76618kJ x 10 \u0394H 477kJ Part 2 1 Choose one of the available unknown metals

# 76618kj x 10 δh 477kj part 2 1 choose one of the

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Change in heat for one mole of reaction is -4.76618kJ x 10 ΔH = -47.7kJ Part 2 1) Choose one of the available unknown metals available and weigh it. 2) Add 200 mL of tap water to a 400 mL beaker and heat it until boiling. 3)Add 50.0 mL of cold water to the calorimeter. Measure the initial temperature of water in the calorimeter. 4) Obtain the mass of calorimeter + water. 5) When the water in the beaker is boiling, carefully place your unknown metal into it and keep it there for five minutes. 6) Quickly transfer the metal from the beaker to the calorimeter, cover it with the lid, mix by gently swirling the cup, and start recording the temperature every 30 seconds (using a timer) for 10 minutes. Mix the solution from time to time. 7)Plot the temperature-time data and perform a linear fit of a linear portion of the graph. Identify your unknown metal.Mass of metal = 62.66gInitial temperature of water = 9.5CMass of calorimeter and water = 79gMass of water = 79 – 8.83 = 70.17gChange in heat of metal = -change in heat of water (surroundings) 1) Add 50.0 mL of deionized water to the calorimeter and obtain the mass of calorimeter + water. 2) Measure the initial temperature of water in the calorimeter. 3) Weigh about 2g of ammonium chloride salt and record the mass. 4) Quickly add the salt to the calorimeter, cover it with the lid, mix by gently swirling the calorimeter, and start recording the temperature of the solution every 30 seconds until the temperature remains constant. Mix the solution from time to time. 5) Plot the temperature-time data and determine the molar heat of solution of ammonium chloride salt. Assume that the heat capacity of the solution is the same as for pure water. Mass of calorimeter and 50mL deionized water = 83.99g Initial temperature of water in calorimeter = 17.8C Mass of ammonium chloride salt = 2g mass of water = 83.99 – 8.33 = 75.11 Final temperature = 23.1C -qsurroundings(water) = -mC T = qNH 4 Cl solution -(49.94)(4.184)(23.1C – 25.8C) = 564.16J per 2g NH 4 Cl Molar mass of NH 4 Cl = 53.491g/mole 2g NH 4 Cl/53.491 = 0.0374mole NH 4 Cl 564.16J 0.0374mole NH 4 Cl = 15088.8J/mole NH 4 Cl Hsolution = 15.1kJ/mole NH 4 Cl Discussion and Conclusion Heat and work are the two most common ways for a system to exchange energy with its surroundings. The work term in reactions that do not involve gases is zero, so all of the energy change results in heat. The amount of heat that flows into or out of the surroundings is determined with a technique called calorimetry (heat measurement). A calorimeter is composed of an insulated container, a thermometer, a mass of water, and the system to be studied. The use of an insulated container (Styrofoam cup in this experiment) allows us to assume that there is no heat transferred through the calorimeter walls. In other words, we can assume that the  #### You've reached the end of your free preview.

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