Given that 2Al(s) + 3 2 O2(g) → Al2O3(s) ΔH o rxn = −1601 kJ/mol and
2Fe(s) + 3 2 O2(g) → Fe2O3(s)
ΔH o rxn = −821 kJ/mol
calculate the standard enthalpy change for the following reaction:
2Al(s) + Fe2O3(s) → 2Fe(s) + Al2O3(s)

ΔH°rxn = -780kJ

Explanation:

It is possible to find enthalpy change of a reaction by the sum of analogue reactions (Hess's law).

Using the reactions:

1. 2Al(s) + 3/2O2(g) → Al2O3(s)  ΔH°rxn = −1601 kJ/mol

2. 2Fe(s) + 3/2O2(g) → Fe2O3(s)  ΔH°rxn = −821 kJ/mol

1 - 2:

2Al(s) + Fe2O3(s) → Al2O3(s) + 2Fe(s)

ΔH°rxn = ΔH°rxn(1) - ΔH°rxn(2)

ΔH°rxn = −1601 kJ/mol - (-821 kJ/mol)

ΔH°rxn = -780kJ

(newtons first law)

conservation of momentum law states :

momentum before collision = momentum after collision

momentum p (kg-m/s)= mass * velocity

say moving car=1 000 kg, velocity = 10 m/s then p = 1 000*10= 10 000 kg-m/s

say still car = 2 000 kg, velocity = 0 m/s then p = 2 000* 0 = 0 kg-m/s

total momentum prior to collision = 10 000 + 0 = 10 000 kg-m/s

momentum after collision = 10 000 kg-m/s

mass now = 1 000 kg+2 000 kg = 3 000 kg

10 000 = 3 000 kg* velocity m/s

10 000/3 000 = velocity

3.33 m/s = velocity after collision

an impulse is a force applied over time , momentum is exclusive of external forces and a perfect collision is implied