Question

What is the change in enthalpy for the following reaction?

2H2O2(aq) 2H2O(l) + O2(g)

Given:
H2O: ∆H= -242 kJ
H2O2: ∆H= -609 kJ

Answer 1

The change in enthalpy is determined by the formula:

`\Delta H_(reaction) = \Sigma \Delta H_(products) - \Sigma \Delta H_(reactants)`

The reaction is:

`2H_2O_2(aq)\rightarrow 2H_2O(l)+O_2(g)`

Given:

`\Delta H_(H_2O) = -242 kJ`

`\Delta H_(H_2O_2) = -609 kJ`

The change in enthalpy for the reaction is calculated as:

`\Delta H_(reaction) = (2H_(H_2O)+H_(O_2)) - 2H_(H_2O_2)`

`H_(O_2) = 0 kJ` (as it is in its standard state)

Substituting the values:

`\Delta H_(reaction) = (2* (-242)+0) - 2* (-609)`

`\Delta H_(reaction) = -484 kJ + 1218 kJ = 734 kJ`

Hence, change in enthalpy for the reaction is
`734 kJ`.

Answer 2

Answer : The reaction will be
`2H_(2)O_(2) ----\ \textgreater \ 2H_(2)O + O_(2)`.

Now as the given enthalpy of
`H_(2)O` is -242 KJ
and the enthalpy of
`H_(2)O_(2)` is -609 KJ.

According to the formula of Hess's summation, ΔHreaction= (ΣΔHproducts)-(ΣΔHreactants)

Therefore, ΔHreaction = [
`H_(2)O` +
`O_(2)`] -
`H_(2)O_(2)`

on substituting the values and omitting the value for oxygen as it is in gaseous state we get,
ΔHreaction = [0 + (2 X -242)] - [2 X -609]

on solving we get, ΔHreaction = 734 KJ.