This chemistry question involves key chemical concepts and calculations. The detailed solution below walks through each step, from identifying the reaction type to computing the final answer.
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Here are the solutions to the questions: Question 5 (Kinetics) The reaction is 2A_(g) + B_(g) 2C_(g). The rate law can be expressed as Rate = k[A]^x[B]^y. (c)(i) (α) What is the order of reaction with respect to A? Step 1: Compare Experiment 1 and Experiment 2. In Experiment 1, [A] = 0.100 mol dm^-3, [B] = 0.100 mol dm^-3, Rate = 2.0 × 10^-4 mol dm^-3s^-1. In Experiment 2, [A] = 0.200 mol dm^-3, [B] = 0.100 mol dm^-3, Rate = 4.0 × 10^-4 mol dm^-3s^-1. When [B] is kept constant and [A] is doubled (from 0.100 to 0.200), the rate doubles (from 2.0 × 10^-4 to 4.0 × 10^-4). This indicates that the reaction is first order with respect to A. The order of reaction with respect to A is 1. (β) What is the order of reaction with respect to B? Step 1: Compare Experiment 2 and Experiment 3. In Experiment 2, [A] = 0.200 mol dm^-3, [B] = 0.100 mol dm^-3, Rate = 4.0 × 10^-4 mol dm^-3s^-1. In Experiment 3, [A] = 0.200 mol dm^-3, [B] = 0.200 mol dm^-3, Rate = 1.6 × 10^-3 mol dm^-3s^-1. When [A] is kept constant and [B] is doubled (from 0.100 to 0.200), the rate increases by a factor of 1.6 × 10^-34.0 × 10^-4 = 4. Since doubling [B] quadruples the rate, the reaction is second order with respect to B. The order of reaction with respect to B is 2. (c)(ii) What is the overall order of reaction? The overall order of reaction is the sum of the individual orders with respect to each reactant. Overall order = (order with respect to A) + (order with respect to B) = 1 + 2 = 3. The overall order of reaction is 3. Question 5 (Thermodynamics & Hybridization) (a) Hess's law of constant heat summation states that the total enthalpy change for a chemical reaction is independent of the pathway taken, as long as the initial and final conditions are the same. It means that if a reaction can be expressed as a sum of several steps, the enthalpy change for the overall reaction is the sum of the enthalpy changes for each step. (b) The thermochemical equations for the given statements are: (i) Standard enthalpy of formation of LiF_(s) is -617 kJ mol^-1. Li_(s) + (1)/(2)F_2(g) LiF_(s) H_f^ = -617 kJ mol^-1 (ii) Atomization energy of Li_(s) is +161 kJ mol^-1. Li_(s) Li_(g) H_atom = +161 kJ mol^-1 (iii) Ionisation energy of Li_(g) is +520 kJ mol^-1. Li_(g) Li^+_(g) + e^- H_ion = +520 kJ mol^-1 (iv) Dissociation energy of F_2(g) is +79 kJ mol^-1. (1)/(2)F_2(g) F_(g) H_diss = +(79)/(2) = +39.5 kJ mol^-1 (v) Electron affinity of F_(g) is -328 kJ mol^-1. F_(g) + e^- F^-_(g) H_EA = -328 kJ mol^-1 (c)(i) Use the thermochemical equations in (b) to draw an energy cycle. The Born-Haber cycle for the formation of LiF_(s) is shown below: c Li_(s) + (1)/(2)F_2(g) \\ H_atom(Li) = +161 \\ Li_(g) + (1)/(2)F_2(g) \\ H_ion(Li) = +520 \\ Li^+_(g) + e^- + (1)/(2)F_2(g) \\ H_diss(F_2) = +39.5 \\ Li^+_(g) + e^- + F_(g) \\ H_EA(F) = -328 \\ Li^+_(g) + F^-_(g) \\ H_lattice \\ LiF_(s) \\ H_f^ = -617 \\ Li_(s) + (1)/(2)F_2(g) (Note: The arrows for H_f^ and H_lattice should point downwards, indicating exothermic processes for formation and lattice energy, respectively, as drawn in the cycle above. The cycle starts from elements in standard states, goes through gaseous ions, and then forms the ionic solid.) (c)(ii) Using Hess's law, write an equation relating all the energy terms in (b). According to Hess's Law, the enthalpy of formation of LiF is equal to the sum of the enthalpy changes in the Born-Haber cycle: H_f^ = H_atom(Li) + H_ion(Li) + H_diss(F_2) + H_EA(F) + H_lattice (c)(iii) Using the equation in (ii) calculate the lattice energy of LiF_(s). Step 1: Substitute the given values into the equation from (ii). -617 kJ mol^-1 = (+161 kJ mol^-1) + (+520 kJ mol^-1) + (+39.5 kJ mol^-1) + (-328 kJ mol^-1) + H_lattice Step 2: Sum the known enthalpy changes. -617 = 161 + 520 + 39.5 - 328 + H_lattice -617 = 392.5 + H_lattice Step 3: Solve for H_lattice. H_lattice = -617 - 392.5 H_lattice = -1009.5 kJ mol^-1 The lattice energy of LiF_(s) is -1009.5 kJ mol^-1. (d)(i) Give the type of hybridization shown by the central atoms in each of the following compounds: • BCl_3: The central atom is Boron (B). Boron has 3 valence electrons and forms 3 single bonds with chlorine atoms. There are no lone pairs on boron. The steric number is 3 (3 bonding pairs + 0 lone pairs), which corresponds to sp^2 hybridization. • NH_3: The central atom is Nitrogen (N). Nitrogen has 5 valence electrons and forms 3 single bonds with hydrogen atoms, leaving 1 lone pair. The steric number is 4 (3 bonding pairs + 1 lone pair), which corresponds to sp^3 hybridization. • H_2O: The central atom is Oxygen (O). Oxygen has 6 valence electrons and forms 2 single bonds with hydrogen atoms, leaving 2 lone pairs. The steric number is 4 (2 bonding pairs + 2 lone pairs), which corresponds to sp^3 hybridization. (d)(ii) Deduce the shape of each of the molecules in (i): • BCl_3: With 3 bonding pairs and 0 lone pairs around the central boron atom, the electron geometry and molecular geometry are both trigonal planar. The shape is trigonal planar. • NH_3: With 3 bonding pairs and 1 lone pair around the central nitrogen atom, the electron geometry is tetrahedral, but the molecular geometry is trigonal pyramidal due to the repulsion from the lone pair. The shape is trigonal pyramidal. • H_2O: With 2 bonding pairs and 2 lone pairs around the central oxygen atom, the electron geometry is tetrahedral, but the molecular geometry is bent (or V-shaped) due to the repulsion from the two lone pairs. The shape is bent. Drop the next question.