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Use the standard enthalpies of formation for the reactants and products to solve for the [tex]\Delta H_{\text{rxn}}[/tex] for the following reaction. (The [tex]\Delta H_f[/tex] of [tex]\text{Ca(OH)}_2[/tex] is [tex]-986.09 \, \text{kJ/mol}[/tex] and liquid [tex]\text{H}_2\text{O}[/tex] is [tex]-285.8 \, \text{kJ/mol}[/tex].)

[tex]\[ \text{Ca (s)} + 2 \text{H}_2\text{O (l)} \rightarrow \text{Ca(OH)}_2\text{(s)} + \text{H}_2\text{(g)} \][/tex]

[tex]\[ \Delta H_{\text{rxn}} = \][/tex]

The reaction is [tex]\square[/tex]



Answer :

GinnyAnswer
To determine the change in enthalpy ([tex]\(\Delta H_{\text{rxn}}\)[/tex]) for the given chemical reaction:

[tex]\[ \text{Ca (s)} + 2\text{H}_2\text{O (l)} \rightarrow \text{Ca(OH)}_2 \text{(s)} + \text{H}_2 \text{(g)} \][/tex]

we will use the standard enthalpies of formation for the reactants and products. The standard enthalpies of formation ([tex]\(\Delta H_f\)[/tex]) are given as:

- [tex]\(\Delta H_f \ \text{of} \ \text{Ca(OH)}_2 \ \text{(solid)} = -986.09 \ \text{kJ/mol}\)[/tex]
- [tex]\(\Delta H_f \ \text{of} \ \text{H}_2\text{O (liquid)} = -285.8 \ \text{kJ/mol}\)[/tex]
- [tex]\(\Delta H_f \ \text{of} \ \text{H}_2 \ \text{(gas)} = 0 \ \text{kJ/mol}\)[/tex] (standard enthalpy of formation for [tex]\(\text{H}_2\)[/tex] gas is usually considered 0)
- [tex]\(\Delta H_f \ \text{of} \ \text{Ca (solid)} = 0 \ \text{kJ/mol}\)[/tex] (standard enthalpy of formation for calcium in its standard state is 0)

The standard enthalpy change of the reaction ([tex]\(\Delta H_{\text{rxn}}\)[/tex]) is determined by the difference between the sum of the standard enthalpies of formation of the products and the sum of the standard enthalpies of formation of the reactants:

[tex]\[ \Delta H_{\text{rxn}} = \sum \Delta H_f (\text{products}) - \sum \Delta H_f (\text{reactants}) \][/tex]

Step-by-Step Calculation:

1. Calculate the sum of the enthalpies of formation of the reactants:
[tex]\[ \Delta H_{\text{reactants}} = \Delta H_f \ (\text{Ca (solid)}) + 2 \times \Delta H_f \ (\text{H}_2\text{O (liquid)}) \][/tex]
Substituting the given values:
[tex]\[ \Delta H_{\text{reactants}} = 0 + 2 \times (-285.8) = 0 + (-571.6) = -571.6 \ \text{kJ/mol} \][/tex]

2. Calculate the sum of the enthalpies of formation of the products:
[tex]\[ \Delta H_{\text{products}} = \Delta H_f \ (\text{Ca(OH)}_2 \ (\text{solid})) + \Delta H_f \ (\text{H}_2 \ (\text{gas})) \][/tex]
Substituting the given values:
[tex]\[ \Delta H_{\text{products}} = -986.09 + 0 = -986.09 \ \text{kJ/mol} \][/tex]

3. Determine the enthalpy change of the reaction:
[tex]\[ \Delta H_{\text{rxn}} = \Delta H_{\text{products}} - \Delta H_{\text{reactants}} \][/tex]
Substituting the calculated sums:
[tex]\[ \Delta H_{\text{rxn}} = -986.09 - (-571.6) = -986.09 + 571.6 = -414.49 \ \text{kJ/mol} \][/tex]

Conclusion:

[tex]\[ \Delta H_{\text{rxn}} = -414.49 \ \text{kJ/mol} \][/tex]

The reaction is exothermic, as indicated by the negative value of [tex]\(\Delta H_{\text{rxn}}\)[/tex].

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