Enthalpy lab report

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Marquette University *

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1001

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Chemistry

Date

Feb 20, 2024

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pdf

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4

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CHEM 1001 ENTHALPY Laboratory Report NAME : Liam Cedeno Agosto DATE : 4/18/23 SECTION : 403 TA : Audrey Vice A. Goals/Purpose of Experiment ( 2 Points, ~50 words ) : The purpose of this experiment is to understand energy changes associated with chemical reactions and be able to measure these chemical reaction changes using a simple calorimeter. B. Theory/Introduction ( 6 Points, ~100 words ) : All chemical reactions have heat/energy exchanges. When energy is gained, the reaction in endothermic, whereas when energy is release or lost, the reaction is exothermic. These are known are thermochemical reactions. We can calculate these reactions through enthalpy change. C. Procedure summary ( 5 Points, ~100 words ) : For this experiment, we were given 2.0 M of HCl and NaOH, and water. It was divided into three parts. Part one dealt with HCl and NaOH reacting. We were to add about 10 mL of each into separate graduated cylinders, and we were to pour one into a Styrofoam cup first. After starting up the measure.net station, we were to wait up to 15 seconds before pressing start, then wait another up to 15 seconds to measure the temperature before pouring the other substance after and wait until the temperature stabilized before stopping and saving the data. We were to repeat these two more times, one with HCl and water, and one with NaOH and water. D. Results and Calculations: ( 12 Points ) : Solution A Solution B Volume solution A (mL) Volume solution B (mL) Total mass ( m , g) Limiting reagent (moles) Initial Temp (ºC) Final Temp (ºC) Change in temp (ΔT , ºC) Q rxn (J) ΔH (kJ/mol)
NaOH HCl 10 mL 10 mL 20 m2g 0.02 20.74 C 25.98 C 5.24 C 438.48 21.92 NaOH water 10 mL 10 mL 20 m2g 0.02 20.09 C 21.66 C 1.57 C 131.37 6.56 HCl water 10 mL 10 mL 20 m2g 0.02 19.78 C 21.86 C 2.08 C 174.05 8.0 ΔH 3 (NaOH + HCl) = ΔH 1 (rxn) - ΔH 2a (NaOH dilution) - ΔH 2b (HCl dilution) ΔH 3 (NaOH + HCl) = 21.92 kJ/mol – 6.56 kJ/mol – 8.0 kJ/mol= 7.36 kJ/mol Δ H Theoretical (NaOH + HCl) = Σ n Δ H f °(products) – Σ m Δ H f °(reactants) (-407.25 kJ/mol) + (-285.83 kJ/mol) - (-470.09kJ/mol) + (-167.15kJ/mol) = -55.84 kJ/mol % 𝑒𝑟𝑟𝑜𝑟 NaOH-HCl = |experimental−theoretical| theoretical × 100= ( -55.84) – 7.36/7.36 * 100= -113.18% Insert all 3 graphs here (no screenshot/picture)
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