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Calorimetry

Calorimetry. The enthalpy change associated with a chemical reaction or process and the specific heat of a substance can be measured experimentally using calorimetry . The experimental measurement of the amount of heat gained or lost Determined by measuring the temperature change that occurs.

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Calorimetry

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  1. Calorimetry • The enthalpy change associated with a chemical reaction or process and the specific heat of a substance can be measured experimentally using calorimetry. • The experimental measurement of the amount of heat gained or lost • Determined by measuring the temperature change that occurs. • Calorimeter: • An instrument used to measure the heat gained or lost

  2. Calorimetry Two types of calorimetry are commonly used: Constant-pressure Calorimetry Specific heat DHrxn or DHsoln Bomb Calorimetry (Constant Volume Calorimetry) DHcombustion

  3. Calorimetry In constant-pressure calorimetry, Measurements are made in an “open” container. Reactions occur at constant pressure Calorimeter is insulated Assume that the amount of heat gained from or lost to the surroundings is negligible any heat gained or lost during a chemical reaction or process comes from or goes into the solution being studied.

  4. Cs = qsubst mass x DT DH = qsubst moles subst. Calorimetry • In order to measure the specific heat of a substance, we need to know three values: • qsubstance • mass • DT • In order to measure a molar enthalpy change for a reaction, we need to know two values: • qsubstance or qreactant • moles of substance or reactant

  5. Calorimetry • In both cases, the “unknown” that must be determined experimentally is the amount of heat gained or lost by the substance being studied (i.e. qsubstance). • In constant-pressure calorimetry, qsubstance is determined indirectly by measuring the heat gained or lost by the liquid present in the calorimetry (whose specific heat is known).

  6. Calorimetry • We then apply the First Law of Thermodynamics: • If heat is lost by the chemicals (or object) during a reaction or process, then it must be gained by the solution (liquid) and vice versa. • The heat gained or lost by the reactants (or object) and the solution (liquid) are equal in magnitude but opposite in sign. qobject = - qsoln or qrxn = -qsoln

  7. Calorimetry Calorimetry Simulation: Measuring the specific heat of a metal How is the experiment performed? What kind of data is collected? How are the data used to determine specific heat? http://group.chem.iastate.edu/Greenbowe/sections/projectfolder/animationsindex.htm

  8. Calorimetry Determining specific heat or molar heat capacity experimentally: Heat an object with a known mass to 95-100oC. Measure initial temperature of object. Place known mass of water into calorimeter and measure its temperature. Add hot object to the water Measure the equilibrium temperature (i.e. final temperature) of the water and object.

  9. Calorimetry Types of data obtained from or used in a calorimetry experiment often include: Sample Data Mass of sample (object or compound) Tinitial (sample) Tfinal (sample) Solution Data Mass of solution Tinitial (solution) Tfinal (solution) Csoln Often use specific heat of water if water is the solvent

  10. Calorimetry Using the data to find specific heat: Calculate the amount of heat gained by the water: qwater = Cwater x mwater x DTwater Calculate the amount of heat lost by the object qobj = - qwater Calculate specific heat of object Cs = qobj mobj x DTobj

  11. Calorimetry Example: A 55.0 g piece of aluminum metal at 100.0oC was added to 51.3 g of water at 20.0oC. The equilibrium temperature of the system was 35.0oC. If the specific heat of water is 4.18 J/g.K, what is the specific heat of aluminum?

  12. Calorimetry

  13. Calorimetry This is the same procedure you will use to calculate the specific heat of the metal sample you use in the lab. Follow this procedure to do the calculations for Expt. 2. You should expect a similar problem on your exam!

  14. Calorimetry Calorimetry Simulation: Measuring the molar heat of solution (DHsoln) How is the experiment performed? What kind of data is collected? How are the data used to determine molar heat of solution?

  15. Calorimetry Determining molar heat of solution (DHsoln per mole of solute) from calorimetry data. Place known mass of water into calorimeter. Measure the initial temperature of the water. Add known mass of solute to water. Record equilibrium temperature of the resulting solution.

  16. Calorimetry Types of data collected when measuring molar heat of solution: Mass of water Initial temperature of water Mass of solute Final temperature of solution

  17. Calorimetry Determining DHsoln per mole of solute from calorimetry data. Assume Csoln = Cwater = 4.18 J/g-K unless told otherwise Calculate the heat gained or lost by the entire solution qsoln = Csoln x mass of solution x DTsoln Mass of solution = mass of H2O + mass of solute

  18. Calorimetry Calculate heat gained or lost by the solute (salt) qsolute = - qsoln Calculate molar heat of solution DHsoln per mole = qsolute mole solute

  19. Calorimetry Example:When a 2.125 g sample of solid ammonium nitrate dissolves in 30.000 g of water in a constant pressure calorimeter, the temperature drops from 22.0oCto 16.9oC. Calculate the molar DHsoln (in kJ/mole) for the dissolution process: NH4NO3 (s)  NH4+ (aq) + NO3- (aq) Assume that the specific heat of the solution is the same as pure water (4.18 J/g.K).

  20. Calorimetry

  21. Calorimetry You should expect a calorimetry problem similar to this on your exam. What would you actually observe when the dissolution process is exothermic? What would you actually observe when the dissolution process is endothermic?

  22. Calorimetry For an exothermic process, the heat produced causes the temperature of the solution to increase. (DTsoln >0) Chemical particle q q q q Heat released by chemical particle q q q

  23. Calorimetry For an endothermic process, the heat gained comes from the reaction mixture and causes the temperature of the solution to decrease (DTsoln < 0) q Chemical particle q q q Heat gained by chemical particle from the rxn mixture q q

  24. Calorimetry Bomb Calorimetry Constant Volume Calorimetry Used to study combustion reactions Measure DHcombustion

  25. Calorimetry As combustion occurs, Heat is released Heat is absorbed by the calorimeter and its contents Temperature of calorimeter & contents increases. The change in temperature of the calorimeter and its contents can be used to determine the heat of combustion.

  26. Calorimetry To calculate the molar heat of combustion from a calorimetry experiment: Calculate the heat absorbed by the calorimeter and its contents using the heat capacity of the calorimeter: qcal = Ccal x DT (Notice that you do not need a mass term because heat capacity has units of J/K) Calculate the heat lost by the reactants: qrxn = -qcal

  27. Calorimetry Calculate molar heat of combustion (DHcomb per mole reactant) DHcomb = qrxn mole reactant

  28. Calorimetry Example: When 2.00 g of methylhydrazine (CH6N2) is burned in a bomb calorimeter, the temperature of the calorimeter increases from 25.00oC to 32.25oC. If the heat capacity of the calorimeter is 7.794 kJ/oC, what is the molar heat of combustion for CH6N2? 2 CH6N2 (l) + 5 O2 (g)  2 N2 (g) + 2 CO2(g) + 6 H2O (g)

  29. Calorimetry

  30. Hess’s Law The heats of reaction (DHrxn) have been measured and tabulated for many chemical reactions. There are two approaches to determining the heat of reaction for a particular chemical reaction: Calorimetry Use tabulated DHrxn to calculate the heat of reaction for another reaction of interest

  31. Hess’s Law The enthalpy change for a reaction or process is a state function: Depends only on the amount of reactants and products used/formed and on their physical state Does not depend on how the reaction was done. one step vs. multiple steps

  32. Hess’s Law Hess’s Law: If a reaction is carried out in a series of steps, DH for the overall (one-step) reaction is equal to the sum of the DH’s for the individual steps.

  33. Hess’s Law IMPORTANT: When applying Hess’s Law, if you need to multiply or divide the coefficients of an equation by a number then the DH must also be multiplied by the same number DH depends on amount of material H2O (g)  H2O (l) DH = - 44 kJ 2H2O (g) 2H2O (l)DH = 2 (- 44 kJ) = - 88 kJ

  34. Hess’s Law IMPORTANT: If a reaction has to be reversed, the magnitude of DH stays the same but the sign must be reversed. H2O (g) H2O (l)DH = - 44 kJ H2O (l) H2O (g)DH = + 44 kJ

  35. Hess’s Law Example: Calculate the DHrxn for the incomplete combustion of C forming CO: C (s) + ½ O2 (g)  CO (g) DH = ??? given the following reactions: C (s) + O2 (g)  CO2 (g) DH = - 393.5 kJ 2 CO (g) + O2 (g)  2 CO2 (g) DH = - 566.0 kJ

  36. Hess’s Law

  37. Hess’s Law Example: Calculate DHrxn for NO (g) + O (g)  NO2 (g) DH = ??? using the following thermochemical equations. NO (g) + O3 (g)  NO2 (g) + O2 (g) DH = -198.9 kJ O3 (g)  3/2 O2 (g) DH = -142.3 kJ O2 (g)  2 O (g) DH = 495.0 kJ

  38. Hess’s Law

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