Answer :
To find the enthalpy change for the chemical reaction where hydrogen peroxide [tex]\(\text{H}_2\text{O}_2\)[/tex] decomposes into water and oxygen, we need to consider the bond energies of the reactants and the products.
1. Identify the bonds in the reactants [tex]\( \text{H}_2\text{O}_2 \)[/tex]:
[tex]\[ H - O - O - H \][/tex]
In one molecule of hydrogen peroxide, there are:
- 2 O-H bonds
- 1 O-O single bond
2. Determine the total bond energy of the reactants:
Using the given bond energies:
- Bond energy of [tex]\(O-H\)[/tex] is 459 kJ/mol
- Bond energy of [tex]\(O-O\)[/tex] single bond is 142 kJ/mol
Total bond energy for the reactants:
[tex]\[ \text{Total energy of reactants} = 2 \times 459 \, \text{kJ/mol} + 142 \, \text{kJ/mol} = 918 \, \text{kJ/mol} + 142 \, \text{kJ/mol} = 1060 \, \text{kJ/mol} \][/tex]
3. Identify the bonds in the products:
- For [tex]\( \text{H}_2 \)[/tex]:
[tex]\[ H - H \][/tex]
with a bond energy of 435 kJ/mol.
- For [tex]\( \text{O}_2 \)[/tex]:
[tex]\[ O = O \][/tex]
with a bond energy of 494 kJ/mol.
4. Determine the total bond energy of the products:
Using the given bond energies:
- Bond energy of [tex]\(H-H\)[/tex] is 435 kJ/mol
- Bond energy of [tex]\(O=O\)[/tex] is 494 kJ/mol
Total bond energy for the products:
[tex]\[ \text{Total energy of products} = 435 \, \text{kJ/mol} + 494 \, \text{kJ/mol} = 929 \, \text{kJ/mol} \][/tex]
5. Calculate the enthalpy change ([tex]\(\Delta H\)[/tex]):
Enthalpy change for the reaction is given by:
[tex]\[ \Delta H = \text{Total energy of products} - \text{Total energy of reactants} \][/tex]
Substituting the values:
[tex]\[ \Delta H = 929 \, \text{kJ/mol} - 1060 \, \text{kJ/mol} = -131 \, \text{kJ/mol} \][/tex]
Therefore, the enthalpy change for the chemical reaction is [tex]\(-131 \, \text{kJ/mol}\)[/tex]. Since none of the given options match exactly, it appears there was a misunderstanding in the given options. The closest value provided in the options should be selected, however, [tex]\(-131 \, \text{kJ/mol}\)[/tex] is the most accurate based on the calculations.
In this case, none of the options are correct. The correct calculated enthalpy change is [tex]\(-131 \, \text{kJ/mol}\)[/tex].
1. Identify the bonds in the reactants [tex]\( \text{H}_2\text{O}_2 \)[/tex]:
[tex]\[ H - O - O - H \][/tex]
In one molecule of hydrogen peroxide, there are:
- 2 O-H bonds
- 1 O-O single bond
2. Determine the total bond energy of the reactants:
Using the given bond energies:
- Bond energy of [tex]\(O-H\)[/tex] is 459 kJ/mol
- Bond energy of [tex]\(O-O\)[/tex] single bond is 142 kJ/mol
Total bond energy for the reactants:
[tex]\[ \text{Total energy of reactants} = 2 \times 459 \, \text{kJ/mol} + 142 \, \text{kJ/mol} = 918 \, \text{kJ/mol} + 142 \, \text{kJ/mol} = 1060 \, \text{kJ/mol} \][/tex]
3. Identify the bonds in the products:
- For [tex]\( \text{H}_2 \)[/tex]:
[tex]\[ H - H \][/tex]
with a bond energy of 435 kJ/mol.
- For [tex]\( \text{O}_2 \)[/tex]:
[tex]\[ O = O \][/tex]
with a bond energy of 494 kJ/mol.
4. Determine the total bond energy of the products:
Using the given bond energies:
- Bond energy of [tex]\(H-H\)[/tex] is 435 kJ/mol
- Bond energy of [tex]\(O=O\)[/tex] is 494 kJ/mol
Total bond energy for the products:
[tex]\[ \text{Total energy of products} = 435 \, \text{kJ/mol} + 494 \, \text{kJ/mol} = 929 \, \text{kJ/mol} \][/tex]
5. Calculate the enthalpy change ([tex]\(\Delta H\)[/tex]):
Enthalpy change for the reaction is given by:
[tex]\[ \Delta H = \text{Total energy of products} - \text{Total energy of reactants} \][/tex]
Substituting the values:
[tex]\[ \Delta H = 929 \, \text{kJ/mol} - 1060 \, \text{kJ/mol} = -131 \, \text{kJ/mol} \][/tex]
Therefore, the enthalpy change for the chemical reaction is [tex]\(-131 \, \text{kJ/mol}\)[/tex]. Since none of the given options match exactly, it appears there was a misunderstanding in the given options. The closest value provided in the options should be selected, however, [tex]\(-131 \, \text{kJ/mol}\)[/tex] is the most accurate based on the calculations.
In this case, none of the options are correct. The correct calculated enthalpy change is [tex]\(-131 \, \text{kJ/mol}\)[/tex].
