At 850°c, caco3 undergoes substantial decomposition to yield cao and co2. assuming that the δh o f values of the reactant and products are the same at 850°c as they are at 25°c, calculate the enthalpy change (in kj) if 68.10 g of co2 is produced in one reaction.

The enthalpy if 68.10 grams of CO2 is produced is  -189.04 kJ

Explanation:

Step 1: Data given

temperature = 850 °C

Mass of 68.10 grams of CO2

ΔH°f (CaO) = -635.6 kJ/mol

ΔH°f (CO2) = -693.5 kJ/mol

ΔH°f (CaCO3) =-1206.9 kJ/mol

Step 2: The balanced equation

CaCO3(s) → CaO(s) + CO2(g)

Step 3:  Calculate ΔH°reaction

ΔH°reaction = ΣΔH°f (products) - ΣΔH°f (reactants)

ΔH°reaction = (ΔH°f (CaO) + ΔH°f (CO2)) -  ΔH°f (CaCO3)

ΔH°reaction = (-635.6 kJ/mol + -693.5 kJ/mol) + 1206.9 kJ/mol

ΔH°reaction = -122.2 kJ /mol

Step 4: Calculate moles of CO2

Moles CO2 = mass CO2 / Molar mass CO2

Moles CO2 = 68.10 grams / 44.01 g/mol

Moles CO2 = 1.547 moles

Step 5: Calculate the enthalpy change for 68.10 grams of CO2

-122.2 kJ/mol * 1.547 moles = -189.04 kJ

The enthalpy if 68.10 grams of CO2 is produced is  -189.04 kJ

The enthalpy change when 68.10 g of CO₂ is produced in the decomposition of CaCO₃ is -189.0 kJ.

Let's consider the decomposition of CaCO₃.

CaCO₃(s) ⇒ CaO(s) + CO₂(g)

Given the standard enthalpies of formation (ΔH°f) of products (p) and reactants (r), we can calculate the standard enthalpy of the reaction (ΔH°rxn) using the following expression.

The molar mass of CO₂ is 44.01 g/mol. The moles corresponding to 68.10 g are:

122.2 kJ are evolved when 1 mole of CO₂ is formed. The heat evolved when 1.547 moles of CO₂ are evolved is:

The enthalpy change when 68.10 g of CO₂ is produced in the decomposition of CaCO₃ is -189.0 kJ.

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