24. Hydrogen peroxide can decompose to water and oxygen according to:
2H2O2 (l) + -----------> 2H2O(l)+ O2 (g) ^ H = -196 kJ
What is the enthalpy change accompanying the reaction of 5.00 g H2O2 (l) according to the above mentioned reaction ?
To find the enthalpy change accompanying the reaction of 5.00 g H2O2, we need to first calculate the number of moles of H2O2 using its molar mass.
1. Calculate the molar mass of H2O2:
H = 1.0079 g/mol x 2 = 2.0158 g/mol
O = 15.999 g/mol x 2 = 31.998 g/mol
Total molar mass = 2.0158 g/mol + 31.998 g/mol = 34.014 g/mol
2. Convert the mass of H2O2 to moles:
moles = mass / molar mass
moles = 5.00 g / 34.014 g/mol ≈ 0.147 moles
3. Use the balanced equation to determine the molar ratio between H2O2 and O2:
According to the balanced equation:
2 moles of H2O2 produce 1 mole of O2
4. Calculate the moles of O2 produced:
moles of O2 = 0.147 moles H2O2 x (1 mole O2 / 2 moles H2O2) = 0.0735 moles O2
5. Calculate the enthalpy change (ΔH) in kJ/mol of O2:
ΔH = (-196 kJ / 2 moles) = -98 kJ/mol
6. Calculate the enthalpy change accompanying the reaction of 5.00 g H2O2:
enthalpy change = ΔH x moles of O2
enthalpy change = -98 kJ/mol x 0.0735 moles O2 ≈ -7.21 kJ
Therefore, the enthalpy change accompanying the reaction of 5.00 g H2O2 is approximately -7.21 kJ.