Question

using the table of standard formation enthalpies that youll find under the aleks data tab, calculate the reaction enthalpy of this reaction under standard conditions:
2naoh(s) + cacl₂(s) → ca(oh)₂(s) + 2nacl(s)
round your answer to the nearest kj.

Explanation:

Step1: Recall the formula for reaction enthalpy

$\Delta H^\circ_{\text{rxn}} = \sum \Delta H^\circ_f(\text{products}) - \sum \Delta H^\circ_f(\text{reactants})$

Step2: List standard formation enthalpies (from ALEKS Data)

$\Delta H^\circ_f(\text{NaOH}(s)) = -425.6\ \text{kJ/mol}$, $\Delta H^\circ_f(\text{CaCl}_2(s)) = -795.8\ \text{kJ/mol}$, $\Delta H^\circ_f(\text{Ca(OH)}_2(s)) = -986.1\ \text{kJ/mol}$, $\Delta H^\circ_f(\text{NaCl}(s)) = -411.2\ \text{kJ/mol}$

Step3: Calculate product enthalpy sum

$[1\times(-986.1) + 2\times(-411.2)] = -986.1 - 822.4 = -1808.5\ \text{kJ}$

Step4: Calculate reactant enthalpy sum

$[2\times(-425.6) + 1\times(-795.8)] = -851.2 - 795.8 = -1647.0\ \text{kJ}$

Step5: Compute reaction enthalpy

$\Delta H^\circ_{\text{rxn}} = -1808.5 - (-1647.0) = -161.5\ \text{kJ}$ (rounded to nearest kJ)

Answer:

-162 kJ