Thisispartoneentalphy 121121135040-phpapp02

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The coffee cup bomb calorimeter By:Dr. Robert D. Craig, Ph.D

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Web quest for chm 02 enthalpy this one

Transcript of Thisispartoneentalphy 121121135040-phpapp02

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The coffee cup bomb calorimeterBy:Dr. Robert D. Craig, Ph.D

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Page 287 only

• A set of nested coffee cups is a good constant pressure calorimeter

• http://smithecta.weebly.com/chapter-9---heat.html

• http://snapguide.com/david-shipp/• http://web.lemoyne.edu/~giunta/chm151L/ca

lorimetry.html

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Today’sAgenda

• To determine the specific heat of a metal• To determine the enthalpy of neutralization

for a strong acid-strong base reaction• To determine the enthalpy of solution for the

dissolution of a salt

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Introduction from Your text

• Accompanying all chemical and physical changes is a transfer of heat (energy); heat may be either evolved (exothermic) or absorbed (endotherminc) .

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heat (energy) evolved (exothermic)

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heat (energy) absorbed (endotherminc) .

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NH4NO3 (c, IV) in water at 298.15 K to be ΔHo = 25.41 kJ mol−1.

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A calorimeter . . . .hold heat well

• A calorimeter is the laboratory apparatus that is used to measure the quantity and direction of heat flow accompanying a chemical or physical change.

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A professional one used in industry

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Measures heat content . .

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Entalphy . . A function of state

• The heat change in chemical reactions is quantitatively expressed as the enthalpy (or Heat) of reaction DH, at constant pressure

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negative for exothermic

• DH values are negative for exothermic reactions and positive for endothermic reactions.

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a state function –independent of path

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a state function –independent of path

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The classic things today . .

• Three quantitative measurements of heat are detailed in this experiment:

• 1. Measurements of the specific heat of a metal,

• 2. the heat change accompanying an acid-base reaction,

• 3. and the heat change associated with the dissolution of a salt in water.

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in joules,J

The energy (heat, expressed in joules,J) required to change the temperature of one gram of a substance by 1oC is the specific heat of that substance SI units:

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The specific is specific to all metals

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Intensive properties-How much ???

• Intensive properties (temperature, pressure) do not depend upon the sample size.

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Extensive-not how much

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Confusing. . .must check units

• Intensiveproperty**

• Specific heat capacityat constant volume ======J/(kg·K)

• extensiveproperty**

• heat capacityat constant volume ======J/K

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heat capacity as an intensive property

• heat capacity as an intensive property, i.e., independent of the size of a sample, are the molar heat capacity, which is the heat capacity per mole of a pure substance, and the specific heat capacity, often simply called specific heat, which is the heat capacity per unit mass of a material.

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(25.2)

• Or rearranging for energy,

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DT is the temperature change

• DT 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.

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• The specific heat of a metal that does not react with water is determined by heating a measured mass of the metal, M to a known (higher) temperature placing it into a measured amount of water at a known (lower) temperature measuring the final equilibrium temperature after the two are combined.

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final equilibrium temperature after the two are combined

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• The following equations, based on the law of conservation of energy (insert NoW-Rob!!!!!!!!!)

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specific heat of a metal

• Show the calculations for determining the specific heat of a metal. Considering the direction of energy flow by the conventional sign notation energy loss being “negative” and energy gain being “positive” then

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(25.3)

--Energy Lost (-J) by the metal = Energy (+J) gained by water

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specific heat of a metal

Substituting and rearranging, We have -specific heat x mass(M) x DT = specific heat (H20) x mass (H20) x DT (H20)

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This is (25.4)

This is equation (25.4) Rearranging equation (25.4) , we solve for the specific heat of the Metal as:

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(25.5)

+specific heat x mass(M) x DT =(-) specific heat (H20) x mass (H20) x DT (H20)• mass(M) x DT

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(25.6)

• In the equation, the temperature change for either substance is defined as the difference between the final temperature Tf, and the initial temperature, Ti, of the substance

DT-= Tf - Ti

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DT-= Tf - Ti

• These equations assume no heat lost to the calorimeter when the metal and the water are combined. The specific heat of water in 4.184 J/g.oC

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.

• In addition to using the calorimeter properly, key techniques for obtaining accurate results are starting with a dry calorimeter, measuring solution volumes precisely, and determining T accurately. Careful experimenters deal with the first two items easily.

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• The last is somewhat more difficult. The change in temperature is determined by measuring the initial temperature, T1, of the reactants, and the maximum temperature, T2, of the contents of the calorimeter during the exothermic reaction.

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• . The determination of a precise value for T2 is complicated by the fact that a small heat exchange occurs between the surroundings and the contents of the calorimeter, both during the reaction and after its completion. The rate of exchange depends on the insulating properties of the calorimeter and on the rate of stirring. A correction for this heat loss is made by an extrapolation of a temperature vs. time curve (see Figure 1 in your lab manual).

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• The rate of exchange depends on the insulating properties of the calorimeter and on the rate of stirring. A correction for this heat loss is made by an extrapolation of a temperature vs. time curve (see Figure 1 in your lab manual).

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Calorimeter video one-peterjackson118

• http://www.youtube.com/watch?v=WfO2sY-GJec&list=UUSnmgHe14lSSjaUfrtFdPGA&index=5&feature=plcp

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Figure 2. Graph of temperature as a function of time for an exothermic reaction in a perfect calorimeter.

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no calorimeter is perfect!!!!

• Unfortunately, no calorimeter is perfect, and instantaneous mixing and reaction are not always achieved (even with efficient mixing). In this case, the graph of temperature as a function of time looks more like the figure above

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.

• We can still find ΔT, but now we must extrapolate back to when the solutions were mixed (time, t, equals zero). This is most easily done by performing a linear regression on the sloped portion of the graph (where, for exothermic reactions, heat is leaking out of the calorimeter) and obtaining Tfinal from the y-intercept.

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Put 3 caution slides here

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Part B