Question

The decomposition of hydrogen peroxide, H2O2, has been used to provide thrust in the control jets of various space vehicles. Using the data in Appendix G, determine how much heat is produced by the decomposition of exactly 1 mole of H2O2 under standard conditions. 2H2 O2 (l) ⟶ 2H2 O(g) + O2 (g)

Answer 1

Answer:

$\Delta H^0 _{reaction} = - 54.04 \ kJ/mol$

Explanation:

The given equation for the chemical reaction can be expressed as;

$2H_2O_{(l)} \to 2H_2O_{(g)} + O_{2(g)}$

Using Hess Law to determine how much heat is produced by the decomposition of exactly 1 mole of H2O2 under standard conditions; we have the expression showing the Hess Law as follows:

$\Delta H^0 _{reaction} = \sum n* \Delta H^0 _{products} - \sum n* \Delta H^0 _{reactants}$

At standard conditions;

the molar enthalpies of the given equation are as follows:

$\Delta H_2O_{(g)} =-241.82\ kJ/mol$

$\Delta H_ O_{2(g)} = 0 \ kJ/mol$

$\Delta H _{H_2O_{(l)}}= -187.78 \ kJ/mol$

Replacing them into above formula; we have:

$\Delta H^0 _{reaction} = (2*(-241.82\ kJ/mol) + 0 \ kJ/mol + (2 *(-187.78 \ kJ/mol))$

$\Delta H^0 _{reaction} =-108.08 \ kJ/mol$

The above is the amount of heat of formation for two moles of hydrogen peroxide; thus for 1 mole hydrogen peroxide ; we have :

$\Delta H^0 _{reaction} = \dfrac{-108.08 \ kJ/mol}{2}$

$\Delta H^0 _{reaction} = - 54.04 \ kJ/mol$

Hence; the heat produced after the decomposition of 1 mole of hydrogen peroxide is -54.04 kJ/mol