ATTEMPT ANY TWO (2) QUESTIONS

Here are the solutions to the questions: 1a) Define the following: i) Rate of reaction: The rate of reaction is the change in the concentration of a reactant or product per unit time. It measures how fast reactants are consumed or products are formed. ii) Half-Life: Half-life (t_1/2) is the time required for the concentration of a reactant to decrease to half of its initial concentration. iii) Activation energy: Activation energy (E_a) is the minimum amount of energy that reactant molecules must possess to undergo a chemical reaction. It represents the energy barrier that must be overcome for a reaction to occur. 1b) i) Write the rate equation for the reaction 2A B + C if the order of the reaction is zero. For a zero-order reaction, the rate is independent of the reactant concentration. Rate = k[A]^0 = k The rate equation is Rate = k. ii) For a chemical reaction A B, it is found that the rate of reaction doubles when the concentration of A is increased four times. The order of the reaction is: Let the rate law be Rate = k[A]^n. If [A] is increased by 4 times, the rate doubles. 2 × Rate_1 = k(4[A]_1)^n 2 = 4^n To solve for n, we can write 4^n = (2^2)^n = 2^2n. 2^1 = 2^2n 1 = 2n n = (1)/(2) The order of the reaction is (1)/(2). iii) For a reaction R P, half-life (t_1/2) is observed to be independent of the initial concentration of reactants. What is the order of reaction? For a first-order reaction, the half-life is independent of the initial concentration (t_1/2 = ( 2)/(k)). For other orders, it depends on the initial concentration. The order of reaction is First order. 1c) CaCO_3 pieces are placed in a flask and HCl is added. Immediately a balloon is placed over the top of the flask and the balloon starts to inflate. i) Explain why the balloon inflates. The reaction between calcium carbonate (CaCO_3) and hydrochloric acid (HCl) produces carbon dioxide gas (CO_2). This gas accumulates inside the flask, increasing the pressure and causing the balloon to inflate. CaCO_3(s) + 2HCl(aq) CaCl_2(aq) + H_2O(l) + CO_2(g) ii) If the same mass of CaCO_3 in a powdered form is used, the balloon inflates faster. Explain why using collision theory. Using powdered CaCO_3 increases the surface area available for the reaction. According to collision theory, a larger surface area leads to more frequent effective collisions between the reactant particles (CaCO_3 and HCl), thus increasing the rate of reaction and producing CO_2 gas faster, which inflates the balloon more quickly. iii) Suggest two other ways to make the balloon inflate faster. • Increase the concentration of HCl: A higher concentration of HCl means more reactant particles per unit volume, leading to more frequent effective collisions and a faster reaction rate. • Increase the temperature: Raising the temperature increases the kinetic energy of the reactant molecules, causing them to move faster and collide more frequently and with greater energy, leading to a higher proportion of effective collisions and a faster reaction rate. 1d) Consider the following reactions. Predict and arrange the reactions as they occur from the most rapidly at room conditions to the slowest. The general principles for reaction rates are: • Ionic reactions in solution are typically very fast. • Reactions involving gases or complex molecules, especially those requiring significant bond breaking and forming, tend to be slower. • Decomposition reactions can vary but are often slow without catalysts. Let's analyze each reaction: i) C_6H_12O_6 + O_2 2CO_2 + 3H_2O (Combustion/oxidation of glucose) - This is a complex organic reaction involving many bond rearrangements. It is generally slow at room temperature without initiation. ii) Fe^2+ + O_2 Fe_2O_3 (Oxidation of iron(II) to iron(III) oxide) - This is a redox reaction involving a gas (O_2) and ions. While faster than complex organic reactions, it's typically slower than simple ionic reactions in solution. iii) H_2O_2 H_2O + O_2 (Decomposition of hydrogen peroxide) - This reaction is relatively slow at room temperature without a catalyst. iv) 2Fe^3+ + Sn^2+ 2Fe^2+ + Sn^4+ (Redox reaction between ions) - This is a reaction between ions in solution, which typically occurs very rapidly due to minimal bond breaking and strong electrostatic attractions. Arranging from most rapidly to slowest: 1. iv) 2Fe^3+ + Sn^2+ 2Fe^2+ + Sn^4+ (Fastest - ionic reaction) 2. ii) Fe^2+ + O_2 Fe_2O_3 (Moderately fast - redox, but involves gas) 3. iii) H_2O_2 H_2O + O_2 (Slow - decomposition without catalyst) 4. i) C_6H_12O_6 + O_2 2CO_2 + 3H_2O (Slowest - complex organic reaction) The order from most rapidly to slowest is: iv) > ii) > iii) > i) 2a) A substance dissociates through a first-order reaction with a half-life of 32 min. If the initial concentration of the substance is 2.7 × 10^11 molecules/L, what will be the concentration 2.24 h later? Step 1: Calculate the rate constant (k) from the half-life (t_1/2). For a first-order reaction: t_1/2 = ( 2)/(k) k = ( 2)/(t_1/2) = (0.693)/(32 min) = 0.021656 min^-1 Step 2: Convert the total time to minutes. Time (t) = 2.24 h × 60 min/h = 134.4 min Step 3: Use the integrated rate law for a first-order reaction to find the concentration at time t. [A]_t = -kt + [A]_0 [A]_t = -(0.021656 min^-1)(134.4 min) + (2.7 × 10^11 molecules/L) [A]_t = -2.907 + 26.319 [A]_t = 23.412 [A]_t = e^23.412 [A]_t = 1.47 × 10^10 molecules/L The concentration 2.24 h later will be 1.47 × 10^10 molecules/L. 2b) i) Express the rate of the following reaction in terms of the formation of ammonia: N_2(g) + 3H_2(g) 2NH_3(g) The rate of reaction can be expressed in terms of the change in concentration of reactants and products, divided by their stoichiometric coefficients. For the formation of ammonia: Rate = +(1)/(2)d[NH_3]dt The rate of reaction in terms of the formation of ammonia is Rate = +(1)/(2)d[NH_3]dt. ii) For the reaction 3A + 2B C + 2D, reactant A is found to disappear at the rate of 4.6 × 10^-3 Ms^-1. xiii) What is the rate of disappearance of the reactant B? The relationship between the rates of disappearance of A and B is given by their stoichiometric coefficients: -(1)/(3)(d[A])/(dt) = -(1)/(2)(d[B])/(dt) We are given -(d[A])/(dt) = 4.6 × 10^-3 Ms^-1. -(1)/(2)(d[B])/(dt) = (1)/(3) × (4.6 × 10^-3 Ms^-1) -(d[B])/(dt) = (2)/(3) × (4.6 × 10^-3 Ms^-1) -(d[B])/(dt) = 3.066... × 10^-3 Ms^-1 The rate of disappearance of reactant B is 3.07 × 10^-3 Ms^-1. xiv) What is the rate of appearance for the product D? The relationship between the rate of disappearance of A and the rate of appearance of D is: -(1)/(3)(d[A])/(dt) = +(1)/(2)(d[D])/(dt) We are given -(d[A])/(dt) = 4.6 × 10^-3 Ms^-1. +(1)/(2)(d[D])/(dt) = (1)/(3) × (4.6 × 10^-3 Ms^-1) +(d[D])/(dt) = (2)/(3) × (4.6 × 10^-3 Ms^-1) +(d[D])/(dt) = 3.066... × 10^-3 Ms^-1 The rate of appearance for product D is 3.07 × 10^-3 Ms^-1. Drop the next question! 📸