Summary
Highlights
Equilibrium occurs in a reversible reaction when the rate of the forward reaction equals the rate of the reverse reaction. At this point, the concentrations of reactants and products remain constant, even though the reaction is still ongoing, making it a dynamic equilibrium.
An analogy using two cities and cars traveling between them demonstrates dynamic equilibrium. If 10 cars travel from city A to city B, and 10 cars travel from B to A every hour, the number of cars in each city remains constant, even though movement is occurring.
A concentration profile graph shows that concentrations of reactants decrease and products increase until they become constant at equilibrium. A rate profile graph illustrates that the rate of the forward reaction decreases and the rate of the reverse reaction increases until they become equal at equilibrium.
By setting the forward and reverse reaction rates equal at equilibrium, the equilibrium constant (K) can be derived as the ratio of the forward rate constant (k1) to the reverse rate constant (k-1). K is also equal to the ratio of product concentrations divided by reactant concentrations, with coefficients becoming exponents.
The video explains how to write equilibrium expressions for Kc (concentration-based) and Kp (partial pressure-based) using the law of mass action. Products are in the numerator and reactants in the denominator, with their stoichiometric coefficients as exponents.
A practice problem demonstrates calculating the value of Kc given the equilibrium concentrations of reactants and products for a balanced chemical equation. It emphasizes correctly setting up the Kc expression and plugging in the given values.
Another practice problem shows how to calculate Kp using equilibrium partial pressures. The process is similar to Kc, but involves partial pressures instead of molar concentrations.
The video derives the relationship between Kp and Kc: Kp = Kc(RT)^Δn. Δn is the difference between the sum of the coefficients of gaseous products and the sum of the coefficients of gaseous reactants. The appropriate R value for partial pressures (0.08206 L·atm/mol·K) and temperature in Kelvin must be used.
Two examples illustrate how to use the Kp = Kc(RT)^Δn formula to convert between Kp and Kc, given one value and the temperature.
The video explains how the equilibrium constant K changes when a reaction is modified: if coefficients are multiplied by a factor (n), K is raised to the nth power (K^n); if the reaction is reversed, K becomes 1/K.
An example demonstrates using an ICE (Initial, Change, Equilibrium) table to find equilibrium concentrations when starting concentrations and one equilibrium concentration are known. This allows for the calculation of Kc.
This section presents a reverse problem: given Kc and some equilibrium concentrations, calculate an unknown equilibrium concentration. This involves setting up the Kc expression and solving for the unknown variable.
A final example shows the application of an ICE table to partial pressures. Given an initial partial pressure and one equilibrium partial pressure, the table helps determine other equilibrium partial pressures and subsequently calculate Kp.