Calculate the ∆H for the following reaction: 2H2O2 🡪 2H2O + O2

Question

You are given these two equations:

2H2  +  O2 🡪   2H2O  ∆H = -572 kJ

H2  + O2 🡪 H2O2  ∆H = -188 kJ 

Answer 1

See Hess' Law

2H2O2 ===> 2H2O + O2 ... TARGET EQUATION

We want to use the other equations that are given in order to make the target equation.

1) 2H2 + O2 ==> 2H2O ∆H = -572 kJ

2) H2 + O2 ==> H2O2 ∆H = -188 kJ

Reverse eq. 2 and multiply by 2 to get 2H2O2 on the left

2H2O2 ==> 2H2 + 2O2 ∆H = +376 kJ (change sign of ∆H and multiply by 2)

Copy eq.1

2H2 + O2 ==> 2H2O ∆H = -572

Add the two to get...

2H2O2 + 2H2 + O2 ==> 2H2 + 2O2 + 2H2O and cancel like terms on each side to get...

2H2O2 ==> 2H2O + O2 <==TARGET EQUATION

∆H = +376 kJ - 572 kJ = -196 kJ