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What is the enthalpy of reaction for the
decomposition of calcium carbonate?
CaCO3(s) → CaO(s) + CO2(g)
kJ

What is the enthalpy of reaction for the decomposition of calcium carbonate CaCO3s CaOs CO2g kJ class=

Respuesta :

znk

Answer:

[tex]\boxed{\text{656.3 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:    -1207.1          -157.3    -393.5

[tex]\begin{array}{rcl}\Delta_{\text{r}}H^{\circ} & = & [-157.3 + (-393.5)] - (-1207.1)\\& = & -550.8 +1207.1\\& = & \textbf{656.3 kJ/mol}\\\end{array}\\\\\text{The enthalpy of decomposition is } \boxed{\textbf{656.3 kJ/mol}}[/tex]

pstnonsonjoku

The enthalpy of decomposition is  656.3  KJ/mol.

Given that the enthalpy of decomposition is obtained from;

ΔHrxn = ∑ΔHf(products) -  ∑ΔHf(reactants)

Now, the species involved in the reaction are;

ΔHf(CaO)(s)) = -157.3 KJ/mol

ΔHf((CO2)(g)) = -393.5  KJ/mol

ΔHf(CaCO3)(s)) = -1207.1 KJ/mol

Substituting values;

ΔHrxn = [(-393.5) + ( -157.3)] - ( -1207.1) KJ/mol

ΔHrxn =(( -550.8) + 1207.1)  KJ/mol

ΔHrxn = 656.3  KJ/mol

Learn more: https://brainly.com/question/7827769

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