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Calorimetry: Measuring Heats of Reaction D rxn H

Calorimetry: Measuring Heats of Reaction D rxn H. we cannot measure absolute enthalpies, only differences in enthalpy or enthalpy changes.

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Calorimetry: Measuring Heats of Reaction D rxn H

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  1. Calorimetry: Measuring Heats of Reaction DrxnH • we cannot measure absolute enthalpies, only differences in enthalpy or enthalpy changes • calorimetry involves carrying out a reaction in carefully constructed “surroundings” that retain the heat and measure the temperature change. Then we relate the quantity of heat released (or absorbed) to the temperature change. First we need to define a physical property of substances called heat capacity. Heat Capacity • the more heat an object absorbs the hotter it gets, or the heat (q) absorbed by an object is proportional to its temperature change, • this constant is called the heat capacity, C, which is the amount of heat required to change an objects temperature by 1 K

  2. q = the heat absorbed or given off • since q has units of J and DT has units of K, C has units of J K-1 Specific Heat Capacity • a related property is the specific heat capacity, s (or c in some texts), which is the quantity of heat required to change the temperature of 1 g of a substance by 1 K. where m is the mass of the substance and s has units of J g-1 K-1

  3. notice from the two equations involving the calculation of heat that when an object gets hotter, it absorbs heat, the final temperature is greater so DT is positive and q is positive. • when the object gives off heat to its surroundings, the final temperature is less than the initial temperature and q is negative.

  4. eg. A layer of copper welded to the bottom of a skillet weighs 125 g. a) How much heat is required to raise the temperature of the copper layer from 10 oC to 100 oC? The specific heat capacity of Cu is 0.387 J g-1 K-1. b) Compare this with the amount of heat required to raise the temperature of the same amount water from 10 oC to 100 oC.

  5. Water has an unusually high specific heat capacity, 4.184 J g-1 K-1. This is almost 6 times as high as the specific heat of rock (0.7 J g-1 K-1). If there was no water on earth temperatures during the day would be very high due to the sun and at night temperatures would plummet as the energy is released.

  6. Calorimetry • the calorimeter is a device used to measure the heat released or absorbed by a physical or chemical process. • there are two common types of calorimetry. The first is a constant-pressure calorimeter which can be constructed from a simple Styrofoam cup. • for a process where the system gives off heat, the heat lost by the system equals the heat gained by the surroundings, in this case the calorimeter, • this comes from the first law of thermodynamics or the law of conservation of energy

  7. if we assume that all of the heat is absorbed by the water and our system is, for example, a hot piece of metal, • subbing in the equation for q involving specific heat from before, • and we can measure or look up the values in this equation and try to determine, for example, the specific heat capacity of the metal

  8. eg. A 24.64 g sample of a solid was heated in a test tube to 100.00 oC in boiling water and carefully added to a Styrofoam cup calorimeter containing 50.00 g of water. The water temperature increased from 25.10 oC to 28.49 oC. What is the specific heat of the solid. (In this example, we assume that all the heat from the solid is absorbed by the water!)

  9. eg. A sheet of gold (s=0.129 J g-1 K-1) weighing 10.0 g and at a temperature of 18.0 oC is placed flat on a sheet of iron (s=0.444 J g-1 K-1) weighing 20.0 g and at a temperature of 55.6 oC. Assuming no heat is lost to the surroundings, what is the final temperature of the combined metals?

  10. Constant Volume Calorimetry: The Bomb Calorimeter • used to determine very accurate heats of combustion • in order to get very accurate values, it is necessary to calibrate the calorimeter

  11. in a previous slide it was stated “assume that the water absorbed all of the heat.” In fact this is not really valid because other parts of the calorimeter like the cup, thermometer, stirrer etc. also must absorb heat. • as well, since the calorimeter is not perfectly insulated, heat is lost to the outside air and is not accounted for. • using only the specific heat capacity of water and mass of water would be an error, therefore, it is necessary to “calibrate” the calorimeter, by using something with a known heat capacity and determining the heat capacity of the calorimeter.

  12. eg. A manufacturer claims that its new dietetic dessert has “fewer than 10 Calories” per serving” To test the claim, a chemist at “MarketPlace” places one serving in a bomb calorimeter and burns it in O2. The heat capacity of the calorimeter was previously determined to be 8.151 kJ K-1. The temperature increases 4.937 oC. Is the manufacturer’s claim correct?

  13. Note that previously we determined that • at constant volume, the heat (released or absorbed) is equal to DU. To obtain DH we must add a correction factor of PDV. Although the volume doesn’t change in a bomb calorimeter, if there is a change in the number of moles of gas, then we say work has been done (by the surroundings) so we must add DnRT, where Dn is the change in the number of moles of gas. • Unfortunately for an example like the previous one, with a complex sample it is virtually impossible to know the change in the moles of gas. Recall, that even when the number of moles of gas changes, DU is very close to DH.

  14. eg. 1.023 g of gasoline, octane, was burned in excess oxygen in a bomb calorimeter. The temperature of two litres of water increased by 5.611 oC. The heat capacity of the calorimeter was determined in a previous experiment to be 8.371 kJ oC-1. Determine the enthalpy or heat of combustion for one mole of octane at 298 K using these experimental data.

  15. Enthalpy Changes Accompanying a Change in State of Matter

  16. A 10. g ice cube at 0 oC is added to a cup of coffee (100. g) at 57.6 oC. Given that the heat of fusion of water is 6.02 kJ mol-1, what is the final temperature of the liquid in the cup?

  17. ? Stoichiometry and Thermochemical Equations • a thermochemical equation is a balanced equation that also states the heat of reaction, DrH exothermic • when writing the thermochemical equation the DrH value is specific for the states of matter in the specific equation for example • if an exothermic reaction is written in reverse, the sign of DrH becomes positive endothermic • DrH is also proportional to the amount of substance reacting,

  18. eg. The major source of aluminum in the world is bauxite (mostly Al2O3). Its thermal decomposition can be represented by If aluminum is produced this way, how many grams of aluminum can be produced when 1.000 x 103 kJ of heat is used?

  19. eg. Nitromethane, CH3NO2, an organic solvent, burns in oxygen according to the following reaction: You place a 1.724 g sample of nitromethane in a calorimeter with oxygen. The nitromethane is ignited and burns in oxygen. The temperature of the calorimeter increases from 22.23 oC to 28.81 oC. In a separate experiment you determine that the heat capacity of the calorimeter and its contents is 3.044 kJ K-1. What is DrH for this reaction?

  20. eg. When a solution containing 8.00 g of NaOH in 50.0 g of water at 25.0 oC is added to a solution of 8.00 g of HCl in 250 g of water at 25.0 oC in an insulated beaker, the temperature of the solution increases to 33.5 oC. Assuming that the specific heat of the solution is that of water, (4.184 J g-1 K-1) and that the beaker absorbs a negligible amount of heat, calculate DH for the reaction,

  21. Hess’s Law of Heat Summation • many reactions are impossible to perform as an individual reaction, for example, reactions which occur as part of a very complex biochemical process. • even if we cannot carry a reaction out in the laboratory, it is still possible to find the heat of reaction, DH, since H is a state function. • Hess’s Law of Heat Summation says that the enthalpy change of an overall process is the sum of the enthalpy changes of its individual steps. For example, if we want to know the enthalpy change for the conversion of graphite into diamond, it would be very difficult to perform this experiment since extremely high temperatures and pressures are required. However, we can look up the heats of combustion for these reactions, or burn graphite and diamond $ A B

  22. Why does the reaction require extreme temperatures and pressures if it is only 2 kJ endothermic?

  23. not to scale!! Cdiamond + O2 2 kJ Cgraphite + O2 396 kJ -394 kJ CO2

  24. eg. Given the following data, 1 2 3 Calculate DH for the following reaction.

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