Summary (CAIE) Cambridge A Level Chemistry (9701) - Chemical Kinetics, Study notes of Chemistry

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Summary (CAIE) Cambridge A Level Chemistry (9701) Summary & 15’s Chemical Kinetics Problems with Solution Chemical Kinetics Solution Concentration Reaction Rate Concept Reaction rate is the rate at which reactants are reduced or the rate at which products are formed v = - [𝑹] 𝒕 or v = + [𝑷] 𝒕 v = reaction rate (M/s or Ms-1) - [𝑅] 𝑡 = the rate of reduction in the molar concentration of one of the MOLARITY M = 𝒏 𝑽 M =  . 10 . % Mr M = solution molarity n = number of moles of solute V = volume of solution  = density (g/mL) % = content (% of mass) Mr = relative molecular mass DILUTION M1V1 = M2V2 M1, M2 = molarity of the solution before and after dilution V1, V2 = volume of the solution before and after dilution MIXING M = M1V1 + M2V2 V1 + V2 M = molarity of the mixed solution M1, M2 = molarity of solutions 1 and 2 V1, V2 = volume of solutions 1 and 2 Factors Affecting the Reaction Rate: 1. Solution Concentration(aq): Concentration Collision Frequency Reaction Rate  2. Particle Size(s): Particle Size  Touch Surface Area  Collision Frequency Reaction Rate  3. Pressure/Volume(g): Volume  / Presure  Collision Frequency Reaction Rate  4. Temperature(s, l, aq, g): Temperature  Kinetic Energy  Enough Reactant Energy  Reaction Rate  5. Catalyst(s, l, aq, g): Add catalyst Activation Energy  Reaction Rate  Effect of Catalysts on Exothermic Reactions 1. A catalyst is a substance that can speed up a reaction, but the substance itself is not changed chemically at the end of the reaction. 2. The way a catalyst speeds up a reaction is by lowering the activation energy Reactant Reactant Product Effect of Catalysts on Endothermic Reactions Catalyst Types 1. Homogeneous catalyst: catalyst which is in the same phase as the reactants. 2. Heterogeneous catalyst: the catalyst has a phase different from the reactants. 3. Autocatalyst: reaction products (products) that act as catalysts. 4. Biocatalyst: catalysts that work in the metabolic processes of living things, are called enzymes Reactant Product Catalyst Properties 1. Does not undergo any lasting change in reactions, but may be involved in reaction mechanisms. 2. Speed up the rate of reaction, but does not change the type or amount of reaction products. 3. Lowers the activation energy, but does not change the enthalpy change for the reaction. 4. Changing the mechanism by providing steps that have a lower. activation energy. 5. Specific, meaning that it can only catalyze certain reactions. 6. Only needed in small quantities. 7. Can be poisoned by certain substances. Reaction Rate Equation On reaction: aA + bB cC + dD v = k[A]x [B]y Reaction Rate Constant: a constant that depends on the type of reactant, pressure, temperature and catalyst. Order of Reaction: How big is the effect of changing the concentration of one of the reactants on the reaction rate. v = reaction rate (M/s or Ms-1) k = reaction rate constants [A], [B] = concentration of A and B x = order of reaction to A y = order of reaction to B x + y = total reaction order Ans. a. M = . 10. % = 1,8. 10. 98 = 18M Mr 98 b. M1. V1 = M2. V2 18. 10 = 0,1. V2 V2 = 1800 mL Vwater = 1800 – 10 = 1790 mL of water added 03. A total of 100 mL of 0.1M HCl solution was mixed with 400 mL of 0.3M HCl solution. What is the molarity of the solution after mixing? Ans. M = M1. V1 + M2. V2 = 0,1. 100 + 0,3. 400 = 130 = 0,26 M V1 + V2 100 + 400 500 04. Ammonia gas (NH3) can be made from the reaction between nitrogen gas (N2) and hydrogen gas (H2). At a certain temperature and pressure, the rate of formation of ammonia gas is 2x10-4 M/s, determine the rate of reduction of nitrogen gas and hydrogen gas. Ans. N2 + 3 H2 2 NH3 - [N2] = - 1 2 . 2. 10-4 = -10-4 M/s t - [H2] = - 3 2 . 2. 10-4 = -3. 10-4 M/s t [NH3] = 2. 10-4 M/s t 05. In a 5L chamber, 0.8 mol of N2O4 gas is heated to a certain temperature, until it decomposes into NO2 gas. If after 4 seconds in the chamber there are 0.6 moles of NO2 gas, determine the average rate of decomposition of N2O4 gas. Ans. N2O4 which reacted as much: 1 2 x 0,6 = 0,3 mole N2O4 2 NO2 S 0,8 - S = start reaction R 0,3 0,6 R = react L 0,5 0,6 L = leftover Molarity of decomposed N2O4: 𝟎,𝟑 5 = 0,06 M The average speed of decomposition of N2O4 gas is - [N2O4] = -0,06 = - 0,015 M/s t 4 06. In a room with a volume of 4 liters, some water vapor is heated until it decomposes into hydrogen gas and oxygen gas. After 15 seconds, it turns out that there are still 0,3 moles of water vapor in the room. If at this temperature the rate of formation of oxygen gas is 0.01M/s, determine: a. The rate of decomposition of water vapor b. The initial number of moles of water vapor in the chamber Ans. 2 H2O 2 H2 + O2 S 1,5 R 1,2 L 0,3 The rate of formation of oxygen gas: [O2] = 0,01 M/s t a. -[H2O] = (2/1) x 0,01 = -0,02 M/s t b. The amount of water vapor that reacts: -[H2O] = -0,02 [H2O] = 0,3 M 15 n H2O = 0,3. 4 = 1,2 moles The initial number of moles of water vapor: 0,3 + 1,2 = 1,5 moles 07. State the ways in which you can increase the rate of the following reactions. Also give a brief reason why this method can increase the rate of reaction based on the collision theory. a. Mg(s) + HCl(aq) MgCl2(aq) + H2(g) b. NH3((g) + O2(g) NO(g) + H2O(g) Ans. a. i. Increase the concentration of HCl: to increase the number of HCl particles and the frequency of collisions between HCl and Mg particles. v 1 152 1 76 1 19 1 19 c. x + y = 2 + 3 = 5 d. v1 = k[A]1 x [B]1 y 1,25.10-2 = k. (0,1)2.(0,1)3 k = 1250 M-4 s-1 e. v = k [A]x [B]y = 1250 [A]2 [B]3 f. v = 1250 [0,4]2 [0,2]3 = 1,6 Ms-1 10. Indicate the reaction rate equation (in k) of the reaction BrO3 - (aq) + Br- (aq) + 6H+ (aq) . 3 Br2(g) + 3 H2O(l) if the experimental results obtained data as follows: [BrO3 -] [M] [Br-] [M] [H+] [M] t (s) 0,4 0,24 0,01 152 0,8 0,24 0,01 76 0,4 0,48 0,02 19 0,8 0,48 0,01 19 time is inversely proportional to speed, so: t = 𝟏 𝒗 v = 𝟏 𝒕 To determine the order of the BrO3 - reaction, we choose the others fixed and a changing BrO3 -so that the change in velocity (v) is purely due to BrO3 -, and vice versa for the other orders of the reaction. 2 2 2 4 2 2 8 Ans. We use the fast method: v = k [BrO3 -]x [Br-]y [H+]z [BrO3 -]x 2x = 2, x = 1 [Br-]y 2y = 4, y = 2 [H+]z 2y.2z = 8 22.2z = 8 2z = 2 z = 1 So, the reaction rate equation: v = k [BrO3 -] [Br-]2 [H+] 11. In the reaction A + B + C D + E, obtained the following data:  If the concentration A is fixed, the concentration of B and C each is raised twice, then the reaction rate is twice as large as.  If the concentration of A and C each is raised twice, while the concentration B is fixed, then the reaction rate is eight times greater.  If the concentration of A and B each is raised twice, while the concentration of C is fixed, the reaction rate is four times greater. Determine the total reaction order from the reaction (1) x – y = 2 (4) x + y = 2 + 2x = 4 x = 2 y = 0 (using equation number 4) z = 1 (using equation number 2) Ans. v = k [A]x [B]y [C]z [A] [B] [C] v 1 a b c v 2 a 2b 2c 2v 3 2a b 2c 8v 4 2a 2b c 4v We use the fast method: 2x.( 1 2 )y = 4 2x-y = 22 x – y = 2….(1) 2y.2z = 2 2y+z = 2 y + z = 1….(2) 2x. ( 1 2 )z = 2 2x-z = 2 x – z = 1….(3) We use the elimination method: (2) y + z = 1 (3) x – z = 1 + x + y = 2….(4) The total reaction order: x + y + z = 2 + 0 + 1 = 3 2 𝟏 𝟐 4 2 2 2 2 2 2 2 14. Use data from table below to establish the order of reaction: 2 HgCl2(aq) + C2O4 2- (aq) 2 Cl- (aq) + 2 CO2(g) + Hg2Cl2(s) with respect to HgCl2 dan C2O4 2- and also the overall order of the reaction. Ans. v = k [HgCl2]m [C2O4 2-]n We need to determine the values of m and n. In comparing experiment 2 and experiment 3, note that [HgCl2] is essentially doubled while [C2O4 2-]n is held constant. Note also that R2 = 2 x R3. Rather than use the actual concentrations and rates in the following rate equation, let’s work initially with their symbolic equivalents. R2 = k x [HgCl2]2 m x [C2O4 2-]2 n = k x (2 x [HgCl2]3)m x [C2O4 2-]3 n R3 = k x [HgCl2]3 m x [C2O4 2-]3 n 2m = 2, so m = 1 To determine the value of n, we can form the ratio R2/R1. Now [C2O4 2-] is double and [HgCl2] is held constant. We have value m = 1 R2 = k x [HgCl2]2 1 x [C2O4 2-]2 n = k x (0,105)1 x (2 x 0,15)n R1 = k x [HgCl2]1 1 x [C2O4 2-]1 n = k x (0,105)1 x (0,15)n 2n = 4, n = 2 First order in HgCl2 (m = 1), second order in C2O4 2- (n = 2) and third order overall (m + n = 1 + 2 = 3). 15. The decomposition of hydrogen peroxide (H2O2) in water occurs according to the reaction: 2 H2O2(aq) 2 H2O(l) + O2(g) From 100 mL of H2O2 solution at a certain time it is observed that the rate of formation of O2 gas is 4.88 mL/s (under the condition that the volume of 1 mol of N2 gas is 22.4 L) at the same time the rate of decomposition of H2O2 is Ans. The Gay Lussac Law: = 2.10-4 moles O2 Reaction rate: 1 ATH,O,] _ 1 A[O,] _ 1, A202] _ 1, AlOz] 2 t 1 t Mier 2 sy = 2x 2.10 moles/s = 4.103 M/s Vo, 1 ™ 1 Decomposition rate of H2O2 = 4.10°° Ms?