Medium MCAT Thermochemistry Practice Questions

Medium MCAT Thermochemistry Practice Questions

Mastering Medium MCAT Thermochemistry Practice Questions is essential for any pre-medical student aiming to excel in the Chemical and Physical Foundations of Biological Systems section. Thermochemistry focuses on the energy changes that accompany chemical reactions and physical transformations, bridging the gap between physics and biology. By understanding how heat transfer, enthalpy, and entropy dictate the spontaneity of biological processes, you can better predict molecular behavior in physiological environments.

Concept Explanation

Thermochemistry is the study of energy and heat associated with chemical reactions and physical transformations, primarily governed by the laws of thermodynamics. In the context of the MCAT, this involves understanding system-surrounding interactions, where the "system" is the specific chemical reaction being studied and the "surroundings" are everything else. Key variables include Enthalpy ($\Delta H$), Entropy ($\Delta S$), and Gibbs Free Energy ($\Delta G$).

According to the First Law of Thermodynamics, energy cannot be created or destroyed, only transferred. This is represented by the equation $\Delta U = Q - W$, where $\Delta U$ is the change in internal energy, $Q$ is heat added, and $W$ is work done by the system. For most MCAT problems, we focus on constant pressure conditions where heat flow equals the change in enthalpy ($\Delta H$).

To succeed on these problems, you must be comfortable with:

• Hess’s Law: The total enthalpy change of a reaction is the sum of the enthalpy changes of its individual steps.
• Calorimetry: Using the equation $q = mc\Delta T$ to calculate heat transfer.
• Gibbs Free Energy: Using $\Delta G = \Delta H - T\Delta S$ to determine reaction spontaneity.
• Phase Changes: Understanding that temperature remains constant during a phase change while potential energy changes.

Effective preparation often involves retrieval practice for medical students to ensure these formulas and concepts are accessible under the time pressure of the actual exam.

Solved Examples

Example 1: Calculating Enthalpy of Reaction
Given the following bond enthalpies: C-H: 413 kJ/mol, Cl-Cl: 242 kJ/mol, C-Cl: 339 kJ/mol, H-Cl: 427 kJ/mol.
Calculate the $\Delta H$ for the reaction: $$\text{CH}_4(g) + \text{Cl}_2(g) \rightarrow \text{CH}_3 \text{Cl}(g) + \text{HCl}(g)$$

1. Identify bonds broken (reactants): 1 C-H bond and 1 Cl-Cl bond. Total energy absorbed = $413 + 242 = 655 \text{ kJ/mol}$.
2. Identify bonds formed (products): 1 C-Cl bond and 1 H-Cl bond. Total energy released = $339 + 427 = 766 \text{ kJ/mol}$.
3. Apply the formula $\Delta H = \text{Bonds Broken} - \text{Bonds Formed}$.
4. $\Delta H = 655 - 766 = -111 \text{ kJ/mol}$. The reaction is exothermic.

Example 2: Calorimetry and Specific Heat
A 50.0 g piece of an unknown metal at 100.0°C is dropped into 100.0 g of water at 25.0°C. The final temperature of the system is 28.0°C. Calculate the specific heat of the metal (Specific heat of water = 4.18 J/g°C).

1. Calculate heat gained by water: $q_{ \text{water}} = mc\Delta T = (100.0 \text{ g})(4.18 \text{ J/g}^\circ \text{C})(28.0 - 25.0) = 1254 \text{ J}$.
2. Set heat lost by metal equal to heat gained by water: $q_{ \text{metal}} = -1254 \text{ J}$.
3. Solve for $c_{ \text{metal}}$: $-1254 = (50.0 \text{ g})(c)(28.0 - 100.0)$.
4. $-1254 = (50.0)(c)(-72.0) \rightarrow -1254 = -3600c$.
5. $c = 0.348 \text{ J/g}^\circ \text{C}$.

Example 3: Gibbs Free Energy and Spontaneity
A reaction has $\Delta H = -120 \text{ kJ/mol}$ and $\Delta S = -400 \text{ J/mol}\cdot \text{K}$. At what temperature does the reaction change from spontaneous to non-spontaneous?

1. The transition occurs when $\Delta G = 0$. Use the equation $\Delta G = \Delta H - T\Delta S$.
2. Convert units so they match: $\Delta S = -0.400 \text{ kJ/mol}\cdot \text{K}$.
3. Set $0 = -120 - T(-0.400)$.
4. $120 = 0.400T$.
5. $T = \frac{120}{0.400} = 300 \text{ K}$. The reaction is spontaneous below 300 K.

Practice Questions

Test your knowledge with these Medium MCAT Thermochemistry Practice Questions. Be sure to pay close attention to units and signs.

1. A reaction is found to have a positive $\Delta H$ and a positive $\Delta S$. Under which of the following conditions will the reaction be spontaneous?
A) At all temperatures
B) At high temperatures
C) At low temperatures
D) It will never be spontaneous

2. How much energy is required to melt 36 g of ice at 0°C? (Heat of fusion for water = 6.01 kJ/mol).
A) 6.01 kJ
B) 12.02 kJ
C) 18.03 kJ
D) 216 kJ

3. Consider the following reaction: $$2 \text{Al}(s) + \frac{3}{2} \text{O}_2(g) \rightarrow \text{Al}_2 \text{O}_3(s) \quad \Delta H = -1675 \text{ kJ/mol}$$
What is the enthalpy change for the decomposition of 2 moles of $\text{Al}_2 \text{O}_3(s)$ into its elements?
A) -1675 kJ
B) +1675 kJ
C) +3350 kJ
D) -3350 kJ

4. If a gas expands from a volume of 2 L to 6 L against a constant external pressure of 3 atm, how much work is done by the system? (1 L·atm = 101.3 J).
A) -1215 J
B) -12.15 J
C) 12.15 J
D) 1215 J

5. Which of the following processes results in a decrease in entropy ($\Delta S < 0$)?
A) Sublimation of dry ice
B) Dissolving salt in water
C) Freezing of liquid water
D) Heating a gas at constant volume

6. Using Hess’s Law, find the enthalpy of the reaction $\text{A} \rightarrow \text{C}$ given:
$\text{A} \rightarrow \text{B} \quad \Delta H = +50 \text{ kJ}$
$\text{C} \rightarrow \text{B} \quad \Delta H = -20 \text{ kJ}$
A) +30 kJ
B) +70 kJ
C) -70 kJ
D) -30 kJ

7. A 200 g sample of water at 20°C is mixed with 100 g of water at 80°C. What is the final temperature of the mixture?
A) 40°C
B) 50°C
C) 60°C
D) 30°C

8. For a certain reaction, $\Delta G^\circ$ is negative. What can be said about the equilibrium constant $K_{ \neq}$?
A) $K_{ \neq} < 1$
B) $K_{ \neq} = 1$
C) $K_{ \neq} > 1$
D) $K_{ \neq} = 0$

9. A bomb calorimeter is used to measure the heat of combustion of a glucose sample. In this closed, rigid container, which of the following is true regarding work ($W$) and heat ($Q$)?
A) $W > 0$
B) $W = 0$
C) $\Delta V > 0$
D) $Q = 0$

10. The bond dissociation energy of $\text{H}_2$ is 436 kJ/mol and for $\text{Cl}_2$ is 243 kJ/mol. If the enthalpy of formation ($\Delta H_f^\circ$) of HCl is -93 kJ/mol, what is the bond energy of the H-Cl bond?
A) 246 kJ/mol
B) 432 kJ/mol
C) 186 kJ/mol
D) 865 kJ/mol

Answers & Explanations

1. B: According to $\Delta G = \Delta H - T\Delta S$, if both $\Delta H$ and $\Delta S$ are positive, the term $-T\Delta S$ becomes more negative as temperature increases. At high temperatures, $|T\Delta S| > |\Delta H|$, making $\Delta G$ negative (spontaneous).
2. B: First, find the moles of water: $36 \text{ g} / 18 \text{ g/mol} = 2 \text{ moles}$. Energy = $\text{moles} \times \Delta H_{fus} = 2 \times 6.01 = 12.02 \text{ kJ}$.
3. C: The original reaction is for the formation of 1 mole of $\text{Al}_2 \text{O}_3$. To find the decomposition of 2 moles, reverse the reaction (change sign of $\Delta H$) and multiply by 2. $+1675 \times 2 = +3350 \text{ kJ}$.
4. A: $W = -P\Delta V$. $W = -(3 \text{ atm})(6 \text{ L} - 2 \text{ L}) = -12 \text{ L}\cdot \text{atm}$. Convert to Joules: $-12 \times 101.3 = -1215.6 \text{ J}$. The negative sign indicates work done by the system.
5. C: Entropy is a measure of disorder. Freezing turns a liquid into a structured solid, which is a decrease in disorder ($\Delta S < 0$). Sublimation and dissolving increase disorder.
6. B: We need $\text{A} \rightarrow \text{C}$. We have $\text{A} \rightarrow \text{B}$ ($\Delta H = 50$) and $\text{B} \rightarrow \text{C}$ (reverse of the second reaction, so $\Delta H = +20$). Adding them: $50 + 20 = 70 \text{ kJ}$.
7. A: $m_1c(T_f - T_1) = -m_2c(T_f - T_2)$. Since $c$ is the same, $200(T_f - 20) = -100(T_f - 80)$. $2(T_f - 20) = -1(T_f - 80) \rightarrow 2T_f - 40 = -T_f + 80 \rightarrow 3T_f = 120 \rightarrow T_f = 40^\circ \text{C}$.
8. C: The relationship is $\Delta G^\circ = -RT \ln K_{ \neq}$. If $\Delta G^\circ$ is negative, $\ln K_{ \neq}$ must be positive, which means $K_{ \neq} > 1$. This indicates products are favored at equilibrium.
9. B: A bomb calorimeter is a constant-volume (isochoric) system. Since $W = P\Delta V$ and $\Delta V = 0$, no pressure-volume work is done ($W = 0$).
10. B: The formation reaction is $\frac{1}{2} \text{H}_2 + \frac{1}{2} \text{Cl}_2 \rightarrow \text{HCl}$. $\Delta H_f = [ \frac{1}{2} \text{BE}( \text{H-H}) + \frac{1}{2} \text{BE}( \text{Cl-Cl}) ] - [ \text{BE}( \text{H-Cl}) ]$.
    $-93 = [ \frac{1}{2}(436) + \frac{1}{2}(243) ] - X$
    $-93 = [ 218 + 121.5 ] - X$
    $-93 = 339.5 - X \rightarrow X = 432.5 \text{ kJ/mol}$.

Frequently Asked Questions

What is the difference between specific heat and heat capacity?

Specific heat is an intensive property representing the heat required to raise 1 gram of a substance by 1°C, while heat capacity is an extensive property representing the heat required for the entire object, regardless of mass. Essentially, heat capacity depends on how much of the substance you have, but specific heat is constant for the material.

Why is enthalpy considered a state function?

Enthalpy is a state function because its value depends only on the current state of the system, such as pressure and temperature, and not on the path taken to reach that state. This allows us to use Hess’s Law to calculate overall energy changes by summing individual steps.

How does the MCAT test the concept of entropy?

The MCAT frequently tests entropy through phase changes (solid to liquid to gas increases entropy) and the number of moles of gas in a reaction. You should also be familiar with how Gibbs Free Energy links entropy and enthalpy to determine if a biological reaction will proceed spontaneously.

What is the sign convention for work in MCAT physics vs. chemistry?

In chemistry and most MCAT thermochemistry contexts, work done by the system is negative ($W = -P\Delta V$) because energy is leaving the system. However, always check the passage context, as some physics conventions use $\Delta U = Q + W$ where work done on the system is positive.

Why is retrieval practice important for mastering thermochemistry?

Because thermochemistry involves multi-step calculations and conceptual integration, using retrieval practice as an evidence-based study method helps solidify the mental pathways needed to recall formulas like $q = mc\Delta T$ and $\Delta G = \Delta H - T\Delta S$ quickly during the exam.