Hydrogen peroxide decomposes according to the following:
H2O2--> H2O(l) + 1/2O2(g); deltaH = -98.2kJ
Calculate the change in enthalpy, deltaH, when 1.00g of Hydrogen peroxide decomposes
98.2 kJ is released for 1 mole H2O2 (and 1 mole is 34 g approximately. You can do it exactly.) Now, how much of a mole is 1 g H2O2? So 98.2 kJ x (1 g/34 g) = ?? kJ energy released for 1 g Hydrogen.
To calculate the change in enthalpy (ΔH) when 1.00g of hydrogen peroxide decomposes, we need to first convert the given mass of hydrogen peroxide to moles. Here's how to do it:
1. Determine the molar mass of hydrogen peroxide (H2O2):
The molar mass of hydrogen is 1.01 g/mol, and there are two hydrogen atoms in hydrogen peroxide. The molar mass of oxygen is 16.00 g/mol, and there are two oxygen atoms in hydrogen peroxide. Therefore, the molar mass of hydrogen peroxide is 1.01 g/mol (hydrogen) + 16.00 g/mol (oxygen) + 16.00 g/mol (oxygen) = 34.02 g/mol.
2. Convert the given mass of hydrogen peroxide to moles:
Using the molar mass calculated in step 1, divide the given mass (1.00 g) by the molar mass (34.02 g/mol):
moles = mass / molar mass
moles = 1.00 g / 34.02 g/mol
moles ≈ 0.0294 mol
Now that we have the number of moles of hydrogen peroxide, we can calculate the change in enthalpy using the stoichiometry of the reaction.
3. Use the stoichiometry of the reaction to find the change in enthalpy:
According to the balanced equation, 1 mole of hydrogen peroxide produces 1 mole of water and 1/2 mole of oxygen gas. Therefore, the molar ratio of hydrogen peroxide to the change in enthalpy is 1:1.
ΔH = (ΔH of the reaction) × (moles of H2O2)
ΔH = -98.2 kJ/mol × 0.0294 mol
Calculating this gives us:
ΔH ≈ -2.88 kJ
Therefore, the change in enthalpy (ΔH) when 1.00g of hydrogen peroxide decomposes is approximately -2.88 kJ.