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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the positions which points up in upward carbon and down in downward carbon are axial

and vice versa holds true for equatorial positions

so the correct answer with labelled diagram will be