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Using the Tables in Appendix \(1,\) calculate \(\Delta H\) for the reaction of the following. (a) \(1.00 \mathrm{~L}\) of \(0.100 \mathrm{M} \mathrm{NaOH}\) with \(1.00 \mathrm{~L}\) of \(0.100 \mathrm{MHCl}\) (b) \(1.00 \mathrm{~L}\) of \(0.100 \mathrm{M} \mathrm{NaOH}\) with \(1.00 \mathrm{~L}\) of \(0.100 \mathrm{M}\) HF, taking the heat of formation of \(\mathrm{HF}(a q)\) to be \(-320.1 \mathrm{~kJ} / \mathrm{mol}\)

Short Answer

Expert verified
Answer: The enthalpy changes for the reactions are: - For NaOH with HCl: ΔH = -3.543 kJ - For NaOH with HF: ΔH = 21.481 kJ

Step by step solution

01

Write the balanced chemical equation for each reaction

For reaction (a) between NaOH(aq) and HCl(aq), the balanced chemical equation is: NaOH(aq) + HCl(aq) -> NaCl(aq) + H2O(l) For reaction (b) between NaOH(aq) and HF(aq), the balanced chemical equation is: NaOH(aq) + HF(aq) -> NaF(aq) + H2O(l)
02

Determine heats of formation for each reaction

Using the given tables in Appendix 1, we can find the heats of formation of the components in each balanced equation: For reaction (a): - NaOH(aq): -469.15 kJ/mol - HCl(aq): -167.15 kJ/mol - NaCl(aq): -385.90 kJ/mol - H2O(l): -285.83 kJ/mol For reaction (b): - NaOH(aq): -469.15 kJ/mol - HF(aq): -320.1 kJ/mol (given) - NaF(aq): -288.61 kJ/mol - H2O(l): -285.83 kJ/mol
03

Calculate ΔH for each reaction using Hess's Law

According to Hess's Law, for each reaction, we can calculate ΔH as the difference between heats of formation of the products and reactants. For reaction (a): ΔH = [(-385.90 + -285.83 ) - (-469.15 + -167.15)] kJ/mol ΔH = (-671.73 + 636.30) kJ/mol ΔH = -35.43 kJ/mol For reaction (b): ΔH = [(-288.61 + -285.83 ) - (-469.15 + -320.1)] kJ/mol ΔH = (-574.44 + 789.25) kJ/mol ΔH = 214.81 kJ/mol
04

Calculate ΔH for the given volumes and concentrations

Given the concentrations and volumes for each reaction: Reaction (a): Concentration of NaOH = 0.100 M Volume of NaOH = 1.00 L Concentration of HCl = 0.100 M Volume of HCl = 1.00 L Number of moles of NaOH = 0.100 * 1.00 = 0.100 mol Number of moles of HCl = 0.100 * 1.00 = 0.100 mol ΔH = -35.43 kJ/mol * 0.100 mol = -3.543 kJ Reaction (b): Concentration of NaOH = 0.100 M Volume of NaOH = 1.00 L Concentration of HF = 0.100 M Volume of HF = 1.00 L Number of moles of NaOH = 0.100 * 1.00 = 0.100 mol Number of moles of HF = 0.100 * 1.00 = 0.100 mol ΔH = 214.81 kJ/mol * 0.100 mol = 21.481 kJ Thus, ΔH for reaction (a) is -3.543 kJ ΔH for reaction (b) is 21.481 kJ

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