A Calorimetry lab asks to measure 30mL of 3.0 mol HCl into a styrofoam cup, and

Question

A Calorimetry lab asks to measure 30mL of 3.0 mol HCl into a styrofoam cup, and record the temperature (22 degrees celsius). It then asks to accurately weigh out between 1.5 and 2.2g of solid lithium carbonate. Then, you are to mix the lithium carbonate and HCl together. In the first trial, 2.042g of Lithium carbonate is added, and the temperature goes from 22 degrees celcius to 27.5 degrees celcius. In the second trial, 2.026g of lithium carbonate is added, and the temperature foes from 21 degrees celcius to 26.5 degrees celcius. In trial three, 2.198 g of lithium carbonate is added, and the temperature goes from 23 to 28 degrees celcius. Okay so given that I understand how to calculate the delta H of the reaction using heat of water = mass of water x heat capcity of water x temperature change of water...and then when you get the heat of water, multiply it by negative to get your delta H of the solid lithium carbonate. How would you go about getting your theoretical delta H of the reaction (using the products - reactants formula)..How would you set it up here as the balanced equation is:

2HCl + 2Li2CO3 --> 2LiCl + H2O + CO2

Would you mutiply the 2 coefficient by the heat of formation for 3Molar HCl? Would that be the only heat of formation that would be different here? If not, please explain how to find the theoretical delta H of this reaction.

Note this is a AP Chemistry course, not "Chem110"

Explanation / Answer

theoretical calculation of dH of reaction can be done in two ways

dH of reaction = dH products - dH reactants

= 2 dH LiCl + dH H2O + dH CO2 - dH Li2CO3 - 2 dH HCl

2nd way is

dH reaction = Bond energy of reactants bonds broken - Bond energy of product bonds formed

= 2 BE ( H-Cl) + 2( C-O) + (C=O) + 2( Li-O) - 2( Li-Cl) - 2(O-H) - 2(C=O)

= 2 BE( H--Cl) + 2BE( C-O) + 2BE(Li-O) - 2BE(Li-Cl) - 2BE(O-H) - BE(C=O)

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