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Experiment 12: Enthalpy of Neutralization, Lab Reports of Chemistry

Northern State UniversityChemistry

Enthalpy is sum of the internal energy added to the product of the pressure and volume of the system

Typology: Lab Reports

2020/2021

Uploaded on 05/11/2021

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Experiment*#12.*Enthalpy*of*Neutralization*
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Introduction*
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!In!the!course!of!most!physical!processes!and!chemical!reactions!there!is!a!change!in!energy.!!In!chemistry!what!
is!normally!measured!is!ΔH!(enthalpy!change),!the!change!in!heat!at!constant!pressure!and!ignoring!any!work!done!by!
the!reacting!system.!!If!the!reaction!is!exothermic,!heat!is!given!off!and!ΔH!has!a!negative!value.!!When!heat!is!absorbed!
from!the!surroundings!during!the!reaction,!the!reaction!is!endothermic!and!ΔH!has!a!positive!value.!!The!change!in!
enthalpy!is!a!measure!of!the!difference!in!energy!between!the!bonds!in!the!products!and!the!bonds!in!the!reactants.!!
The!absolute!energy!of!a!compound!cannot!be!measured!directly,!but!the!change!in!enthalpy!that!occurs!during!a!
chemical!reaction!can!be!measured.!!*
!
The!reaction!in!today’s!experiment!is!acid!/!base!neutralization!which!is!an!exothermic!process.!!!
!
!!HCl*(aq)****+*****NaOH*(aq)**→*****H2O*(l)***+****NaCl*(aq)*****+*****heat*
!
The!heat!released!by!the!reaction!will!be!absorbed!by!the!surroundings!(aqueous!solution).!!Coffee!Cup!
Calorimetry!will!be!employed!to!determine!the!amount!of!heat!lost!by!the!reaction!and!gained!by!the!salt!water!
solution.!!A!calorimeter!is!simply!a!container!used!to!measure!the!heat!change.!!Coffee!Cup!Calorimetry!just!means!that!
we!will!be!measuring!heat!at!constant!pressure,!ΔH.!!The!heat!lost!by!the!reaction!will!actually!be!transferred!to!both!
the!salt!water!and!its!surroundings!(the!calorimeter.)!!The!heat!capacity!of!the!calorimeter!(in!this!case,!two!Styrofoam!
cups)!usually!would!be!calculated!first;!however,!we!have!found!that!the!heat!capacity!of!the!cups!is!so!small!that!it#can#
be#neglected.!!!
!
As!the!First!Law!of!Thermodynamics!applies,!!
*
Heat!(q)!lost!by!the!reaction!+!heat!(q)!gained!by!the!solution!=!0!!!!
or!**qrxn*=*–*qsoln!
!
To!find!the!heat!lost!by!the!reaction!we!use:!
!
!q*=*m**x**Cs**x**ΔT!! ! where:! !m!=!mass!of!the!solution!(The!density!of!the!solution!is!1.04!g/mL)!
!!!!!!Cs!=!specific!heat!capacity!of!the!solution!(3.87!J/g°C)!
!!!!!!ΔT=!change!in!temperature!of!the!solution!(Tfinal!–!Tinitial)!
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Experiment #12. Enthalpy of Neutralization

Introduction In the course of most physical processes and chemical reactions there is a change in energy. In chemistry what is normally measured is ΔH ( enthalpy change), the change in heat at constant pressure and ignoring any work done by the reacting system. If the reaction is exothermic, heat is given off and ΔH has a negative value. When heat is absorbed from the surroundings during the reaction, the reaction is endothermic and ΔH has a positive value. The change in enthalpy is a measure of the difference in energy between the bonds in the products and the bonds in the reactants. The absolute energy of a compound cannot be measured directly, but the change in enthalpy that occurs during a chemical reaction can be measured. The reaction in today’s experiment is acid / base neutralization which is an exothermic process. HCl (aq) + NaOH (aq) → H 2 O (l) + NaCl (aq) + heat The heat released by the reaction will be absorbed by the surroundings (aqueous solution). Coffee Cup Calorimetry will be employed to determine the amount of heat lost by the reaction and gained by the salt water solution. A calorimeter is simply a container used to measure the heat change. Coffee Cup Calorimetry just means that we will be measuring heat at constant pressure, ΔH. The heat lost by the reaction will actually be transferred to both the salt water and its surroundings (the calorimeter.) The heat capacity of the calorimeter (in this case, two Styrofoam cups) usually would be calculated first; however, we have found that the heat capacity of the cups is so small that it can be neglected. As the First Law of Thermodynamics applies, Heat (q) lost by the reaction + heat (q) gained by the solution = 0 or qrxn = – qsoln To find the heat lost by the reaction we use: q = m x Cs x ΔT where: m = mass of the solution (The density of the solution is 1.04 g/mL) Cs = specific heat capacity of the solution (3.87 J/g°C) ΔT= change in temperature of the solution (Tfinal – Tinitial)

Equipment 2 Styrofoam cups (2) Thermometers 50 or 100 mL graduated cylinder Stirring rod 150 mL beaker Chemicals 2.00 M HCl 2.00 M NaOH Procedure

  1. Rinse and dry the Styrofoam cups (calorimeter). Place one cup inside the other. Measure out 25.0 mL of 2.00 M NaOH and pour it into the calorimeter. In a clean dry 150 mL beaker, take 25.0 mL of 2.00 M HCl.
  2. Determine the temperature of both the acid and the base to the nearest 0.1°C. Average the temperatures. Record this average as the initial temperature Tinitial.
  3. Pour the acid into the base quickly and carefully with gentle stirring. Start monitoring the temperature as soon as the two are mixed. Continue to stir and monitor the temperature. Record the maximum temperature that the solution reaches. This is Tfinal
  4. Repeat Steps 1 -­‐ 3 (above) for a second trial. Disposal All reactants and products may be disposed of into the sink.

Heat of Neutralization: Data Sheet Name ________________________ Calculations (Show all your work completely and clearly. Include proper units)

1. Calculate the mass of the solution for each trial (given the density of the solution is 1.04 g/ml) 2. Calculate the heat absorbed by the solution in each trial.(qsoln) qsoln (Trial 1) ____________ qsoln (Trial 2) _____________ Average qsoln _________ 3. Calculate q (^) rxn = ____________________________ 4. Calculate the number of moles of water formed (This is the same as moles of HCl): moles of water = moles of HCl = __________________ Trial 1 Trial 2 Volume of 2.00 M HCl Volume of 2.00 M NaOH Volume of solution T (^) initial T (^) final ΔT (T (^) final — T (^) initial) Specific heat capacity of NaCl solution (^) 3.87 J/g °C 3.87 J/g °C