Question

Calculate the Δ H°rxn for the decomposition of calcium carbonate to calcium oxide and carbon dioxide. Δ H°f [CaCO3(s)] = –1206.9 kJ/mol; Δ H°f [CaO(s)] = –635.1 kJ/mol; Δ H°f [CO2(g)] = –393.5 kJ/mol

Answer 1

Answer:

$\boxed{\text{178.3 kJ/mol}}$

Explanation

The formula for calculating the enthalpy change of a reaction by using the enthalpies of formation of reactants and products is

$\Delta_{\text{r}}H^{\circ} = \sum \Delta_{\text{f}} H^{\circ} (\text{products}) - \sum\Delta_{\text{f}}H^{\circ} (\text{reactants})$

                         CaCO₃(s) ⟶ CaO(s) + CO₂(g)

ΔH°f/kJ·mol⁻¹:    -1206.9         -635.1    -393.5

$\begin{array}{rcl}\Delta_{\text{r}}H^{\circ}& = & [(-635.1 + (-393.5)] - (-1206.9)\\& = & -1028.6 +1206.9\\& = & \mathbf{178.3}\\\end{array}\\\text{The enthalpy of decomposition is } \boxed{\textbf{178.3 kJ/mol}}$l}}