Reaction Kinetics Cheat Sheet

Kinetics H2 CHEMISTRY ORDERS OF REACTION RATE EQUATION DEFINITIONS CONTINUOUS METHOD QUESTIONS Question: Consider the reaction 2A + B → C. A couple of experiments wereconducted to study the reaction kinetics. B is kept in excess for all experiments.Find the order of reaction with respect to A and B. Observations Order of reaction When [A] is doubled, the rate ofreaction doubles Rate ∝ [A]Reaction is first order with respect to [A] When [A] is doubled, the rate ofreaction quadruples Rate ∝ [A]²Reaction is second order with respect to [A] When [A] is doubled, the rate ofreaction does not change Rate ∝ [A] ⁰Reaction is zero order with respect to [A] Term Definition Rate equation An experimentally-determined equation that relates the rate of reactionto the concentrations of the reactants raised to the appropriate orders. Order of Reaction The power on a reactant’s concentration term in the rate equation. Thispower is determined experimentally. Overall Order ofReaction The sum of the individual orders. k The proportionality constant in the experimentally-determined rateequation. INITIAL RATE METHOD QUESTIONS Consider the reaction 2A + B → C. A couple of experiments were conducted to find the order of reaction with respect to A and B. Exp V of A /cm³ V of B /cm³ V of H ₂ O/cm³ Initial rate 1 5 5 10 1.0 2 5 10 5 2.0 3 10 10 0 8.0 Therefore, rate = k[A]²[B] This is to ensure that total volume remains constant. Whentotal volume is constant, the concentration of reagents isproportional to volume ( [A] ∝ V of A. When volume doubles,concentration doubles). What is the rate equation? Comparing experiments 1 and 2, where [A] is constant, when [B]doubles, the initial rate doubles. Therefore, reaction is first orderwith respect to [B].Comparing experiments 2 and 3, where [B] is constant, when [A]doubles, the initial rate increases by 4 times. Therefore, reactionis second order with respect to [A]. Why must water be added to experiments 1 and 2? If order ofreaction w.r.tA is... Zero Order First Order The[reactant]-time graphwould looklike... Answeringtechnique The [A] decreases linearly overtime. This suggests that thedecrease in [A] has no effect onthe rate of reaction. *find two half-lives: 100% → 50% and 50%→ 25%*Since, half-life is constant at 10 min,reaction is first order w.r.t. [A]. The[product]-time graphwould looklike... Answeringtechnique The [C] increases linearly overtime. This suggests that thedecrease in [A] has no effect onthe rate of reaction. *find final [C] (100%) + two half-lives: 0%→ 50% and 50% → 75%*Since, half-life is constant at 10 min,reaction is first order w.r.t. [A]. [A] time [A] time 100% 50% 25% 1010 [C] time [C] time 100% 50%75% 1010 If order ofreaction w.r.t Bis... Zero Order First Order The [reactant]-time graphwould look like... The [product]-time graphwould look like... Answeringtechnique When the [B] halves, the initialrate of the reaction did notchange. This suggests that thedecrease in [B] has no effect onthe rate of reaction. *draw a tangent at t=0 for eachgraph and calculate gradient*When [B] halves, the initial ratehalves. Reaction is first order withrespect to [B]. [A] time Exp 1: [B] = 1.0 mol dm ⁻ ³Exp 2: [B] = 0.5 mol dm ⁻ ³ [A] timeExp 1: [B] = 1.0 mol dm ⁻ ³Exp 2: [B] = 0.5 mol dm ⁻ ³ [C] time Exp 1: [B] = 1.0 mol dm ⁻ ³Exp 2: [B] = 0.5 mol dm ⁻ ³ [C] time Exp 1: [B] = 1.0 mol dm ⁻ ³ Exp 2: [B] = 0.5 mol dm ⁻ ³ For an overall first – order reaction, half-life is t ₁ / ₂ = ln 2 / k, where k is the rate constant % remaining = (1/2) ⁿ, where n is the number of half-lives When 25 mg dm ⁻ ³ melphalan is dissolved in blood, the decrease inits concentration has a constant half–life of 90 minutes. What isthe concentration of melphalan in the blood six hours later? n = 360 / 9 = 4amount remaining = initial amount (1/2) ⁿamount remaining = 25 (1/2)⁴ amount remaining = 1.5625 mg dm ⁻ ³ HALF-LIFE is the time taken for the concentration of areactant to decrease to half its initial value. To experience why 90% of MACRO students score A/B in A Levels, text us at +65 83662396 :)

1. TEMPERATURE 2.CATALYST No of particles Energy No of particles withenergy > E ₐ at T ₁ E ₐ T ₁ T ₂ No of particles withenergy > E ₐ at T ₂ Average kinetic energy of reactant particles increases whentemperature increasesT ₁ to T ₂ More reactant particles possess energies greater than or equal toactivation energy Frequency of effective collisions increases Rate of reaction increases No of particles withenergy > E ₐ₁ No of particles withenergy > E ₐ₂ No of particles EnergyE ₐ₁E ₐ₂In the presence of a catalyst, activation energy of the reaction lowersE ₐ₁ to E ₐ₂More reactant particles possess energies greater than or equal to E ₐ₂ Frequency of effective collisions increases Rate of reaction increases Kinetics H2 CHEMISTRY FACTORS AFFECTING RATE OF REACTION A. HOMOGENEOUS CATALYSIS B. HETEROGENEOUS CATALYSIS Explain the role of oxides of nitrogen in the formation of acid rain Step 1: SO ₂ (g) + NO ₂ (g) → SO ₃ (g) + NO(g) Step 2: NO(g) + 1⁄2O ₂ (g) → NO ₂ (g)Overall equation: SO ₂ (g) + 1⁄2O ₂ (g) → SO ₃ (g) Step 3: SO ₃ (g) + H ₂ O(l) → H ₂ SO ₄ (aq) Explain how a catalytic converter removes harmful exhaust gases C ₓ H (g) + (x + y/4) O ₂ (g) → xCO ₂ (g) + H ₂ O(g)2CO(g) + O ₂ (g) → 2CO ₂ (g) 2NO(g) + 2CO(g) → N ₂ (g) + 2CO ₂ (g) y Explain the catalytic role of Fe² ⁺ in the I ⁻ /S ₂ O ₈ ² ⁻ reaction Step 1: 2Fe² ⁺ (aq) + S ₂ O ₈ ² ⁻ (aq) → 2Fe³ ⁺ (aq) + 2SO ₄ ² ⁻ (aq) Step 2: 2Fe³ ⁺ (aq) + 2I ⁻ (aq) → 2Fe² ⁺ (aq) + I ₂ (aq)Overall equation: 2I ⁻ (aq) + S ₂ O ₈ ² ⁻ (aq) → I ₂ (aq) + 2SO ₄ ² ⁻ (aq) Explain the mode of action of Fe in catalysing theHaber Process. Process explanation Adsorption N ₂ and H ₂ diffuse towards the surface andadsorbs onto the active sites of the catalyst byformation of weak attraction forces At the activesite BOWAF:Molecules are brought closer(B) together, andin the right orientation (O). N ≡ N and H-H covalent bonds within themolecules are weakened (W). Activation energy(A) of the reaction decreasesFrequency(F) of effective collision increases Desorption NH ₃ desorbs and diffuses away from thesurface The active sites are available for furtherreaction C. AUTOCATALYSIS 2MnO ₄⁻ + 5C ₂ O ₄ ² ⁻ + 16H ⁺ → 2Mn² ⁺ + 10CO ₂ + 8H ₂ O Given that Mn² ⁺ behaves as a catalyst, explain how the [MnO ₄⁻ ]changes over time. [MnO ₄⁻ ] time POINTS explanation (1) Rate of reaction is slow as there are limited amounts of catalystMn² ⁺ at the start. (2) Rate of reaction increases sharply as more catalyst Mn² ⁺ isproduced as reaction proceeds (3) Rate of reaction slows down as there are less reactants D. BIOLOGICAL CATALYSTS Explain how to concentration of enzyme affectsthe rate of a biologic reaction. Rate [Enzyme] first-order reaction zero-order reaction POINTS explanation At low [substrate] There are still many active sites available forsubstrate to bind to. Rate of reaction increases proportionally withincreasing substrate concentration. Hence reaction is first order w.r.t the substrate. At high[substrate] All active sites are filled. Substrates moleculeshave to wait to bind to the active sites. Any further increase in substrate concentrationwill have no effect on the rate of reaction. Hence reaction is zero order w.r.t the substrate. (1) (2) (3) www.macroacademy.org