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Chemistry Chapter 14
Chemical Equilibrium
| Term | Definition |
|---|---|
| Raising Temp in Gas Phase Reactions | Increased temperature or pressure for a gas-phase reaction increases the number of collisions. (14.1) |
| Reversible Reactions | Reversible reactions proceed in both the forward and the reverse directions at the same time. Represented by double arrow. (14.1) |
| Equilibrium | Forward Rate = Reverse Rate. No net change in product/reactant concentration regardless of time. Eventually, the increasing rate of the reverse reaction matches the slowing rate of the forward reaction. Reactants are produced and consumed at the same rate. (14.1) |
| Chemical Equilibria are... | Dynamic! Reaction is continuously occurring, but its speed is equal and opposite both forward and backward. (14.1) |
| Law of Mass Action | Any reversible reaction eventually reaches a state in which the ratio of the concentrations of products to reactants, with each value raised to a power corresponding to the coefficient for that substance in the balanced chemical equation for the reaction, has a characteristic value at a given temperature. (14.2) |
| Equilibrium Constant Expression | Ratio of mass action, described by the temperature-dependent constant K. This constant is always positive, as a ratio of positive values. (14.2) |
| K Value Relationship with Product/Reactant Concentration | K values much greater than 1 are associated with reaction mixtures that consist of mostly products at equilibrium. K values much less than 1 are associated with reaction mixtures that contain few products at equilibrium. (14.2) |
| Coefficients & Equilibrium Expressions | A substance’s coefficient always becomes its exponent in the equilibrium constant expression for a reaction. (14.2) |
| Kc | Represents that the (products/reactants) ratio is being expressed in concentrations. A unitless constant. (14.2) |
| Kp | When all components are gasses: represents that the (products/reactants) ratio is being expressed in partial pressures. A unitless constant. (14.2) |
| Kc = ([C]^c[D]^d) ÷ ([A]^a[B]^b) | Equilibrium constant expressed via concentrations. (14.2) |
| Kp = ([P_C]^c[P_D]^d) ÷ ([P_A]^a[P_B]^b) | Equilibrium constant expressed via partial pressures. (14.2) |
| Does Kc = Kp? | Depends on whether the number of moles of gaseous reactants is the same as the number of moles of gaseous products. (14.2) |
| Assumption of Ideal Behavior | When directly using concentrations and partial pressures in equilibrium equations, we assume they exhibit normal, ideal behavior. (14.2) |
| ∑ⁿ₍ᵢ₌₁₎Pᵢ | The total pressure exerted by a mixture of gases is the sum of the partial pressures of component gases. Multiplying this sum by (RT/V) will yield the pressure of the entire mixture. (Review) |
| PV=nRT | Ideal gas law relating pressure, temperature, moles, volume, and the ideal gas constant (8.314 L ∙ Mol⁻ ∙ K⁻). (Review) |
| Range of K-values | 0<K<∞ often meaning even "large" values are considered relatively close to one, an equivalent amount of reactants and products when the reaction reaches equilibrium. (14.2) |
| Equilibria that "lie far to the right" | Many reactions reach equilibrium only after essentially all of the reactants have formed products. Equilibria is very close to the product side. Ex) H₂ combustion at ~ room temperature. (14.2) |
| Equilibria that "lie far to the left" | When little product is formed before equilibrium is reached. Equilibria is very close to the reactant side. Ex) decomposition of CO₂ to CO at 25°C essentially does not occur. We can infer this because of how stable CO₂ is, and how small the K-value is. (14.2) |
| Kp=Kc if... | Δn = 0 since any term (RT) raised to zero will equal one. (14.3) |
| Δn | Sum of Product Moles - Sum of Reactant Moles (14.3) |
| Kp=Kc(RT)ᵟⁿ | Relationship between Kc and Kp derived from ideal gas law. (14.3) |
| K for Reverse Reactions | K(forward) = 1 ÷ K(reverse). In other word, the forward reaction's K value is the reciprocal of the reverse reaction's K value. Reactants become the top while products move to the bottom of the ratio! (14.4) |
| Dividing Coefficients by 2 | Dividing all the coefficients in the original reaction equation by 2 produces a Kc expression (and a K value) that is the square root of the original. (14.4) |
| General Division/Multiplication Law | If the balanced chemical equation of a reaction is multiplied by some factor n, then the value of K for the new reaction is the original K value raised to the nth power. Similarly, dividing all coefficients by n raises the value of K to the 1/n power. (14.4) |
| K(series of ᶰ rxts) = K₁ x K₂ x K₃... x Kᶰ | The overall equilibrium constant for a sum of two or more reactions is the product of the equilibrium constants of the individual reactions. (14.4) |
| Manipulation Similar to Hess's Law | We may need to reverse or divide/multiply reactions to create the correct product equation. To do so, each K value must be raised to the correct n value or its reciprocal taken, BEFORE multiplying all respective K values to create K(overall) for series. (14.4) |
| Mass Action Expression | Applies to concentrations/partial pressures of products and reactants that may or may NOT be at equilibrium. Instead, they may be undergoing the process of reaching it. (14.5) |
| Reaction Quotient, Qc or Qp | Found using similar computational methods to solving for K, but concentration/partial pressures are from a time during which rxt isn't in equilibrium. Forward and reverse reactions are not yet equal, although they are approaching that point. (14.5) |
| Example: Qc=6 and Kc=21 | To achieve equilibrium, some of the reactants must form products. Mathematically, that will increase the numerator and decrease the denominator in Qc, thereby increasing the value of Qc until it matches the value of Kc and equilibrium has been achieved. (14.5) |
| Q > K | Recall that these values are ratios of product : reactant. If ratio is higher than at equilibrium, then too much product is present. Reverse reaction will dominate until Q falls to K. (14.5) |
| Q < K | Recall that these values are ratios of product : reactant. If ratio is lower than at equilibrium, then too much reactant is present. Forward reaction will dominate until Q rises to K. (14.5) |
| Concentration-Based Equilibrium Expression and Solid Species | The concentrations of solids do not change: as long as these solids are present in the reaction mixture, their concentrations are defined by their densities, which do not change during the reaction. Therefore, solids and pure liquids are removed from the equation altogether. For this reason, H₂O is often removed from K/Q expressions even when it participates in reactions. (14.6) |
| Le Châtelier's Principle | If a system at equilibrium is perturbed (or subjected to an external stress), the position of the equilibrium shifts in either the forward or reverse direction as a response to reduce that stress (14.7) |
| Removing a Product | Decreases the number of particles and thus the number of collisions between product molecules. This decreased number of collisions decreases the speed of the reverse reaction. Reaction "shifts to the right" when Q falls below K. (14.7) |
| Adding Reactants | Adding more reactants increases their concentration in the reaction mixture, which increases the rate of the forward reaction. Some of the added reactants are converted into additional products. This approach works even when only one of multiple reactants is added, as long as there are sufficient quantities of all the other reactants. (14.7) |
| Boyle's Law | P₁V₁ = P₂V₂ (Review) |
| Halving Volume in NO₂ ⇌ N₂O₄ System | Doubles Partial Pressure. Qp becomes ½Kp. Forward reaction dominates. (14.7) |
| Compressing a Reaction | Whenever a reaction mixture at equilibrium is compressed—and the partial pressures of its gas-phase reactants and products increase—the equilibrium shifts toward the side of the reaction equation with fewer moles of gases. (14.7) |
| Increasing Volume of Reaction System | At constant temperature results in fewer collisions, lowers the partial pressures of all the gaseous reactants and products, and shifts the equilibrium toward the side of the reaction equation with more moles of gases. (14.7) |
| P🠕 or V🠗 | Equilibrium moves toward side with less moles. (14.7) |
| P🠗 or V🠕 | Equilibrium moves toward side with more moles. (14.7) |
| Energy as a Product | If we think of energy as a product of an exothermic reaction, then lowering the temperature of the reaction mixture favors the forward reaction (because a product is effectively removed). (14.7) |
| T🠕 Exothermic | A product is added. Q > K (14.7) |
| T🠗 Exothermic | A product is removed. Q < K (14.7) |
| T🠕 Endothermic | A reactant is added. Q < K (14.7) |
| T🠗 Endothermic | A reactant is removed. Q > K (14.7) |
| Changes in Temperature | In general, the value of K decreases as temperature increases for exothermic reactions and increases as temperature increases for endothermic reactions. (14.7) |
| Haber-Bosch Process | Industrial production of ammonia was developed by two scientists in the early 20th century. This process is commercially feasible because of the use of iron-based catalysts. (14.7) |
| Catalyst Effect on Equilibrium | No effect on the equilibrium constant or on the composition of an equilibrium reaction mixture. However, it reduces the amount of energy and time required to reach equilibrium. (14.7) |
| Assuming X is negligibly small | In equilibrium calculations, we can ignore the −x or +x component of an equilibrium concentration or partial pressure term if the value of x is less than 5% of the initial value. (14.8) |
| The 5% Check | The ratio of x to the initial partial pressure of a reactant decreasing by x must be less than 5% to ignore x in the denominator. (14.8) |
| RICE | Reaction, Initial, Change, Equilibrium (14.8) |
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bluonk
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