#Equilibrium: Kinetic vs. Thermodynamic Views ⚖️

Hey there, future AP Chem master! Let's break down equilibrium, connecting kinetics and thermodynamics. This is a big topic, so let's make sure you've got it down cold for tomorrow!

#Kinetic and Thermodynamic Definitions of Equilibrium

# Kinetic Definition of Equilibrium 🏃

• Equilibrium is when the forward and reverse reaction rates are equal.
• This means that the concentrations of reactants and products remain constant over time.

# Thermodynamic Definition of Equilibrium ⚡

• Equilibrium is the point of minimum free energy (G).

• A reaction proceeds spontaneously (ΔG < 0) until it reaches this equilibrium point.

• After equilibrium, the reaction requires energy input (ΔG > 0) to continue forming products.

#Visualizing Free Energy 📉

Let's look at those graphs again. They're super important for understanding how free energy changes during a reaction:

#Spontaneous Reaction (ΔG° < 0)

• ΔG is the rate of change of the graph, not the y-position.
• On the left, we have 100% reactants, and on the right, 100% products.
• The difference in free energy between reactants and products is the standard free energy change (ΔG°).
• For a spontaneous reaction, ΔG° < 0, and the graph slopes down until it hits the minimum point (equilibrium).

#Nonspontaneous Reaction (ΔG° > 0)

• Here, ΔG° > 0, meaning the reaction is nonspontaneous.
• The equilibrium point is shifted towards the reactants.

#Relationship Between ΔG°, ΔG, and K

#Connecting ΔG° and ΔG 🔗

• ΔG is the free energy change under nonstandard conditions, while ΔG° is at standard conditions.

• The relationship is given by:

  $$ΔG = ΔG° + RTln(Q)$$

  Where:

  • R is the gas constant (8.314 J/mol·K)

  • T is the temperature in Kelvin

  • Q is the reaction quotient (non-equilibrium concentrations or pressures)

![](https://zupay.blob.core.windows.net/resources/files/0baca4f69800419293b4c75aa2870acd_e0d2ce_1838.png?alt=media&token=07bf70a7-38bd-492a-a87b-81cd4eae90d5)

#Deriving the Relationship with K 💡

• At equilibrium, ΔG = 0 and Q = K.

• Substituting into the above equation, we get:

  $$0 = ΔG° + RTln(K)$$

• Rearranging, we have:

  $$ΔG° = -RTln(K)$$

• Solving for K:

  $$K = e^{-ΔG°/RT}$$

#Qualitative Insights 🧠

• A larger ΔG° means a smaller K (and vice versa).
• If ΔG° > 0, then K < 1 (nonspontaneous, reactant-favored).
• If ΔG° < 0, then K > 1 (spontaneous, product-favored).

#Final Exam Focus 🎯

#Key Topics:

• Understanding both kinetic and thermodynamic definitions of equilibrium.
• Interpreting free energy diagrams and their relationship to spontaneity.
• Using the equations relating ΔG°, ΔG, Q, and K.
• Relating the sign of ΔG° to the magnitude of K.

#Common Question Types:

• MCQs asking for the relationship between ΔG and spontaneity.
• FRQs that require you to calculate ΔG, K, or Q.
• Questions that mix concepts from thermodynamics and equilibrium.

#Last-Minute Tips:

• Time Management: Quickly identify the key concepts in each question.
• Common Pitfalls: Double-check your units and signs in calculations.
• Strategies: Draw diagrams to visualize free energy changes.

#

Practice Question

Practice Questions

#Multiple Choice Questions

1. For a reaction at equilibrium, which of the following statements is always true? (A) The concentrations of reactants and products are equal. (B) The rates of the forward and reverse reactions are equal. (C) The change in free energy, ΔG, is positive. (D) The equilibrium constant, K, is equal to 1. 2. A reaction has a positive ΔG°. Which of the following is true about the equilibrium constant, K, and the spontaneity of the reaction? (A) K > 1, spontaneous (B) K > 1, nonspontaneous (C) K < 1, spontaneous (D) K < 1, nonspontaneous

2. Which of the following best describes the relationship between ΔG and ΔG°? (A) ΔG = ΔG° + RT (B) ΔG = ΔG° - RT (C) ΔG = ΔG° + RTln(Q) (D) ΔG = ΔG° - RTln(Q)

#Free Response Question

Consider the following reaction:

$$N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$$

At 298 K, the standard free energy change, ΔG°, for this reaction is -33.0 kJ/mol.

(a) Calculate the equilibrium constant, K, for this reaction at 298 K. (b) If the partial pressures of N2 and H2 are each 1.0 atm and the partial pressure of NH3 is 0.5 atm, calculate ΔG for the reaction at 298 K. (c) Is the reaction spontaneous under these conditions? Explain your answer.

#Scoring Breakdown:

(a) 3 points - 1 point for using the correct formula:

    <math-block>ΔG° = -RTln(K)</math-block>
- 1 point for correct substitution:

    <math-block>-33000 \frac{J}{mol} = -(8.314 \frac{J}{mol \cdot K})(298 K)ln(K)</math-block>
- 1 point for correct answer: K = 7.1 x 10^5

(b) 3 points - 1 point for calculating Q:

    <math-block>Q = \frac{(P\_{NH\_3})^2}{(P\_{N\_2})(P\_{H\_2})^3} = \frac{(0.5)^2}{(1.0)(1.0)^3} = 0.25</math-block>
- 1 point for using the correct formula:

    <math-block>ΔG = ΔG° + RTln(Q)</math-block>
- 1 point for correct answer: ΔG = -36.5 kJ/mol

(c) 1 point - 1 point for correct answer and explanation: Yes, the reaction is spontaneous because ΔG < 0.