delta S$, if $\Delta H$ is negative (exothermic) and $\Delta S$ is positive (increased disorder), $\Delta G$ will always be negative regardless of the value of $T$.
4. Heat and work are NOT state functions. A state function depends only on the current state of the system, not the path taken (e.g., Enthalpy, Entropy, Internal Energy). Heat and work are path functions because their values vary depending on the specific process used to change states.
5. $-891 \text{ kJ/mol}$. Use the formula $\Delta H^\circ_{rxn} = \sum \Delta H^\circ_{f, \text{products}} - \sum \Delta H^\circ_{f, \text{reactants}}$. For $\text{CH}_4 + 2 \text{O}_2 \rightarrow \text{CO}_2 + 2 \text{H}_2 \text{O}$: $[(-394) + 2(-286)] - [-75 + 2(0)] = [-394 - 572] + 75 = -966 + 75 = -891 \text{ kJ/mol}$.
6. Zero. At a phase change (equilibrium), the change in Gibbs Free Energy ($\Delta G$) is exactly zero because the two phases coexist in equilibrium at constant temperature and pressure.
7. -10 L·atm. Work done by a gas is $w = -P\Delta V$. Here, $P = 2 \text{ atm}$ and $\Delta V = 10 - 5 = 5 \text{ L}$. Thus, $w = -(2)(5) = -10 \text{ L·atm}$. The negative sign indicates work is done by the system.
8. It must increase. The Second Law of Thermodynamics states that the entropy of the universe is constantly increasing for any spontaneous process ($\Delta S_{univ} > 0$).
9. High temperatures. In the equation $\Delta G = \Delta H - T\Delta S$, if both terms are positive, the $-T\Delta S$ term must be large enough to outweigh the positive $\Delta H$. This occurs when $T$ is large.
10. Sublimation. Entropy measures disorder. Sublimation (solid to gas) represents the largest increase in molecular randomness compared to deposition (gas to solid) or condensation (gas to liquid), which decrease entropy.

Quick Quiz

Frequently Asked Questions

What is the difference between heat and temperature?

Temperature is a measure of the average kinetic energy of the particles in a substance, whereas heat is the total energy transferred between systems due to a temperature gradient. Temperature is an intensive property, while heat transfer is a process that changes the internal energy of a system.

How do you determine if a reaction is spontaneous?

A reaction is spontaneous if the change in Gibbs Free Energy ($\Delta G$) is negative. This is calculated using the formula $\Delta G = \Delta H - T\Delta S$, where enthalpy, entropy, and absolute temperature are the determining factors.

What does a positive ΔH indicate?

A positive $\Delta H$ indicates an endothermic reaction, meaning the system has absorbed energy from its surroundings. These reactions usually feel cold to the touch because they draw heat away from the environment.

What is Hess's Law?

Hess's Law states that the total enthalpy change for a chemical reaction is the same regardless of whether the reaction occurs in one step or several steps. This allows scientists to calculate enthalpy changes for complex reactions using known values from simpler intermediate reactions.

Why is the specific heat of water important on the MCAT?

The high specific heat of water ($4.18 \text{ J/g}\cdot^\circ \text{C}$) is a frequent topic because it explains why water resists temperature changes, which is vital for thermoregulation in biological systems. It is also the standard reference for calorimetry problems.

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