What is the change in enthalpy for the following reaction?   2h2o2(aq)    2h2o(l)  + o2(g)  given: h2o:   ∆h= -242 kjh2o2:   ∆h= -609 kja.-286 kjb.  -572 kjc.  572 kjd.  286 kj

The given chemical reaction is:

The standard heats of formation of are:

Δ) = -285.8kJ/mol[/tex]

Δ) = -187.6 kJ/mol[/tex]

Calculating the change in heat:

Δ=∑ΔH-∑ΔH

                         = [{2 * (-285.8 kJ/mol)} -{2*(-187.6 kJ/mol)}]

                          = -196.4 kJ/mol

Therefore, the change in enthalpy for the given reaction is -196.4 kJ/mol

i believe it is a.) released into

explanation: