Question

The decomposition of hydrogen peroxide, H2O2, has been used to provide thrust in the control jets of various space vehicles. Determine how much heat (in kJ) is produced by the decomposition of 1.71 mol of H2O2 under standard conditions.

Answer 1

`Q=-361.56kJ`

Step-by-step explanation:

Hello,

In this case, the decomposition of hydrogen peroxide is given by:

`2H_2O_2\rightarrow 2H_2O+O_2`

Which occurs in gaseous phase, therefore the enthalpy of reaction is:

`\Delta _rH=2\Delta _fH_(H_2O)-2\Delta _fH_(H_2O_2)`

Oxygen is not included as it is a pure element. The enthalpies of formation for both hydrogen peroxide and water are -136.11 and -241.83 kJ/mol respectively, so we compute the enthalpy of reaction:

`\Delta _rH=2(-241.83kJ/mol)-2(-136.11kJ/mol)=-211.44kJ/mol`

Then, the total heat that is released for 1.71 mol of hydrogen peroxide is:

`Q=n*\Delta _rH=1.71mol*-211.44kJ/mol\\\\Q=-361.56kJ`

Whose sign means a released heat.

Regards.