# Chemical Equilibrium Chemistry I · Energy, Equilibrium and Electrochemistry · https://tryals.app/learn/chemistry-i/chemical-equilibrium ## A Balance of Rates, Not of Amounts A reversible reaction reaches **equilibrium** when forward and reverse rates become equal. Concentrations then stop changing, but both reactions continue, equilibrium is dynamic, not static. For $aA + bB \rightleftharpoons cC + dD$ the **equilibrium constant** is $$K = \frac{[C]^c[D]^d}{[A]^a[B]^b}$$ Pure solids and pure liquids are omitted, because their concentrations cannot change. $K$ depends on temperature and on nothing else, not on starting amounts, not on pressure, not on the presence of a catalyst. The **reaction quotient** $Q$ has the identical form but is evaluated at any moment, and comparing the two predicts the direction of change: | Comparison | What happens | |---|---| | $Q < K$ | Net forward reaction | | $Q = K$ | At equilibrium | | $Q > K$ | Net reverse reaction | **Le Chatelier's principle** says a system at equilibrium responds to a disturbance in the direction that partially offsets it. Adding a reactant drives the reaction forward; removing a product does the same. Raising the pressure on a gaseous equilibrium shifts it toward the side with fewer gas molecules. Raising the temperature shifts an endothermic reaction forward, and temperature is the only disturbance that actually *changes* $K$; everything else merely moves the system to a different point on the same constant. The temperature dependence is the **Van 't Hoff equation**: $$\ln \frac{K_2}{K_1} = -\frac{\Delta H^\circ}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right)$$ Quantitative problems use an **ICE table** — Initial, Change, Equilibrium — where the changes are in the ratio of the stoichiometric coefficients. > **Common pitfall:** thinking a catalyst increases yield. A catalyst speeds the forward and reverse reactions by identical factors, so $Q$ still meets the same $K$. It changes only how soon equilibrium arrives, never where it sits, which is why industrial catalysts are paired with pressure and temperature choices that do move the equilibrium. ## Practice questions 8 of this lesson's 12 practice questions, with answers. The full set is in the app. ### 1. For a gaseous equilibrium at fixed temperature, which statement best explains why an added catalyst leaves the yield untouched? A. The reaction enthalpy shifts to offset rate changes B. It selectively lowers forward activation energy C. Only the time taken to reach equilibrium changes D. The equilibrium constant K increases with pressure **Answer:** C. Only the time taken to reach equilibrium changes **Why:** A catalyst multiplies the forward and reverse rate constants equally, so $K$, their ratio, cannot move. Equilibrium simply arrives sooner, at exactly the same composition. Page: https://tryals.app/practice/chemistry-i/chemical-equilibrium/for-a-gaseous-equilibrium-at-fixed-temperature-which-statement-best ### 2. The equilibrium constant of a reaction changes when the temperature changes. **Answer:** True **Why:** True, temperature is the sole variable that changes $K$ itself. Concentration and pressure changes move the system to a new position consistent with the *same* $K$, which is a distinction worth being strict about. Page: https://tryals.app/practice/chemistry-i/chemical-equilibrium/the-equilibrium-constant-of-a-reaction-changes-when-the-temperature ### 3. Chemical equilibrium is defined by an equality of rates, not an equality of concentrations. What does this distinction imply for a reaction vessel whose concentrations have ceased changing? A. The forward and reverse reaction processes have halted B. Both reactions proceed constantly at identical rates C. The reactants and products exist in equivalent amounts D. The reaction quotient has dropped to its minimum value **Answer:** B. Both reactions proceed constantly at identical rates **Why:** Equal rates keep concentrations steady without equalising them or stopping molecular exchange. Macroscopic constancy masks microscopic dynamism, meaning chemical bonds are still breaking and forming continually rather than freezing at an absolute standstill. Page: https://tryals.app/practice/chemistry-i/chemical-equilibrium/chemical-equilibrium-is-defined-by-an-equality-of-rates-not-an ### 4. At equilibrium a reaction $\mathrm{A \rightleftharpoons B}$ has $[A] = 0.20$ M and $[B] = 0.80$ M. Set the equilibrium constant. **Answer:** 4 (within ±0.4) **Why:** $K = [B]/[A] = 0.80/0.20 = 4.0$. Any value above 1 means products dominate at equilibrium, though it says nothing whatever about how quickly that composition is reached. Page: https://tryals.app/practice/chemistry-i/chemical-equilibrium/at-equilibrium-a-reaction-a-leftharpoons-b-has-a-0-20-m-and-b ### 5. Which changes shift a gaseous equilibrium toward the side with fewer gas molecules? A. Adding a catalyst B. Adding an inert gas at constant volume C. Increasing the total pressure by reducing the volume D. Decreasing the volume of the container **Answer:** C. Increasing the total pressure by reducing the volume; D. Decreasing the volume of the container **Why:** Compressing the container raises every partial pressure and the system relieves it by favouring fewer molecules. An inert gas added at constant volume changes the total pressure but no partial pressures, and a catalyst never moves the position at all. Page: https://tryals.app/practice/chemistry-i/chemical-equilibrium/which-changes-shift-a-gaseous-equilibrium-toward-the-side-with-fewer ### 6. Sort each species by whether it appears in the equilibrium constant expression. **Answer:** - Appears in the expression: A gas taking part in the reaction, A dissolved ion taking part in the reaction - Omitted from the expression: A pure solid reactant, A pure liquid solvent such as water **Why:** Only species whose concentrations can change appear. A pure solid or liquid has a concentration fixed by its density, so it is folded into the constant rather than written out. Page: https://tryals.app/practice/chemistry-i/chemical-equilibrium/sort-each-species-by-whether-it-appears-in-the-equilibrium-constant ### 7. Arrange the steps of an ICE-table calculation in the order you would perform them. **Answer:** 1. Write the balanced equation and the equilibrium expression 2. Enter the initial concentrations 3. Express the changes in terms of x, using the stoichiometric ratios 4. Write the equilibrium concentrations and substitute into K 5. Solve for x and back-substitute to get each concentration **Why:** The balanced equation fixes both the expression and the coefficient ratios used for the changes. Only once the equilibrium row is substituted into $K$ does a solvable equation in $x$ appear. Page: https://tryals.app/practice/chemistry-i/chemical-equilibrium/arrange-the-steps-of-an-ice-table-calculation-in-the-order-you-would ### 8. Match each comparison or disturbance to its consequence. **Answer:** - $Q$ less than $K$ → Net forward reaction until they are equal - $Q$ greater than $K$ → Net reverse reaction until they are equal - Raising the temperature → Changes the value of $K$ itself - Adding more reactant → Moves the position without changing $K$ **Why:** The system always moves so that $Q$ approaches $K$. Concentration changes relocate the system on the same constant, while temperature is unique in changing the constant itself. Page: https://tryals.app/practice/chemistry-i/chemical-equilibrium/match-each-comparison-or-disturbance-to-its-consequence