Lab.6. Thermodynamics, Calorimetry
Key words:
Heat, energy, exothermic & endothermic reaction, calorimeter, calorimetry, enthalpy of reaction, specific heat, chemical & physical change, enthalpy of neutralization, law of conservation of energy, final temperature, initial temperature, lattice energy, hydratation energy, enthalpy of solution
Literature:
J. A. Beran; Laboratory Manual for Principles of General Chemistry, pp. 245-256.
J.E. Brady, F. Senese: Chemistry – Matter and its Changes, 4th ed. Wiley 2003, Chapters 7, 20, .
M. Hein and S. Arena: Introduction to Chemistry, 13th ed. Wiley 2011; pp. 157 - 161
J. Crowe. T. Bradshaw, P. Monk, Chemistry for the Biosciences. The essential concepts., Oxford University Press, 2006; pp. 416 - 450.
J. Brady, N. Jespersen, A. Hysop, Chemistry, International Student Version, 7thed. Wiley, 2015, Chapter 18, 855 – 890.
Theoretical background
Accompanying all chemical and physical changes is a transfer of heat (energy); heat may be either evolved (exothermic) or absorbed (endothermic). A calorimeter (Fig. 1) is the laboratory apparatus that is used to measure the quantity and direction of heat flow accompanying a chemical or physical change. The calorimeter is well-insulated so that, ideally, no heat enters or leaves the calorimeter from the surroundings. For this reason, any heat liberated by the reaction or process being studied must be picked up by the calorimeter and other substances in the calorimeter. The heat change in chemical reactions is quantitatively expressed as the enthalpy (or heat) of reaction, H, at constant pressure. H values are negative for exothermic reactions and positive for endothermic reactions. Lab.6. Thermodynamics, Calorimetry
Fig. 1. A set nested coffee cups is a good constant pressure calorimeter.
Two quantitative measurements of heat are detailed in his experiment: measurements of the heat accompanying an acid-base reaction, and the heat associated with dissolution of a salt in water.
Specific Heat
The energy (heat, expressed in joules, J) required to change the temperature of one gram of a substance by 1⁰C is the specific heat of that substance:
J energy(J ) specific heat = (1) g • C mass (g) T (C) or, rearranging for energy,
J energy (J ) = specific heat mass (g) T (C) (2) g • C
T is the temperature change of the substance. Although the specific heat of a substance changes slightly with temperature, for our purposes, we assume it is constant over the temperature changes of this experiment.
Enthalpy (Heat) of neutralization of an Acid-Base Reaction
The reaction of a strong acid with a strong base is an exothermic reaction that produces water and heat as products. Lab.6. Thermodynamics, Calorimetry
Definition: Enthalpy of neutralization: energy released per mole of water formed in an acid- base reaction-an exothermic quantity.
+ − H 3O (aq) + OH (aq) → 2 H 2O (l) + heat
The enthalpy (heat) of neutralization, Hn , is determined by (1) assuming the density and specific heats of the acid and base solutions are equal to that of water and (2) measuring the temperature change, T , when the two are mixed.
H = − specific heat combined masses T (3) n H 2O acid +base
Hn is generally expressed in units of kJ/mol of acid (or base) reacted. The mass (grams) of the solution equals the combined masses of the acid and base solutions.
Enthalpy (Heat) of Solution for the Dissolution of a Salt
Lattice energy: energy required to vaporize one mole of salt into its gaseous ions-an endothermic quantity.
Hydratation energy: energy released when one mole of a gaseous ion is attracted to and surrounded by water molecules forming one mole of hydrated ion in aqueous solution-an exothermic quantity.
When a salt dissolves in water, energy is either absorbed or evolved depending upon the magnitude of the salt’s lattice energy and the hydration energy of its ions. For the dissolution of KI:
H2O + − KI (s) ⎯ ⎯⎯ → K (aq) + I (aq) Hs =+13 kJ/mol
The lattice energy (an endothermic quantity) of a salt, HLE, and the hydration energy (an exothermic quantity), HHyd, of its composite ions account for the amount of heat evolved or absorbed when one mole of the salt dissolves in water. The enthalpy (heat) of solution, Hs, is the sum of two terms (Fig. 2):
H s = H LE + H Hyd (4)
Whereas HLE and HHyd are difficult to measure in the laboratory, Hs is easily measured. Lab.6. Thermodynamics, Calorimetry
Fig.2. Energy changes in the dissolving of solid KI in water.
The enthalpy of solution for the dissolution of a salt, Hs, is determined experimentally by adding the heat changes of the salt and water when the two are mixed. Hs is expressed in units of kilojoules per mole of salt.
H = (− heat change )+ (heat change ) (5) s H 2O salt
specific heat mass T specific heat mass T H = − H2O H2O H2O + − salt salt salt (6) S mole salt molesalt
A temperature rise for the dissolution of a salt, indicating an exothermic process, means that the HHyd is greater than the HLE for the salt; conversely, a temperature decrease in the dissolution of the salt indicates that HLE is greater than HHyd and Hs is positive.
Lab.6. Thermodynamics, Calorimetry
Table 1. Specific Heat of Some Salts
Salt Formula Specific Heat (J/g•⁰C)
Ammonium chloride NH4Cl 1.57
Ammonium nitrate NH4NO3 1.74
Ammonium sulfate (NH4)2SO4 1.409
Calcium chloride CaCl2 0.657
Sodium carbonate Na2CO3 1.06
Sodium hydroxide NaOH 1.49
Sodium sulfate Na2SO4 0.903
Sodium thiosulfate pentahydrate NaS2O3•5H2O 1.45
Potassium bromide KBr 0.439
Potassium nitrate KNO3 0.95
Plot the Data.
Plot the temperature ( y axis) versus time ( x axis) on the top half of a sheet of linear graph paper. The maximum temperature is the intersection point of two lines: (1) the best line drawn through the data points on the cooling portion of the curve and (2) a line drawn perpendicular to the time axis at the mixing time [when the acid is added to the base or when a salt is added to the water]. (Fig.3).
Lab.6. Thermodynamics, Calorimetry
Fig. 3. Extrapolation of temperature vs. time data (not scale) for an exothermic process.
EXPERIMENTAL PART
PART 1
Enthalpy (Heat) of Neutralization for an Acid-Base Reaction
Calculate the enthalpy of neutralization reaction:
q = m∙ c ∙ΔT = m∙ c∙ ( Tf - Ti ) ∆H = q /n, where c - specific heat of mixture, n – moles of water created
Step 1
Measure 50.0 mL of 1.1 M HCl in a clean, graduated cylinder. Measure temperature with temperature sensor and record it in Table 2.
Step 2
Using another clean, graduated cylinder transfer 50.0 mL of a standard 1.0 M NaOH solution to the dry calorimeter. Record the temperature and exact molar concentration of the NaOH solution in Table 2. Lab.6. Thermodynamics, Calorimetry
Step 3
Carefully, but quickly add the acid to the base, replace the calorimeter lid, and swirl gently. Read and record the temperature and time every 5 seconds for 1 minute and thereafter every 15 seconds for other 2 minutes.
Step 4
Calculate the thermodynamic data and fill the proper places in Table 3.
Step 5
Repeat steps 1-4, for 1.1 M HNO3.
Table 2. Data for acid–base neutralization
HCl + NaOH HNO + NaOH 3 Trial 1 Trial 2 Trial 1 Trial 2 1. Volume of acid (mL) 2. Temperature of acid ⁰C 3. Volume of NaOH (mL) 4. Temperature of NaOH (⁰C) 5. Exact molar concentration of NaOH (mol/L) 6. Maximum temperature from graph( ⁰C)
Table 3. Calculation for Enthalpy (Heat) of Neutralization for an Acid-Base Reaction
HCl + NaOH HNO3 + NaOH Trial Trial Trial 1 Trial 2 1 2 Average initial temperature of acid and base ( ⁰C )
Temperature change, T ( ⁰C )
Volume of final mixture (mL)
Mass of final mixture (g) (Assume the density of the solution is 1.0 g/mL)
Specific heat of mixture 4.18J/g∙◦C 4.18J/g∙◦C
Heat evolved (J)
Amount of OH- reacted, the limiting reactant (mol) Lab.6. Thermodynamics, Calorimetry
Amount of H2O formed (mol)
Heat evolved per mole of H2O, ∆Hn (kJ/mol H2O)
Average ∆Hn (kJ/mol H2O)
Answer to the question: is the enthalpy different for both neutralization reactions, whether or is the same (in frame of experimental error)?
Comment on your answer.
Disposal: Rinse the calorimeter twice with deionized water beforewater before next step
PART 2.
Enthalpy (Heat) of Solution for the Dissolution of a Salt
Calculation for Enthalpy:
specific heat mass T specific heat mass T H = − H2O H2O H2O + − salt salt salt S mole salt molesalt
Step 6.
Prepare the salt. On weighing paper, measure about 5.0 g of the assigned salt. Record the name of the salt in Table 4.
Step 7
Using your clean, graduated cylinder, add 20 g (20.0 mL) of deionized water to the calorimeter and record its temperature in Table 4. Secure the thermal sensor below the water surface.
Lab.6. Thermodynamics, Calorimetry
Step 8
Carefully add (do not spill) the salt to the calorimeter, replace the lid, and swirl gently. Read and record the temperature and time at 5-second intervals for 1 min and thereafter every 15 seconds for 2 min. Record the data.
Step 9
Calculate thermodynamic data for salt and fill the Table 4.
Step 10
Repeat steps 6-9, for other salt assigned by TA.
Disposal: Discard the salt solution into the “Waste Salts” container, followed by additional tap water.
Table 3. Enthalpy (Heat) of solution for the Dissolution of a Salt
Name of the salt ......
Mass of salt (g)
Moles of salt
Mass of water (g)
Initial temperature of water ( ⁰C )
Final temperature of mixture
Table 4. Calculation for Enthalpy (Heat of Solution) for the Dissolution of a Salt
Name of salt ......
Change in temperature of solution
Heat loosed or gained by water (J)
Heat loosed or gained by salt (J)
Total enthalpy change
∆Hs (J/mol salt) Lab.6. Thermodynamics, Calorimetry
Average ∆Hs (J/mol salt)
Additional information
Table 1. Specific Heat of Some Salts
Salt Formula Specific Heat (J/g•⁰C)
Ammonium chloride NH4Cl 1.57
Ammonium nitrate NH4NO3 1.74
Ammonium sulfate (NH4)2SO4 1.409
Calcium chloride CaCl2 0.657
Sodium carbonate Na2CO3 1.06
Sodium hydroxide NaOH 1.49
Sodium sulfate Na2SO4 0.903
Sodium thiosulfate pentahydrate NaS2O3•5H2O 1.45
Potassium bromide KBr 0.439
Potassium nitrate KNO3 0.95