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Sagot :
To calculate the standard enthalpy change of the reaction [tex]\(\Delta H_{\text{rxn}}^0\)[/tex] for the reaction:
[tex]\[2 \text{H}_2\text{O}_2(l) \rightarrow 2 \text{H}_2\text{O}(l) + \text{O}_2(g),\][/tex]
we need to apply the following formula:
[tex]\[ \Delta H_{\text{rxn}}^0 = \sum \Delta H_f^0 (\text{products}) - \sum \Delta H_f^0 (\text{reactants}). \][/tex]
1. Identify the standard enthalpies of formation ([tex]\(\Delta H_f^0\)[/tex]) for each compound involved:
- [tex]\(\Delta H_f^0 \text{(H}_2\text{O}_2(l)) = -187.8 \ \text{kJ/mol}\)[/tex]
- [tex]\(\Delta H_f^0 \text{(H}_2\text{O}(l)) = -285.8 \ \text{kJ/mol}\)[/tex]
- [tex]\(\Delta H_f^0 \text{(O}_2(g)) = 0 \ \text{kJ/mol}\)[/tex]
2. Calculate the sum of [tex]\(\Delta H_f^0\)[/tex] for the products:
- The products are [tex]\(2 \text{H}_2\text{O}(l)\)[/tex] and [tex]\(\text{O}_2(g)\)[/tex].
- Therefore:
[tex]\[ \sum \Delta H_f^0 (\text{products}) = 2 \times \Delta H_f^0 \text{(H}_2\text{O}(l)) + 1 \times \Delta H_f^0 \text{(O}_2(g)). \][/tex]
- Substituting the values:
[tex]\[ \sum \Delta H_f^0 (\text{products}) = 2 \times (-285.8 \ \text{kJ/mol}) + 1 \times (0 \ \text{kJ/mol}) = -571.6 \ \text{kJ/mol}. \][/tex]
3. Calculate the sum of [tex]\(\Delta H_f^0\)[/tex] for the reactants:
- The reactants are [tex]\(2 \text{H}_2\text{O}_2(l)\)[/tex].
- Therefore:
[tex]\[ \sum \Delta H_f^0 (\text{reactants}) = 2 \times \Delta H_f^0 \text{(H}_2\text{O}_2(l)). \][/tex]
- Substituting the value:
[tex]\[ \sum \Delta H_f^0 (\text{reactants}) = 2 \times (-187.8 \ \text{kJ/mol}) = -375.6 \ \text{kJ/mol}. \][/tex]
4. Calculate [tex]\(\Delta H_{\text{rxn}}^0\)[/tex]:
[tex]\[ \Delta H_{\text{rxn}}^0 = \sum \Delta H_f^0 (\text{products}) - \sum \Delta H_f^0 (\text{reactants}). \][/tex]
- Substituting the values:
[tex]\[ \Delta H_{\text{rxn}}^0 = -571.6 \ \text{kJ/mol} - (-375.6 \ \text{kJ/mol}) = -571.6 \ \text{kJ/mol} + 375.6 \ \text{kJ/mol} = -196.0 \ \text{kJ/mol}. \][/tex]
Therefore, the standard enthalpy change for the reaction [tex]\(2 \text{H}_2\text{O}_2(l) \rightarrow 2 \text{H}_2\text{O}(l) + \text{O}_2(g)\)[/tex] is:
[tex]\[ \Delta H_{\text{rxn}}^0 = -196.0 \ \text{kJ/mol}. \][/tex]
So, the correct answer is:
a. [tex]\(-196.4 \ \text{kJ/mol}\)[/tex]
(Note: The exact number is [tex]\(-196.0 \ \text{kJ/mol}\)[/tex], but considering the options given, [tex]\( \)[/tex] matches closest to the abstract value.)
[tex]\[2 \text{H}_2\text{O}_2(l) \rightarrow 2 \text{H}_2\text{O}(l) + \text{O}_2(g),\][/tex]
we need to apply the following formula:
[tex]\[ \Delta H_{\text{rxn}}^0 = \sum \Delta H_f^0 (\text{products}) - \sum \Delta H_f^0 (\text{reactants}). \][/tex]
1. Identify the standard enthalpies of formation ([tex]\(\Delta H_f^0\)[/tex]) for each compound involved:
- [tex]\(\Delta H_f^0 \text{(H}_2\text{O}_2(l)) = -187.8 \ \text{kJ/mol}\)[/tex]
- [tex]\(\Delta H_f^0 \text{(H}_2\text{O}(l)) = -285.8 \ \text{kJ/mol}\)[/tex]
- [tex]\(\Delta H_f^0 \text{(O}_2(g)) = 0 \ \text{kJ/mol}\)[/tex]
2. Calculate the sum of [tex]\(\Delta H_f^0\)[/tex] for the products:
- The products are [tex]\(2 \text{H}_2\text{O}(l)\)[/tex] and [tex]\(\text{O}_2(g)\)[/tex].
- Therefore:
[tex]\[ \sum \Delta H_f^0 (\text{products}) = 2 \times \Delta H_f^0 \text{(H}_2\text{O}(l)) + 1 \times \Delta H_f^0 \text{(O}_2(g)). \][/tex]
- Substituting the values:
[tex]\[ \sum \Delta H_f^0 (\text{products}) = 2 \times (-285.8 \ \text{kJ/mol}) + 1 \times (0 \ \text{kJ/mol}) = -571.6 \ \text{kJ/mol}. \][/tex]
3. Calculate the sum of [tex]\(\Delta H_f^0\)[/tex] for the reactants:
- The reactants are [tex]\(2 \text{H}_2\text{O}_2(l)\)[/tex].
- Therefore:
[tex]\[ \sum \Delta H_f^0 (\text{reactants}) = 2 \times \Delta H_f^0 \text{(H}_2\text{O}_2(l)). \][/tex]
- Substituting the value:
[tex]\[ \sum \Delta H_f^0 (\text{reactants}) = 2 \times (-187.8 \ \text{kJ/mol}) = -375.6 \ \text{kJ/mol}. \][/tex]
4. Calculate [tex]\(\Delta H_{\text{rxn}}^0\)[/tex]:
[tex]\[ \Delta H_{\text{rxn}}^0 = \sum \Delta H_f^0 (\text{products}) - \sum \Delta H_f^0 (\text{reactants}). \][/tex]
- Substituting the values:
[tex]\[ \Delta H_{\text{rxn}}^0 = -571.6 \ \text{kJ/mol} - (-375.6 \ \text{kJ/mol}) = -571.6 \ \text{kJ/mol} + 375.6 \ \text{kJ/mol} = -196.0 \ \text{kJ/mol}. \][/tex]
Therefore, the standard enthalpy change for the reaction [tex]\(2 \text{H}_2\text{O}_2(l) \rightarrow 2 \text{H}_2\text{O}(l) + \text{O}_2(g)\)[/tex] is:
[tex]\[ \Delta H_{\text{rxn}}^0 = -196.0 \ \text{kJ/mol}. \][/tex]
So, the correct answer is:
a. [tex]\(-196.4 \ \text{kJ/mol}\)[/tex]
(Note: The exact number is [tex]\(-196.0 \ \text{kJ/mol}\)[/tex], but considering the options given, [tex]\( \)[/tex] matches closest to the abstract value.)
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