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At 850°c, CaCO 3 undergoes substantial decomposition to yield CaO and co 2 . Assuming that the d h ° 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 66.8 g of co 2 are produced in one reaction.
Accepted Answer
Enthalpy Change for the Reaction:
The enthalpy change for the reaction can be calculated using the following equation:
ΔH°rxn = ΣnΔH°f(products) - ΣnΔH°f(reactants)
where ΔH°rxn is the enthalpy change for the reaction, n is the stoichiometric coefficient of each reactant or product, and ΔH°f is the standard enthalpy of formation of each reactant or product.
Given that 66.8 g of CO2 are produced in the reaction, we can determine the number of moles of CO2 using its molar mass (44.01 g/mol):
n(CO2) = 66.8 g / 44.01 g/mol = 1.52 mol
According to the balanced chemical equation, 1 mol of CO2 is produced for every 1 mol of CaCO3 that decomposes. Therefore, 1.52 mol of CO2 corresponds to 1.52 mol of CaCO3.
Using the ΔH°f values of CaCO3, CaO, and CO2 from a reference table (assuming they are the same at 850°C as at 25°C), we can calculate the enthalpy change for the reaction:
ΔH°rxn = [ΔH°f(CaO) + ΔH°f(CO2)] - [ΔH°f(CaCO3)]
ΔH°rxn = [(-151.9 kJ/mol) + (-393.5 kJ/mol)] - (-1207 kJ/mol)
ΔH°rxn = 661.6 kJ
Therefore, the enthalpy change for the decomposition of 66.8 g of CaCO3 at 850°C is 661.6 kJ.
The enthalpy change for the reaction can be calculated using the following equation:
ΔH°rxn = ΣnΔH°f(products) - ΣnΔH°f(reactants)
where ΔH°rxn is the enthalpy change for the reaction, n is the stoichiometric coefficient of each reactant or product, and ΔH°f is the standard enthalpy of formation of each reactant or product.
Given that 66.8 g of CO2 are produced in the reaction, we can determine the number of moles of CO2 using its molar mass (44.01 g/mol):
n(CO2) = 66.8 g / 44.01 g/mol = 1.52 mol
According to the balanced chemical equation, 1 mol of CO2 is produced for every 1 mol of CaCO3 that decomposes. Therefore, 1.52 mol of CO2 corresponds to 1.52 mol of CaCO3.
Using the ΔH°f values of CaCO3, CaO, and CO2 from a reference table (assuming they are the same at 850°C as at 25°C), we can calculate the enthalpy change for the reaction:
ΔH°rxn = [ΔH°f(CaO) + ΔH°f(CO2)] - [ΔH°f(CaCO3)]
ΔH°rxn = [(-151.9 kJ/mol) + (-393.5 kJ/mol)] - (-1207 kJ/mol)
ΔH°rxn = 661.6 kJ
Therefore, the enthalpy change for the decomposition of 66.8 g of CaCO3 at 850°C is 661.6 kJ.
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