Chemistry - Thermochemistry Question with Solution | TestHub
The standard molar enthalpies of formations of H2O(l) and H2O2(l) are -286 and -188 kJ/mol, respectively. Molar enthalpies of vaporization of H2O and H2O2 are 44 and 53 kJ, respectively. The bond dissociation enthalpy of O2(g) is 498 kJ/mol. Calculate the bond dissociation enthalpy (in kJ/mol) of O-O bond in H2O2, assuming that the bond dissociation enthalpy of O-H bond is same in both H2O and H2O2.
Answer:
Solution:
The standard enthalpy of formation for gaseous water, Δf H(H2O, g), is calculated as -286 + 44 = -242 kJ/mol. For gaseous hydrogen peroxide, Δf H(H2O2, g), it is -188 + 53 = -135 kJ/mol.
Consider the reaction:
H2O2(g) tarrow H2O(g) + 1/2O2(g)
The enthalpy change for this reaction, Δ H, is calculated as -242 - (-135) = -107 kJ.
In this reaction, the bonds broken are one O–O bond (let its energy be x) and two O–H bonds. The bonds formed are two O–H bonds and one-half of an O=O bond, which has an energy of 249 kJ/mol. The O–H bond terms cancel out. Therefore, we can set up the equation:
x - 249 = -107
Solving for x, we find x = 142 kJ/mol.
Thus, the bond energy for the O–O bond is 142 kJ/mol.
The answer is 0142.
