Question

The decomposition of hydrogen peroxide, H2O2, has been used to provide thrust in the control jets of various space vehicles. Using the Supplemental Data, determine how much heat (in kJ) is produced by the decomposition of 1.05 mol of H2O2 under standard conditions.
2 H2O2(l) → 2 H2O(g) + O2(g)

Answer 1

Answer:

The heat released is 56.7 kJ.

Explanation:

To solve this problem, first we need to find out the standard enthalpy of reaction, that is, the energy released at constant pressure in standard conditions (P=1bar, T=298.15K). We can find it using the expression:

ΔH°r = Σn(p).ΔH°f(p) - Σn(r).ΔH°f(r)

where,

n refers to the number of moles of reactants and products in the balanced equation

ΔH°f refers to standard enthalpies of formation (which can be found in tables).

Given the equation:

2 H₂O₂(l) → 2 H₂O(g) + O₂(g)

We can replace with the proper data in the equation:

ΔH°r = Σn(p).ΔH°f(p) - Σn(r).ΔH°f(r)

ΔH°r= [2 mol . ΔH°f H₂O(g) + 1 mol . ΔH°f O₂(g)] - [2 mol .  ΔH°f H₂O₂(l)]

ΔH°r= [2 mol . (-241.8 kJ/mol) + 1 mol . 0 kJ/mol] - [2 mol . (-187.8 kJ/mol)]

ΔH°r = -108.0 kJ

Since enthalpy is an extensive property, it depends on the amount of reagents. In this case, 108.0 kJ of heat are released every 2 moles of H₂O₂(l) decomposed. Then, for 1.05 mol of H₂O₂(l):

$1.05 mol.\frac{-108.0kJ}{2mol} =-56.7kJ$

By convention, the negative sign means that heat is released.