Use thermodynamic data to calculate the standard enthalpy of decomposition of liquid hydrogen peroxide into water and oxygen gas: H2O2(l)-->H2O(g) + 1/2 O2(g).
Calculate the energy released by the decomposition of 1.0 g of 90% hydrogen peroxide solution. Molar mass of hydrogen peroxide = 34.0 g/mol. Estimate the final temperature of the product gases of the decomposition. Assume that the constant-volume heat capacities of water and oxygen are respectively 35, and 25 J/mol/K.
enthalpy of H2O2(l) = -187.8 kJ/mol
enthalpy of H20(g) = -241.8 kJ/mol
enthalpy of O2(g) = 0 kJ/mol
Since mass of H2O2 = 1.0*90/100 = 0.9 gm
The number of moles of H2O2 = 0.9/34 = 0.02647 mol
This is because 1 mole of H2O2 gives out 1 mole of H2O and 1/2 mole of O2.
Hence, 0.02647 mol of H2O2 will produce 0.02647 mol ...
The solution uses thermodynamic data to find the standard enthalpy of decomposition of liquid hydrogen peroxide into water and oxygen gas. The solution also calculates the energy released and the final temperature from the decomposition.