Calculate the standard molar enthalpy of formation for na2o(s), given that the standard enthalpy of formation for na2o2(s) is -505 kj/mol and the enthalpy change for the following reaction is -89.0 kj/mol. na2o(s) + 1/2 o2(g) → na2o2(s)

- 416 kJ/mol

Explanation:

The standard enthalpy of the reaction (Δ H ∘ rxn) is independent of the pathway, so it can be calculated by the enthalpy of formation of the reactants and the products:

Δ H ∘ rxn = ∑n*Δ H ∘f products - ∑n*Δ H ∘f reactants

Where n is the number of moles in the balanced reaction. So, for the reaction given:

Na₂O(s) + 1/2O₂(g) → Na₂O₂(s)

Because O₂ is formed by only one elements, its Δ H ∘f is 0 kJ/mol:

-89.0 = (1*(-505) - (1*Δ H ∘fNa₂O)

Δ H ∘fNa₂O = -505 + 89

Δ H ∘fNa₂O = - 416 kJ/mol

first let us lay out the formula for potential

and kinetic energy:

pe = mgh

ke = (mv^2)/2

where: m=mass, v=velocity, g=gravitational

acceleration, h=height

calculating for mass using the known values of

pe:

340j = m (9.81 m/s^2) (50ft) (1m/3.28ft)

m = 2.27 kg

the law of conservation of energy states that

energy is neither created nor destroyed. therefore, the change in kinetic

energy from 50 ft to 35 ft would be equal to the change in potential energy

from 50 ft to 35 ft. however it would be opposite in signs since one is losing

while other is gaining.

ke(35ft) – ke(50ft) = - [ pe(35ft) – pe(50ft]

ke(50ft) = 0 since the brick is initially at rest

ke(35ft) = 340j - 2.27kg (9.81 m/s^2) (35ft)

(1m/3.28 ft)

ke(35ft)

= 102.38 j