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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You're on a roll — here's the solution for Item 2: a) Understand why temperature changes occur when anhydrous sodium carbonate is added to water. When anhydrous sodium carbonate (Na_2CO_3) is added to water, it dissolves. This dissolution process involves two main energy changes: the energy required to break the ionic bonds in the solid lattice of Na_2CO_3 and the energy released when the ions are hydrated (surrounded by water molecules). If the energy released during hydration is greater than the energy absorbed to break the lattice, the overall process is exothermic, causing the temperature of the solution to increase. Conversely, if more energy is absorbed, the process is endothermic, and the temperature decreases. In the case of anhydrous sodium carbonate, the process is exothermic, leading to a temperature rise. b) Determine the heat of solution of sodium carbonate. Step 1: Calculate the change in temperature ( T). The initial temperature (T_initial) is 20.5^ C (at Time = 0 s). The temperature stabilizes at 27^ C (from Time = 7 s to 9 s). T = T_final - T_initial = 27^ C - 20.5^ C = 6.5^ C = 6.5 K Step 2: Calculate the mass of the solution (m_solution). Mass of anhydrous Na_2CO_3 = 8.5 g. Volume of distilled water = 50 cm^3. Assuming the density of water is 1 g cm^-3, the mass of water is 50 g. m_solution = Mass of Na_2CO_3 + Mass of water = 8.5 g + 50 g = 58.5 g Step 3: Calculate the heat absorbed by the solution (Q_solution). The specific heat capacity of the solution (c) is 4.2 J g^-1 K^-1. Q_solution = m_solution × c × T Q_solution = 58.5 g × 4.2 J g^-1 K^-1 × 6.5 K Q_solution = 1597.05 J Step 4: Calculate the moles of Na_2CO_3. Molar mass of Na_2CO_3: (2 × 22.99 g/mol for Na) + (1 × 12.01 g/mol for C) + (3 × 16.00 g/mol for O) = 105.99 g/mol Moles of Na_2CO_3 = MassMolar mass = 8.5 g105.99 g/mol ≈ 0.080196 mol Step 5: Determine the heat of solution ( H_sol). Since the temperature increased, the dissolution is an exothermic process, so H_sol will be negative. H_sol = -Q_solutionMoles of Na_2CO_3 H_sol = -1597.05 J0.080196 mol H_sol ≈ -199140 J/mol Converting to kilojoules per mole: H_sol ≈ -199.14 kJ/mol Rounding to two significant figures (based on T and mass of Na_2CO_3): H_sol ≈ -200 kJ/mol What's next?