Respuesta :
Answer:
[tex]\boxed{\text{178.3 kJ/mol}}[/tex]
Explanation
The formula for calculating the enthalpy change of a reaction by using the enthalpies of formation of reactants and products is
[tex]\Delta_{\text{r}}H^{\circ} = \sum \Delta_{\text{f}} H^{\circ} (\text{products}) - \sum\Delta_{\text{f}}H^{\circ} (\text{reactants})[/tex]
CaCO₃(s) ⟶ CaO(s) + CO₂(g)
ΔH°f/kJ·mol⁻¹: -1206.9 -635.1 -393.5
[tex]\begin{array}{rcl}\Delta_{\text{r}}H^{\circ}& = & [(-635.1 + (-393.5)] - (-1206.9)\\& = & -1028.6 +1206.9\\& = & \mathbf{178.3}\\\end{array}\\\text{The enthalpy of decomposition is } \boxed{\textbf{178.3 kJ/mol}}[/tex]l}}
A decomposition reaction is a chemical reaction in which the reactants breaks into two or more products. The enthalpy of the decomposition reaction is 178.3 kJ/mol.
What is the enthalpy of reaction?
The enthalpy of reaction [tex]\rm (\Delta Hrxn)[/tex] is the difference between the product enthalpy and reactant enthalpy and is given in the units kilojoules per mole.
It can be calculated by the formula:
[tex]\rm \Delta H rxn = \sum \Delta H_{f}(products) - \sum \Delta H_{f}(reactants)[/tex]
The reaction can be shown as:
[tex]\rm CaCO_{3}(s) \rightarrow CaO(s) + CO_{2}(g)[/tex]
Given,
The enthalpy of formation of reactant (calcium carbonate) = -1206.9
The enthalpy of formation of product (calcium oxide) = -635.1
The enthalpy of formation of product (carbon dioxide) = -393.5
Substituting values in the above equation:
[tex]\begin{aligned}\rm \Delta H rxn &= [(-635.1 + (-393.5)] - (-1206.9)\\\\&= -1028.6 + 1206.9\\\\&= 178.3\end{aligned}[/tex]
Therefore, the enthalpy of the decomposition is 178.3 kJ/mol.
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