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

H2(g) + O2(g) → H2O2(ℓ)

Calculate the enthalpy change using the following data:

2H2O2(ℓ) → 2H2O(ℓ) + O2(g); ∆H = -196.0 kJ

H2(g) + 1/2O2(g) → H2O(l ); ∆H = -285.8 kJ

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

H2(g) + O2(g) → H2O2(ℓ)

Calculate the enthalpy change using the following data:

2H2O2(ℓ) → 2H2O(ℓ) + O2(g); ∆H = -196.0 kJ

H2(g) + 1/2O2(g) → H2O(l ); ∆H = -285.8 kJ

anonymous · Answer

I thought it would be (196.0)/2 + (-285.8)

anonymous · Answer

-196*

anonymous · Answer

I think it is (196.0kj/2)+(-285.8kj)

anonymous · Answer

196 or -196?

anonymous · Answer

196 becayse you need to reverse the first reaction so the H

anonymous · Answer

becomes positive

nikato · Answer

Yup @rzrhdes ur first one was correct

jfraser · Answer

the first answer is correct. Flip the first reaction backwards and cut it in half, so the $\Delta H$ becomes positive and halved. Then add the two reactions together