Question

I'm having issues with thes questions on Hess' law

to determine the enthalpy of the reaction CaO(s) + CO2(g) → CaCO3(s)

The following procedure was carried out

1. Measure 1g of calcium oxide using the top loading balance and place in a calorimeter.

2. Measure 50 cm3 of 1M HCl and allow three minutes to pass and then measure and record its temperature.

3. Pour the acid quickly but carefully into the calorimeter and stir the mixture with the thermometer and record its final temperature.

4. Repeat steps 1-3 using calcium carbonate instead of calcium oxide.

Hence:
1. Calculate the # of moles of
i) calcium oxide used          ii) calcium carbonate used

2. Calculate the enthalpy change of the reaction involving
i) calcium oxide        ii) calcium carbonate
Given that the heat capacity of the solution is 4.18 J g-1 °C-1 and that 1 cm3 of aqueous solution has a mass of 1g

3.   Using your answers to questions 1 and 2, determine the enthalpy change of reaction for
i) one mole of calcium oxide         ii) one mole of calcium carbonate

4. Draw an energy cycle for the enthalpy change of the reaction CaO(s) + CO2(g) → CaCO3(s) using the TWO BALANCED chemical equations for the reactions you have conducted.

5. Use the energy cycle and your answers from question 3 to determine the enthalpy change of the reaction CaO(s) + CO2(g) → CaCO3(s) for one mole of calcium carbonate.

**The answers can be calculated without results from the experiment using known standard enthalpy changes

Answer 1

1. you have 1 gram of CaO, you want to calculate the number of moles

moles = mass / molar mass, molar mass of CaO is 56 g/gmol

moles of CaO = 1 / 56 = 0.01786 moles of CaO

1 gram of CaCO₃, molar mass is 100 g/gmol

moles of CaCO₃ = 1 / 100 = 0.01 moles

so as you can see you have less moles of CaCO3 than CaO , this is your limiting reactant the CaCO3 you must perform calculation with this compound

the reaction is CaO + CO₂ ===== CaCO₃

Enthalpy of formation of product

CaCO₃ = - 1207 KJ / mol

Enthalpy of reactants

CaO = -635.6 KJ / mol

CO₂ = -393.5 KJ / mol

you need to apply the formula

Enthalpy of reaction = Enthalpy of products - Enthalpy of reatants

Enthalpy = -1207 - (-635.6 + (-393.5)) = -177.9 KJ / mol

We can use some equations to match them and get CaCO3 and use the hess law, im finding this equations:

Ca(s) + CO2(g) + 1/2O2(g) → CaCO3(s) ΔH = -812.8 kJ.... equation 1
Ca(s) + 1/2 O₂(g) → CaO(s) ΔH = -634.9 kJ...... equation 2

The best thing to do is to inverse the second equation, remember that when you do this the negative sign becomes positive so

CaO(s)→ Ca(s) + 1/2 O₂(g)  ΔH = 634.9 kJ

now add this equation with the first equation

Ca(s) + CO2(g) + 1/2O2(g) → CaCO3(s) ΔH = -812.8 kJ

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CaO + Ca + CO₂ + 1/2 O₂  →  CaCO3(s) + Ca(s) + 1/2 O₂(g)

remove the elements that are repeated

CaO + CO₂ →  CaCO3(s)

Add the individual enthalpies -812.8 + 634.9 = -177.9 KJ / mol

Since the reaction is 1 to 1, 1 mole produces 1 mole the enthalpy for one mole of CaO is equal to the enthalpy of 1 mole of CaCO3 === -177.9 KJ