UST General Santos Calorimetry Lab PDF

PRAYER BEFORE CLASS Holy Spirit, Divine Creator, true source of light and fountain of wisdom! Pour forth your brilliance upon my dense intellect, dissipate the darkness which covers me, that of sin and of ignorance. Grant me a penetrating mind to understand, a retentive memory, method and ease of le...

PRAYER BEFORE CLASS Holy Spirit, Divine Creator, true source of light and fountain of wisdom! Pour forth your brilliance upon my dense intellect, dissipate the darkness which covers me, that of sin and of ignorance. Grant me a penetrating mind to understand, a retentive memory, method and ease of learning, the lucidity to comprehend, and abundant grace in expressing myself. Guide the beginning of my work, direct its progress and bring it to successful completion. This I ask through Jesus Christ, true God, and true man, living and reigning with You and the Father, forever and ever. Amen.  UST General Santos Post-laboratory Discussion Experiment 5: Calorimetry Chemistry for Engineers Laboratory Unit Intended Learning Outcome: At the end of the unit, the student must be able to: ❖ to determine the amount of heat absorbed or released when solid dissolves in water to form a solution ❖ to determine and compare the amount of heat involved in the three related reactions Calorimetry - is the science associated with determining the change in energy of a system by measuring the amount of heat released or absorbed (heat transfer) during chemical reaction - we cannot know the exact enthalpy of the reactants and products Calorimeter - allows us to measure ∆H - apparatus used to measure the quantity and direction of heat flow accompanying a chemical or physical change RECALL!!! CHANGE IN ENTHALPY or HEAT OF REACTION or ∆H - heat change in chemical reactions at constant pressure - ∆H: negative for exothermic reactions - ∆H: positive for endothermic reactions Two Main Types of Calorimeters 1. Bomb Calorimeter - a calorimeter that maintains constant volume during the process 2. Constant Pressure Calorimeter - a calorimeter that maintains constant pressure but volume changes may occur EX: coffee cup calorimeter Constant Pressure Calorimeter uses the first law of thermodynamics - if heat is lost by the chemicals during reaction, then it must be gained by the solution (liquid) and vice versa qsubstance = - qsolution qsolution = - qsubstance Constant Pressure Calorimeter - The enthalpy (H) of a system accounts for the heat flow of the system at constant pressure. The heat absorbed or released by a reaction at constant pressure Is the same as the change in enthalpy, symbolized as ΔH. Constant Pressure Calorimeter The value of ΔH of a reaction can be determined by measuring the heat flow of the reaction at constant pressure. terms heat and enthalpy change are used interchangeably. In other words, q = ΔH. Constant Pressure Calorimeter SPECIFIC HEAT CAPACITY - amount of energy (heat expressed in joules, J) required to raise the temperature of one (1) gram of a substance by one (1) degree celsius Constant Pressure Calorimeter To measure the enthalpy change for a reaction in aqueous solution in a foam cup calorimeter, dissolve the reacting chemicals (the system) in known volumes of water (the surroundings). Constant Pressure Calorimeter Measure the initial temperature of each solution, and mix the solutions in the foam cup. After the reaction is complete, measure the final temperature (which is the constant temperature) of the mixed solutions. Constant Pressure Calorimeter You can calculate the heat absorbed or released by the surroundings (qsurr) using the formula for the specific heat, the initial and final temperatures, and the heat capacity of water. qsurr = m x C x ΔT q: heat transfer m: mass of the substance ΔT = Final temp - Initial Temp c: specific heat capacity ΔT: change in the temperature Constant Pressure Calorimeter The heat absorbed by the surroundings is equal to, but has the opposite sign of, the heat released by the system. qsurr = –qsys Constant Pressure Calorimeter The enthalpy change for the reaction (ΔH) can be written as follows: CALCULATIONS! EXAMPLE: A 0.45 gram piece of copper (C= 0.385 J/g degree celsius) at temperature of 87 degrees celsius is placed in a calorimeter. The calorimeter contained water (C= 4.184 J/g degree celsius) at a temperature of 23 degree celsius. The solution came to equilibrium at 24 degrees celsius. What was the mass of water in the calorimeter? CALCULATIONS! EXAMPLE: Copper Water m: 0.45 grams m: ?? C of copper: 0.385 J/g degree celsius C of water: 4.184 J/g degree celsius T initial: 87 degrees celsius T initial: 23 degrees celsius T final: 24 degrees celsius T final: 24 degrees celsius CALCULATIONS! Copper qsurr = m x C x ΔT m: 0.45 grams C of copper: 0.385 J/g degree celsius T initial: 87 degrees celsius T final: 24 degrees celsius CALCULATIONS! Copper qsurr = m x C x ΔT m: 0.45 grams C of copper: 0.385 J/g degree celsius q= (0.45)(0.385)(24-87) T initial: 87 degrees celsius T final: 24 degrees celsius q= -10.91 J the heat lost is -10.91 J so the water must have gained 10.91 J therefore qwater = 10.91 J CALCULATIONS! Water qsurr = m x C x ΔT m: ?? C of water: 4.184 J/g degree celsius T initial: 23 degrees celsius T final: 24 degrees celsius qwater: 10.91 J CALCULATIONS! Water qsurr = m x C x ΔT m: ?? C of water: 4.184 J/g degree celsius 10.91= m(4.184)(24-23) T initial: 23 degrees celsius T final: 24 degrees celsius m= 2.61 g qwater: 10.91 J since we solved the q for copper, we can use this number (with opposite sign) to get the q for water since heat lost by one is gained by the other CALCULATIONS! EXAMPLE: A piece of titanium metal with a mass of 20.8 grams is heated in boiling water to 99.5 degrees celsius and then dropped into a calorimeter containing 75 grams water (C= 4.184 J/g degree celsius) at 21.7 degrees celsius. The final temperature in the calorimeter is 24.3 degrees celsius. What is the specific heat of titanium? CALCULATIONS! EXAMPLE: Titanium Water m: 20.8 grams m: 75 grams C of titanium: ?? C of water: 4.184 J/g degree celsius T initial: 99.5 degrees celsius T initial: 21.7 degrees celsius T final: 24.3 degrees celsius T final: 24.3 degrees celsius CALCULATIONS! Water qsurr = m x C x ΔT m: 75 grams C of water: 4.184 J/g degree celsius T initial: 21.7 degrees celsius T final: 24.3 degrees celsius CALCULATIONS! Water qsurr = m x C x ΔT m: 75 grams C of water: 4.184 J/g degree celsius q= 75(4.184)(24.3-21.7) T initial: 21.7 degrees celsius T final: 24.3 degrees celsius q= 815.88 J if water is gaining 815.88 J, then the titanium has lost 815.88 J so make it negative and use it for qtitanium CALCULATIONS! Titanium qsurr = m x C x ΔT m: 20.8 grams C of titanium: ?? T initial: 99.5 degrees celsius T final: 24.3 degrees celsius CALCULATIONS! Titanium qsurr = m x C x ΔT m: 20.8 grams C of titanium: ?? -815.88= 20.8(C)(24.3-99.5) T initial: 99.5 degrees celsius T final: 24.3 degrees celsius -815.88= -1564.16(C) C= 0.522 J/g degrees celsius  UST General Santos Activity Experiment 5: Calorimetry Determination of the Heat Capacity of the Calorimeter, Cs Example: Mass of water: 100g CALCULATION: Cp: 4.184 J/g oC (mw)(Cp)(t1-t2) + Cs (t1-t2) = (mi)(Hf) + Cs: specific heat of calorimeter mi(Cp) (t2-tice) Weight of the beaker, filter paper, ice (m1): 170.4 g (100)(4.184)(32-16) + Cs (32-16) = Weight of the beaker, filter paper (m2): 154 g (16.4)(333.5) + (16.4)(4.184)(16-0) Weight of the ice (m1-m2): 16.4 g Initial temperature (t1): 32 degrees celsius Final temperature (t2): 16 degrees celsius 6,694.4 + Cs (16) = 6,433.88 Temp of ice: 0 degree celsius Cs (16) = 6,433.88 - 6,694.4 Hf: heat of fusion of ice (333.5 J/g) Cs = -260.52/ 16 Cs = -16.28 J/g degrees celsius Determination of Heat of Reaction 1 (∆H1) Example: Volume of water used: 100 mL CALCULATION: Initial temperature of water (t1): 32 degrees celsius Highest temperature of NaOH soln (t2): 44 degree celsius ∆H = (100g x 4.184 -16.28) (32-44) Weight of NaOH: 1g ∆H = -4825.44 J Cs -16.28 exothermic reaction Determination of Heat of Reaction 2 (∆H2) Example: Volume of HCl used: 20 mL CALCULATION: Initial temperature of HCl (t1): 38 degrees celsius Highest temperature of HCl + NaOH (t2): 45 degree celsius ∆H = (20g x 4.184 - 16.28) (38-45) Weight of HCl solution: 20 g ∆H = -471.8 J Weight of NaOH: 1.2 g exothermic reaction Determination of Heat of Reaction 3 (∆H3) Example: Volume of 6M HCl used: 10 mL CALCULATION: Volume of 6M NaOH used: 10 mL Initial temperature of HCl (t1): 33 degree celsius ∆H = (20g x 4.184 - 16.28) (33-42) Initial temperature of NaOH (t1): 33 degree celsius ∆H = -606.6 J Highest temperature of HCl + NaOH (t2): 42 degrees celsius exothermic reaction  UST General Santos Post-laboratory Discussion Experiment 5: Calorimetry Chemistry for Engineers Laboratory End of discussion Do you have any questions? PRAYER AFTER CLASS Grant me, O Lord my God, a mind to know you, a heart to seek you, wisdom to find you, conduct pleasing to you, faithful perseverance in waiting for you, and a hope of finally embracing you, Amen.