calculate the change in enthalpy when 1.00g of hydrogen peroxide decomposes. H2O2(l) yields H2O(l)+ 1/2O2(g) (traingle)H- -98.2kj
I will be happy to critique your thinking on this.
help me?
You know the delta H is for one mole decomposition. So figure the fraction of 1gram/one mole of H2O2, and multiply that by the delta H given.
To calculate the change in enthalpy (ΔH) when 1.00g of hydrogen peroxide (H2O2) decomposes, we need to determine the fraction of 1 gram relative to one mole of H2O2 and then multiply it by the given ΔH value.
First, we need to find the molar mass of H2O2. Hydrogen has a molar mass of approximately 1g/mol, and oxygen has a molar mass of approximately 16g/mol. Since hydrogen peroxide (H2O2) contains two hydrogen atoms and two oxygen atoms, its molar mass is:
2(1g/mol) + 2(16g/mol) = 34g/mol
Next, we can calculate the fraction of 1 gram relative to one mole of H2O2 using the molar mass:
Fraction of 1 gram = 1g / (34g/mol) ≈ 0.0294 mol
Now, we can calculate the change in enthalpy using the fraction of 1 gram and the given ΔH value of -98.2 kJ/mol:
Change in enthalpy = 0.0294 mol * (-98.2 kJ/mol) ≈ -2.88 kJ
Therefore, the change in enthalpy when 1.00g of hydrogen peroxide decomposes is approximately -2.88 kJ.