Química Geral · Thermodynamics and kinetics
Energia, equilíbrio e velocidades
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O que você vai aprender
- Calculate reaction enthalpies with Hess's law, enthalpies of formation, bond enthalpies and calorimetry.
- Use the signs of ΔH, ΔS and ΔG, ΔG = ΔH − TΔS and ΔG° = −RT ln K to decide whether and how far a reaction goes.
- Write equilibrium expressions, compare Q with K and predict Le Chatelier shifts.
- Deduce rate laws, orders and rate constants from data and use first-order half-lives.
- Apply the Arrhenius equation and explain how a catalyst changes the rate but not the equilibrium.
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Ler a aula gratuita (6 partes)
Enthalpy and Hess's law
The enthalpy change ΔH is the heat a reaction exchanges with its surroundings at constant pressure. A negative ΔH means heat is released (exothermic); a positive ΔH means heat is absorbed (endothermic). Burning methane, CH₄ + 2 O₂ → CO₂ + 2 H₂O(l), has ΔH° ≈ −890 kJ/mol. Enthalpy is a state function: ΔH depends only on the starting and final states, so the ΔH values of steps that add up to a reaction add up to its ΔH (Hess's law). Reversing an equation changes the sign of ΔH, and multiplying it by a factor multiplies ΔH by the same factor. Three routes give ΔH°: from enthalpies of formation, ΔH° = Σ n ΔH°f(products) − Σ n ΔH°f(reactants), with ΔH°f = 0 for elements in their standard states; from average bond enthalpies, ΔH ≈ Σ(bonds broken) − Σ(bonds formed), which is only approximate; and from calorimetry, where q = m × c × ΔT for the solution and ΔH = −q(solution) ÷ moles reacted.
Entropy, free energy and spontaneity
Entropy S grows with the number of ways energy and matter can be arranged: solids < liquids < gases, and more gas molecules or larger molecules mean more entropy. A reaction that makes fewer moles of gas, such as N₂ + 3 H₂ → 2 NH₃, has ΔS < 0. The Gibbs energy change ΔG = ΔH − TΔS (with ΔS converted to kJ) decides spontaneity at constant temperature and pressure: ΔG < 0 means the reaction proceeds forward. With ΔH < 0 and ΔS > 0 a reaction is spontaneous at every temperature; with ΔH > 0 and ΔS < 0 at none; when both have the same sign, temperature decides, and the crossover is T = ΔH ÷ ΔS. The standard value links to the equilibrium constant: ΔG° = −RT ln K with R = 8.314 J/(mol·K) and T in kelvin, so ΔG° < 0 means K > 1. Away from equilibrium, ΔG = ΔG° + RT ln Q. ΔG says which way a reaction goes and how far, never how fast.
Equilibrium constants and Le Chatelier's principle
For aA + bB ⇌ cC + dD, K = [C]^c[D]^d ÷ ([A]^a[B]^b); pure solids and liquids are left out. Reversing the equation inverts K, and multiplying the coefficients by n raises K to the power n. The size of K tells you the equilibrium mixture: acetaldehyde in water adds water reversibly to form its hydrate with K ≈ 1, so about half of it is hydrated, whereas formaldehyde (K ≈ 2 × 10³) is almost fully hydrated. The reaction quotient Q has the same form as K but uses the current concentrations: if Q < K the mixture reacts forward, if Q > K it reacts backward, and if Q = K it is at equilibrium. Le Chatelier's principle predicts the response to a disturbance. Adding a reactant or removing a product drives the reaction forward. Compressing a gas mixture favours the side with fewer moles of gas. Adding an inert gas at constant volume changes no partial pressure and causes no shift. Only a temperature change alters K: heating favours the endothermic direction. A catalyst changes none of these positions.
Rate laws, order and half-life
A rate law such as rate = k[A]^m[B]^n is measured, not read from the balanced equation. Compare initial rates in experiments where only one concentration changes: if doubling [A] doubles the rate, m = 1; if it quadruples the rate, m = 2; if the rate is unchanged, m = 0. The overall order is m + n, and the units of k follow from it: s⁻¹ for first order, M⁻¹ s⁻¹ for second order, M⁻² s⁻¹ for third order. Only for an elementary step do the exponents equal the molecularity. The one-step SN2 reaction of hydroxide with bromoethane drawn here is first order in each reactant, rate = k[CH₃CH₂Br][OH⁻]; in the SN1 hydrolysis of 2-chloro-2-methylbutane, the slow ionization is the rate-determining step, so the rate depends only on the concentration of the alkyl chloride and water does not appear. For a first-order reaction ln([A]₀ ÷ [A]) = kt, and the half-life t½ = ln 2 ÷ k ≈ 0.693 ÷ k does not depend on concentration: after n half-lives a fraction (½)^n remains.
Activation energy, Arrhenius and catalysis
On an energy profile, the activation energy Ea is the climb from the reactants to the highest transition state, and the reverse barrier is Ea − ΔH. In a multistep reaction each intermediate sits in a valley between two transition states, and the step with the highest transition state is rate-determining. The Arrhenius equation k = A e^(−Ea/RT) shows why rates rise steeply with temperature: the fraction of collisions with enough energy grows exponentially. For two temperatures, ln(k₂ ÷ k₁) = (Ea ÷ R)(1/T₁ − 1/T₂); use Ea in J/mol and T in kelvin. A catalyst opens a different pathway with a lower barrier and is regenerated, as H₃O⁺ is in the hydration of 2-methylpropene drawn here. Because it lowers the forward and reverse barriers equally, a catalyst speeds up the approach to equilibrium but leaves ΔH, ΔG° and K unchanged.
What to carry forward
Thermodynamics (ΔH, ΔS, ΔG, K) tells you which direction a reaction goes and how far; kinetics (rate laws, Ea, catalysts) tells you how fast it gets there. Keep units consistent: kJ versus J, kelvin, and the units of k that match the order.
Problemas de prática (40)
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Desenhe o mecanismo 4
Equilibrium and Le Chatelier
- Acid-catalysed hydration of acetaldehydeDraw the reversible acid-catalysed addition of water to acetaldehyde and use the hydration equilibrium constant to predict how much hydrate is present.
Rate laws and half-life
- SN1 hydrolysis of 2-chloro-2-methylbutaneDraw the stepwise SN1 hydrolysis of 2-chloro-2-methylbutane and identify the slow ionization as the step that sets the first-order rate law.
- One-step SN2: hydroxide and bromoethaneDraw the single concerted step in which hydroxide displaces bromide from bromoethane, and connect the one-step mechanism to its second-order rate law.
Activation energy and catalysis
- Acid-catalysed hydration of 2-methylpropeneDraw the acid-catalysed hydration of 2-methylpropene and show that H₃O⁺ is consumed in the first step and regenerated in the last, so it acts as a catalyst.
Preveja o produto 5
Free energy and spontaneity
- Which process lowers the entropy?Predict the sign of ΔS from changes in phase and in the number of moles of gas.
- Q, K and the sign of ΔGPredict the sign of ΔG and the direction of net reaction by comparing Q with K.
Equilibrium and Le Chatelier
- Compressing the ammonia equilibriumPredict the effect of compressing a gas-phase equilibrium from the moles of gas on each side.
- Adding argon at constant volumePredict that an inert gas added at constant volume does not shift a gas-phase equilibrium.
- Which way does the NO₂/N₂O₄ mixture shift?Predict the direction of net reaction by calculating Q and comparing it with K.
Classificar 4
Enthalpy and Hess's law
- What a negative ΔH meansClassify a reaction as exothermic from the sign of ΔH and state what that sign does not tell you.
Free energy and spontaneity
- When is an exothermic, entropy-lowering reaction spontaneous?Classify the temperature range of spontaneity from the signs of ΔH and ΔS.
Equilibrium and Le Chatelier
- Equilibrium expression with solidsWrite the equilibrium expression of a heterogeneous equilibrium, omitting pure solids.
Rate laws and half-life
- Overall order from the units of kClassify the overall order of a reaction from the units of its rate constant.
Comparar 2
Free energy and spontaneity
- Rank standard molar entropiesOrder substances by standard molar entropy using phase and molecular complexity.
Rate laws and half-life
- Rank first-order half-lives with mixed unitsRank first-order reactions by half-life after converting rate constants to common units.
Calcular 17
Enthalpy and Hess's law
- Hydrogenation enthalpy from bond enthalpiesEstimate a reaction enthalpy from average bond enthalpies by counting the bonds broken and formed.
- Enthalpy of neutralization from a coffee-cup calorimeterConvert a calorimeter temperature rise into a molar enthalpy of neutralization with the correct sign.
- Combustion enthalpy from enthalpies of formationCompute a reaction enthalpy from standard enthalpies of formation, including stoichiometric coefficients.
- Hess's law for carbon monoxideCombine two thermochemical equations, reversing one, to obtain the enthalpy of formation of CO.
- Reverse and scale a thermochemical equationChange the sign and scale of ΔH when a thermochemical equation is reversed and multiplied by a factor.
Free energy and spontaneity
- ΔG° for acetic acid ionizationCalculate ΔG° from a small equilibrium constant and interpret its positive sign.
- ΔG° for ammonia synthesisCalculate ΔG° from ΔH° and ΔS° with consistent units at 298 K.
- Equilibrium constant from ΔG°Calculate an equilibrium constant from ΔG° using ΔG° = −RT ln K.
- When does limestone decomposition become favourable?Find the temperature at which ΔG° changes sign from ΔH° and ΔS°.
Equilibrium and Le Chatelier
- Equilibrium [HI] from an ICE tableSolve an ICE table with a perfect-square equilibrium expression to find an equilibrium concentration.
- Reverse and halve an equilibriumTransform an equilibrium constant when the equation is reversed and its coefficients are halved.
Rate laws and half-life
- Rate constant from initial-rate dataDetermine the orders from initial-rate data and then calculate the rate constant with its units.
- Half-life of a first-order reactionCalculate a first-order half-life from the rate constant.
- Time for 90 % reactionUse the integrated first-order rate law to find the time for a given fraction of reaction.
Activation energy and catalysis
- Activation energy from two rate constantsCalculate an activation energy from rate constants measured at two temperatures.
- How much faster at 35 °C?Use the two-temperature Arrhenius equation to calculate the change in rate constant for a 10 K rise.
- Rate enhancement from a lower barrierCalculate the rate enhancement produced by a catalyst that lowers Ea, using the Arrhenius equation at body temperature.
Encontre o erro 1
- Are diamonds turning into graphite?Diagnose the error of using ΔG° to predict how fast a reaction occurs.
Interprete as evidências 3
Enthalpy and Hess's law
- Reading the reverse activation energyRead the forward barrier, reverse barrier and ΔH from the energies on a one-step reaction profile.
Rate laws and half-life
- Rate law from initial ratesDeduce the order in each reactant from initial-rate experiments that change one concentration at a time.
Activation energy and catalysis
- Reading a two-step energy profileIdentify the rate-determining step, intermediates and overall barrier from a two-step Gibbs energy profile.
Planeje uma síntese 1
- Conditions for the highest equilibrium yield of ammoniaChoose temperature and pressure that maximise the equilibrium yield of an exothermic, gas-reducing reaction.
Explique por quê 3
Enthalpy and Hess's law
- Why Hess's law worksExplain Hess's law in terms of enthalpy being a state function.
Activation energy and catalysis
- How a catalyst speeds up a reactionExplain how a catalyst increases the rate without changing ΔH or the equilibrium constant.
- Why heating speeds up reactions so muchExplain the steep temperature dependence of rate in terms of the fraction of collisions that exceed Ea.
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