Question

Hydrogen peroxide, H2O2, is a colorless liquid whose solutions are used as a bleach and an antiseptic. H2O2 can be prepared in a process whose overall change is the following.
H2(g) + O2(g) H2O2(l)

Calculate the enthalpy change using the following data.
H2O2(l) H2O(l) + 1/2 O2(g) ÄH = -98.0 kJ
2 H2(g) + O2(g) 2 H2O(l) ÄH = -571.6 kJ

Answer 1

Step reactions Given:
H2O2(l) –> H2O(l) + 1/2 O2(g) ∆H = -98.0 kJ
2 H2(g) + O2(g) –> 2 H2O(l) ∆H = -571.6 kJ

Reverse the first reaction and change the sign of the ∆H. Divide all the coefficients of the 2nd reaction and the ∆H by 2:
H2O(l) + 1/2 O2(g) –> H2O2(l) ∆H = +98.0 kJ
H2(g) + 1/2 O2(g) –> H2O(l) ∆H = -285.8 kJ

If you add the last two reactions you get:
H2(g) + O2(g) –> H2O2(l) (∆Hrxn = Sum of the two ∆H’s)