Medium MCAT Thermodynamics Practice Questions

Medium MCAT Thermodynamics Practice Questions

Mastering thermodynamics is essential for success on the MCAT, as it governs every biochemical reaction in the human body and every phase change in a beaker. This guide provides comprehensive Medium MCAT Thermodynamics Practice Questions to help you bridge the gap between basic definitions and complex application. By understanding how energy, entropy, and enthalpy interact, you will be better prepared for the Chemical and Physical Foundations of Biological Systems section of the exam.

Concept Explanation

MCAT thermodynamics is the study of energy, heat, and work transformations within physical and chemical systems to determine the spontaneity and equilibrium of reactions. At the heart of this subject are the three laws of thermodynamics. The First Law (Conservation of Energy) states that energy cannot be created or destroyed, expressed by the equation $\Delta U = Q - W$, where $\Delta U$ is the change in internal energy, $Q$ is heat added to the system, and $W$ is work done by the system. The Second Law introduces entropy ($S$), asserting that the total entropy of an isolated system always increases over time, favoring disorder. The Third Law establishes that the entropy of a perfect crystal at absolute zero (0 K) is zero.

For the MCAT, you must be proficient in calculating Gibbs Free Energy ($\Delta G$), which combines enthalpy ($\Delta H$) and entropy to predict reaction spontaneity. The relationship is defined as:

A negative $\Delta G$ indicates a spontaneous (exergonic) reaction, while a positive value indicates a non-spontaneous (endergonic) reaction. Understanding these concepts is often a prerequisite for more advanced topics like Medium MCAT Electrochemistry Practice Questions, where cell potential is directly linked to free energy. Additionally, you should be familiar with calorimetry and Hess’s Law, which allow for the calculation of total enthalpy changes by summing the enthalpies of individual reaction steps. These principles are closely related to Medium MCAT Thermochemistry Practice Questions, which focus specifically on heat transfer during chemical changes.

Solved Examples

Review these worked examples to understand the logic required for medium-difficulty thermodynamics problems.

1. Example: Calculating Gibbs Free Energy
   A specific protein folding reaction has an enthalpy change of $-50 \ \text{ kJ/mol}$ and an entropy change of $-0.15 \ \text{ kJ/mol} \cdot \ \text{K}$. Is this reaction spontaneous at $300 \ \text{ K}$?
   1. Identify the given values: $\Delta H = -50 \ \text{ kJ/mol}$, $\Delta S = -0.15 \ \text{ kJ/mol} \cdot \ \text{K}$, and $T = 300 \ \text{ K}$.
   2. Apply the formula: $\Delta G = \Delta H - T\Delta S$.
   3. Substitute the values: $\Delta G = -50 - (300 \ \times -0.15)$.
   4. Calculate: $\Delta G = -50 - (-45) = -5 \ \text{ kJ/mol}$.
   5. Conclusion: Since $\Delta G$ is negative, the reaction is spontaneous at this temperature.
2. Example: Work Done by a Gas
   A gas in a piston expands from $2 \ \text{ L}$ to $5 \ \text{ L}$ against a constant external pressure of $3 \ \text{ atm}$. Calculate the work done by the gas in Joules (Note: $1 \ \text{ L} \cdot \ \text{atm} \approx 101.3 \ \text{ J}$).
   1. Use the formula for pressure-volume work: $W = P\Delta V$.
   2. Find the change in volume: $\Delta V = 5 \ \text{ L} - 2 \ \text{ L} = 3 \ \text{ L}$.
   3. Calculate work in $\ \text{L} \cdot \ \text{atm}$: $W = 3 \ \text{ atm} \ \times 3 \ \text{ L} = 9 \ \text{ L} \cdot \ \text{atm}$.
   4. Convert to Joules: $9 \ \times 101.3 \approx 911.7 \ \text{ J}$.
3. Example: Hess's Law Application
   Find the enthalpy of reaction for $A \ \rightarrow C$ given:
   1) $A \ \rightarrow B \quad \Delta H = +20 \ \text{ kJ}$
   2) $C \ \rightarrow B \quad \Delta H = +5 \ \text{ kJ}$
   1. We need to arrange the reactions to get $A \ \rightarrow C$.
   2. Keep reaction (1) as is: $A \ \rightarrow B \quad (\Delta H = +20 \ \text{ kJ})$.
   3. Reverse reaction (2) to get $B \ \rightarrow C$. When reversing, change the sign of $\Delta H$: $B \ \rightarrow C \quad (\Delta H = -5 \ \text{ kJ})$.
   4. Sum the reactions: $A + B \ \rightarrow B + C$, which simplifies to $A \ \rightarrow C$.
   5. Sum the enthalpies: $20 \ \text{ kJ} + (-5 \ \text{ kJ}) = +15 \ \text{ kJ}$.

Practice Questions

Test your knowledge with these Medium MCAT Thermodynamics Practice Questions. Ensure you pay close attention to units and signs.

1. A reaction is found to have a positive $\Delta H$ and a positive $\Delta S$. Under what conditions will this reaction be spontaneous?
2. Calculate the change in internal energy ($\Delta U$) for a system that absorbs $500 \ \text{ J}$ of heat and performs $200 \ \text{ J}$ of work on its surroundings.
3. According to the Second Law of Thermodynamics, which of the following must be true for any spontaneous process in an isolated system?

4. A $50.0 \ \text{ g}$ sample of copper (specific heat $c = 0.385 \ \text{ J/g} \cdot ^{\circ}\ \text{C}$) is heated from $20^{\circ}\ \text{C}$ to $80^{\circ}\ \text{C}$. How much heat energy was absorbed?
5. Which thermodynamic state function is used to measure the heat content of a system at constant pressure?
6. For the reaction $N_2(g) + 3H_2(g) \ \rightarrow 2NH_3(g)$, predict the sign of $\Delta S$ and explain why.
7. If a reaction has a $K_{ \neq} > 1$ at a specific temperature, what can be said about the standard Gibbs free energy ($\Delta G^{\circ}$)?
8. Compare the entropy of a liquid to its gaseous state at the same temperature. Which is higher and why?
9. A system undergoes an adiabatic process. If the gas performs work on the surroundings, what happens to the temperature of the system?
10. Using Hess's Law, calculate the total enthalpy for a reaction that occurs in three steps with $\Delta H$ values of $-120 \ \text{ kJ}$, $+50 \ \text{ kJ}$, and $-30 \ \text{ kJ}$.

Answers & Explanations

1. Answer: At high temperatures.
   According to $\Delta G = \Delta H - T\Delta S$, if both $\Delta H$ and $\Delta S$ are positive, the term $-T\Delta S$ must be large enough in magnitude to outweigh the positive $\Delta H$. This occurs when $T$ is large.
2. Answer: $+300 \ \text{ J}$.
   Using the First Law: $\Delta U = Q - W$. Here, $Q = +500 \ \text{ J}$ (heat absorbed) and $W = +200 \ \text{ J}$ (work done by the system). Thus, $\Delta U = 500 - 200 = 300 \ \text{ J}$.
3. Answer: The entropy of the system must increase ($\Delta S > 0$).
   The Second Law states that for any spontaneous process, the total entropy of the universe increases. For an isolated system, this means the entropy of the system itself must increase.
4. Answer: $1,155 \ \text{ J}$.
   Use the calorimetry formula $q = mc\Delta T$. $q = (50.0 \ \text{ g}) \ \times (0.385 \ \text{ J/g} \cdot ^{\circ}\ \text{C}) \ \times (80 - 20 ^{\circ}\ \text{C}) = 50 \ \times 0.385 \ \times 60 = 1,155 \ \text{ J}$.
5. Answer: Enthalpy ($H$).
   Enthalpy is defined as $H = U + PV$. At constant pressure, the change in enthalpy ($\Delta H$) is equal to the heat exchanged ($q_p$).
6. Answer: $\Delta S < 0$ (Negative).
   The reaction starts with 4 moles of gas (1 $N_2$ + 3 $H_2$) and produces 2 moles of gas (2 $NH_3$). A decrease in the number of gas moles results in a decrease in disorder, hence negative entropy.
7. Answer: $\Delta G^{\circ} < 0$ (Negative).
   The relationship is $\Delta G^{\circ} = -RT \ln K_{ \neq}$. If $K_{ \neq} > 1$, then $\ln K_{ \neq}$ is positive, making $\Delta G^{\circ}$ negative. This is a common theme in Medium MCAT Equilibrium Practice Questions.
8. Answer: The gas has higher entropy.
   Gases have much more translational freedom and occupy a larger volume than liquids, leading to more microstates and higher disorder (entropy).
9. Answer: The temperature decreases.
   In an adiabatic process, $Q = 0$. Therefore, $\Delta U = -W$. If the system does work ($W > 0$), the internal energy ($\Delta U$) must decrease. Since internal energy is proportional to temperature, the temperature drops.
10. Answer: $-100 \ \text{ kJ}$.
    Simply sum the enthalpy changes of the steps: $(-120) + 50 + (-30) = -100 \ \text{ kJ}$.

Frequently Asked Questions

What is the difference between ΔG and ΔG°?

$\Delta G$ represents the free energy change under any specific set of conditions, while $\Delta G^{\circ}$ is the free energy change under standard conditions (1 M concentration, 1 atm pressure, and 298 K). They are related by the equation $\Delta G = \Delta G^{\circ} + RT \ln Q$.

Can a reaction with positive ΔH be spontaneous?

Yes, an endothermic reaction can be spontaneous if the entropy change ($\Delta S$) is positive and the temperature is high enough such that the $T\Delta S$ term exceeds the $\Delta H$ term. This makes the overall Gibbs free energy negative.

What does a negative work value mean in thermodynamics?

In the convention $\Delta U = Q - W$, a negative work value ($W < 0$) indicates that work is being done on the system by the surroundings, such as during the compression of a gas. This increases the internal energy of the system.

Why is entropy often called the "arrow of time"?

Entropy is called the arrow of time because the Second Law of Thermodynamics dictates that the total entropy of the universe must increase over time, providing a clear direction for the progression of natural events. This distinguishes the future from the past in physical processes.

How do catalysts affect thermodynamic variables like ΔG?

Catalysts do not affect thermodynamic variables such as $\Delta G$, $\Delta H$, or $\Delta S$ because these are state functions that depend only on the initial and final states. Catalysts only lower the activation energy to increase the reaction rate (kinetics).