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Calculate the heat of decomposition for this process at constant pressure and 25°C: CaCO3(s) → CaO(s) + CO2(g) The standard enthalpies of formation for the reactant are: ΔHf CaCO3(s) = −1206.9 kJ/mol ; ΔHf CaO(s) = −635.6 kJ/mol; ΔHf CO2(g) −393.5 kJ/mol

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quasarJose

Answer:

177.8kJ/mol

Explanation:

In this reaction, the heat of decomposition is the same as the heat of formation. This is a decomposition reaction.

Given parameters:

ΔHf CaCO₃ = -1206.9kJ/mol

ΔHf CaO = −635.6 kJ/mol

ΔHf CO₂ = −393.5 kJ/mol

The heat of decomposition =

                     Sum of ΔHf of products - Sum of ΔHf of reactants

The equation of the reaction is shown below:

     CaCO₃ → CaO + CO₂

The heat of decomposition = [ -635.6 + (-393.5)] - [−1206.9 ]

                                             = -1029.1 + 1206.9

                                             = 177.8kJ/mol

fichoh

The heat of decomposition given by the difference in the ΔH of product and reactant is 177.8 kj/mol

Given the equation for the decomposition process :

  • CaCO3(s) → CaO(s) + CO2(g)

The heat for the decomposition process is the difference of the sum of heat of the product and the heat of the reactant :

Reactant :

  • CaCO3(s) = - 1206.9 kj/mol

Product :

  • CaO(s) + CO2(g) = - 635. 6 + (-393.5) = - 1029.1 kj/mol

Heat of decomposition :

  • ΔHproduct - ΔHreactant

Heat of decomposition = - 1029.1 - (- 1206.9)

Heat of decomposition = -1029.1 + 1206.9 = 177.8 kj/mol

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