Question

Problem: Solve for the following enthalpy changes 1) When 5.97 g NaOH(s) is reacted with 200.0 mL of HCl(aq), the temperature increases 17.85 Celsius degrees NaOH(s) + H+ (aq) --> Na(aq) + H2O() AH = ? 2) When 3.18 g NaOH(s) is dissolved in 200.0 mL of water, the temperature increases 4.11 Celsius degrees. NaOH(s) --> Na*(aq) + OH"(aq) AH = ?

Answer 1

Q1. mass NaOH = 5.97 g

moles NaOH = (mass NaOH) / (molar mass NaOH)

moles NaOH = (5.97 g) / (40.0 g/mol)

moles NaOH = 0.14925 mol

volume HCl = 200.0 mL

mass HCl = (volume HCl) * (density HCl)

mass HCl = (200.0 mL) * (1.00 g/mL)

mass HCl = 200.0 g

mass of solution = (mass HCl) + (mass NaOH)

mass of solution = (200.0 g) + (5.97 g)

mass of solution = 205.97 g

Heat change of solution = (mass of solution) * (specific heat solution) * (temperature rise)

Heat change of solution = (205.97 g) * (4.184 J/g.^oC) * (17.85 ^oC)

Heat change of solution = 15382.7 J

Heat change of solution = 15.4 kJ

heat change of reaction = -(Heat change of solution)

heat change of reaction = -(15.4 kJ)

heat change of reaction = -15.4 kJ

$\Delta$H = (heat change of reaction) / (moles NaOH)

$\Delta$H = (-15.4 kJ) / (0.14925 mol)

$\Delta$H = -103 kJ/mol

Q2.

mass NaOH = 3.18 g

moles NaOH = (mass NaOH) / (molar mass NaOH)

moles NaOH = (3.18 g) / (40.0 g/mol)

moles NaOH = 0.0795 mol

volume water = 200.0 mL

mass water = (volume water) * (density water)

mass water = (200.0 mL) * (1.00 g/mL)

mass water = 200.0 g

mass of solution = (mass water) + (mass NaOH)

mass of solution = (200.0 g) + (3.18 g)

mass of solution = 203.18 g

Heat change of solution = (mass of solution) * (specific heat solution) * (temperature rise)

Heat change of solution = (203.18 g) * (4.184 J/g.^oC) * (4.11 ^oC)

Heat change of solution = 3493.9 J

Heat change of solution = 3.494 kJ

heat change of reaction = -(Heat change of solution)

heat change of reaction = -(3.494 kJ)

heat change of reaction = -3.494 kJ

$\Delta$H = (heat change of reaction) / (moles NaOH)

$\Delta$H = (-3.494 kJ) / (0.0795 mol)

$\Delta$H = -43.95 kJ/mol