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 — let's tackle these questions, Ignatius. Question 17 What is meant by the term enthalpy of neutralisation? The enthalpy of neutralisation is the heat energy released when one mole of water is formed from the reaction of an acid with a base under standard conditions. i) Write an equation for the reaction that occurred. The reaction is between sulfuric acid (H_2SO_4) and sodium hydroxide (NaOH). H_2SO_4(aq) + 2NaOH(aq) Na_2SO_4(aq) + 2H_2O(l) ii) Calculate the enthalpy of neutralisation of sodium hydroxide (specific heat capacity of solution = 4.2 J g^-1 K^-1, density of water = 1 g cm^-3). Step 1: Calculate the moles of reactants. Moles of H_2SO_4: Moles = Concentration × Volume = 1 M × (50)/(1000) L = 0.050 mol Moles of NaOH: Moles = Concentration × Volume = 2 M × (50)/(1000) L = 0.100 mol Step 2: Determine the limiting reactant and moles of water formed. From the balanced equation, 1 mol of H_2SO_4 reacts with 2 mol of NaOH to produce 2 mol of H_2O. If 0.050 mol of H_2SO_4 reacts, it would require 2 × 0.050 = 0.100 mol of NaOH. Since we have exactly 0.100 mol of NaOH, both reactants are completely consumed. The moles of water formed will be 2 × 0.050 mol = 0.100 mol. Step 3: Calculate the total mass of the solution. Total volume of solution = 50 cm^3 (H_2SO_4) + 50 cm^3 (NaOH) = 100 cm^3. Assuming the density of the solution is 1 g cm^-3: Mass of solution = Volume × Density = 100 cm^3 × 1 g cm^-3 = 100 g Step 4: Calculate the heat absorbed by the solution (q). Given T = 13.6^ (or 13.6 K) and specific heat capacity (c) = 4.2 J g^-1 K^-1. q = mc T q = 100 g × 4.2 J g^-1 K^-1 × 13.6 K q = 5712 J Convert to kilojoules: q = (5712)/(1000) kJ = 5.712 kJ Step 5: Calculate the enthalpy of neutralisation ( H_neut). Since heat is released (temperature rose), H is negative. H_neut = -Heat releasedMoles of water formed H_neut = -5.712 kJ0.100 mol H_neut = -57.12 kJ mol^-1 Question 18 i) What does point S represent? Point S represents the equivalence point (or neutralization point) of the reaction, where the acid and base have completely reacted with each other, and the maximum temperature of the solution is reached. ii) Why does the temperature rise from R to S? The temperature rises from R to S because the reaction between HCl (acid) and NaOH (base) is an exothermic reaction. As more NaOH is added, more heat is released, causing the temperature of the solution to increase until the equivalence point (S) is reached. iii) Determine the molarity of the HCl acid. Step 1: Identify the volume of NaOH at point S. From the graph, point S occurs when 15 cm^3 of NaOH has been added. Step 2: Write the balanced chemical equation. The reaction is between HCl and NaOH: HCl(aq) + NaOH(aq) NaCl(aq) + H_2O(l) The mole ratio of HCl to NaOH is 1:1. Step 3: Calculate the moles of NaOH at the equivalence point. Given concentration of NaOH = 0.5 M and volume = 15 cm^3. Moles of NaOH = Concentration × Volume Moles of NaOH = 0.5 M × (15)/(1000) L = 0.0075 mol Step 4: Calculate the moles of HCl. Since the mole ratio is 1:1, at the equivalence point: Moles of HCl = Moles of NaOH = 0.0075 mol Step 5: Calculate the molarity of HCl. Given volume of HCl = 20 cm^3. Molarity of HCl = Moles of HClVolume of HCl Molarity of HCl = 0.0075 mol(20)/(1000) L = 0.0075 mol0.020 L Molarity of HCl = 0.375 M Question 19 a) Give the name of H_2. The name of H_2 is hydrogen gas (or simply hydrogen). b) Determine the value of H_3. Step 1: Apply Hess's Law to the energy cycle. Hess's Law states that the total enthalpy change for a reaction is independent of the pathway taken. From the diagram, the direct path from H_2(g) + O_2(g) to H_2O_2(l) is H_1. The indirect path is from H_2(g) + O_2(g) to H_2O(l) + (1)/(2)O_2(g) (which is H_2), and then from H_2O(l) + (1)/(2)O_2(g) to H_2O_2(l) (which is H_3). Therefore, according to Hess's Law: H_1 = H_2 + H_3 Step 2: Substitute the given values and solve for H_3. Given H_1 = -188 kJ mol^-1 and H_2 = -286 kJ mol^-1. -188 kJ mol^-1 = -286 kJ mol^-1 + H_3 H_3 = -188 kJ mol^-1 - (-286 kJ mol^-1) H_3 = -188 kJ mol^-1 + 286 kJ mol^-1 H_3 = 98 kJ mol^-1 That's all of them! What's next?