Question

calculate the enthalpy for the reaction below.
2al(s) + 3cl₂(g) → 2alcl₃(s)
∑ products − ∑ reactants
compound δh°_f (kj/mol)
al(s) 0.0
cl₂(g) 0.0
alcl₃(s) -704.2
δh_rxn = ? kj
enter either a + or - sign and the magnitude.
hint: include the moles for the reactants and products in your calculation.

Explanation:

Step1: Recall the formula for enthalpy of reaction

The formula for the enthalpy of a reaction ($\Delta H_{rxn}$) using standard enthalpies of formation ($\Delta H_f^\circ$) is $\Delta H_{rxn}=\sum n\Delta H_f^\circ(\text{products})-\sum n\Delta H_f^\circ(\text{reactants})$, where $n$ is the stoichiometric coefficient.

Step2: Identify moles and $\Delta H_f^\circ$ for products

For the product $\text{AlCl}_3(s)$, the stoichiometric coefficient $n = 2$ and $\Delta H_f^\circ=- 704.2\space kJ/mol$. So the sum for products is $n\times\Delta H_f^\circ=2\times(-704.2)$.

Step3: Identify moles and $\Delta H_f^\circ$ for reactants

For reactants $\text{Al}(s)$ and $\text{Cl}_2(g)$, their $\Delta H_f^\circ$ are $0.0\space kJ/mol$ and stoichiometric coefficients are $2$ and $3$ respectively. So the sum for reactants is $2\times0.0 + 3\times0.0=0$.

Step4: Calculate $\Delta H_{rxn}$

$\Delta H_{rxn}=2\times(-704.2)-0=- 1408.4$

Answer:

-1408.4