Purpose
To demonstrate that reactions can either give off heat (exothermic) or absorb heat (endothermic), emphasizing the concepts of systems and surroundings.
Materials
- Barium hydroxide octahydrate (32 g)
- Ammonium chloride (11 g)
- Small beaker
- Stir rod
- Black temperature probe (for temperatures < 0°C)
- Piece of wood
- Water (for freezing the beaker)
Procedure
Exothermic Reaction Demonstration
- Introduction:
- Focus on heat flow and perform an exothermic reaction to show heat transfer to the environment.
- Thermite, Hydrogen Balloon, adding NaOH to water.
Endothermic Reaction: Barium Hydroxide and Ammonium Chloride
- Reaction Setup:
- Mix 32 g of barium hydroxide octahydrate with 11 g of ammonium chloride in a small beaker using a stir rod.
- Conduct this demonstration on a document camera for visibility.
- Observations:
- Encourage students to make observations, which may include:
- The odor of ammonia.
- Development of a liquid phase.
- Frosting of the glass.
- Freezing the Beaker:
- Drip a small amount of water onto a piece of wood.
- Place the beaker on the drop of water for a few seconds. The beaker will freeze solidly to the wood, allowing it to be picked up.
- Caution: Moving the beaker prematurely will ruin the experiment.
- Discussion:
- Use the waiting time to write the reaction equation on the board and explain the observations.
Chemical Equations
- Exothermic Dissociation of Water and Sodium Hydroxide:
NaOH (s) → Na+ (aq) + OH− (aq) - Endothermic Reaction Equation:
Ba(OH)2⋅8 H2O (s) + 2 NH4Cl (s) → BaCl2⋅2 H2O (s) + 2 NH3 (g) + 8 H2O (l)
Thermodynamic Calculations
Change in Enthalpy (ΔH∘):
ΔHº = { [ 1 mol • ΔHfº(BaCl2• 2H2O)(s) ] + [ 2 mol • ΔHfº(NH3(aq)) ] + [ 8 mol • ΔHfº(H2O(l))]} – {[1 mol ΔHfº(Ba(OH)2• 8H2O(s))] + [2 mol • ΔHfº(NH4Cl)(aq) ] }
ΔHº = {[1 mol • -1460.1 kJ/mol] + [2 mol • -80.29 kJ/mol] + [8 mol • -285.83 kJ/mol]} – {[1 mol • -3342 kJ/mol] + [2 mol • -314.4 kJ/mol]}
ΔHº = 63 kJ
Change in Entropy (ΔSº):
ΔSº = {[1 mol • Sº (BaCl2• 2H2O(aq))] + [2 mol • Sº (NH3(aq))] + [8 mol • Sº (H2O(l))]} – {[1 mol • Sº (Ba(OH)2• 8H2O(s))] + [2 mol • Sº (NH4Cl(s)) ]}
ΔSº = {[1 mol • 203 J/mol•K ]+ [2 mol • 111 J/mol•K] + [8 mol • 69.91 J/mol•K ]}- {[1 mol • 427 J/mol•K ] +[ 2 mol • 94.6 J/mol•K ]}
ΔSº = 368 J/K
Change in Universal Entropy (ΔSºuniv):
ΔSºuniv = ΔSº – ΔHº/T
ΔSºuniv = [ (368 J/K – 63 kJ)/(273.15 + 25)K • [1000 J/1 kJ] = 157 J/K > 0
Comparison to Ice Melting
- Note that when ice melts to liquid water, there is a similar increase in entropy that offsets its increase in enthalpy:
H2O (s) → H2O (l) - Thus, this reaction is also product-favored at 25 ºC.
Conclusion
This experiment effectively illustrates the concepts of exothermic and endothermic reactions, demonstrating how reactions can either release or absorb heat, and emphasizes the importance of understanding systems and surroundings.