Question

The standard state molar enthalpy of reaction for the formation of water as both a liquid and a gas have been measured: H_2 (g)+1/2 O_2 (g) → H_2 O (l) ∆H^(∘ )= -285.83 kJ⁄mol H_2 O H_2 (g)+1/2 O_2 (g) → H_2 O (g) ∆H^(∘ )= -241.82 kJ⁄mol H_2 O Use these data and Hess’s law to calculate ∆H^(∘ ) for the following reaction: H_2 O (l)→ H_2 O (g

Answer 1

Answer:

mark ME BRAINLIESTS..................

Answer 2

Answer:

Given = H_2 (g)+1/2 O_2 (g) → H_2 O (l) ∆H^(∘ )= -285.83 kJ⁄mol

            H_2 O H_2 (g)+1/2 O_2 (g) → H_2 O (g) ∆H^(∘ )= -241.82 kJ⁄mol

to find = ∆H^

solution =

H₂ + 1/2 O₂  →  H₂O ; ΔH = -285.77KJ/Mol........1

H₂ + 1/2 O₂  → H₂O ; ΔH = -241.84 KJ/mol..........2

H₂O  → H₂O ; ΔH =?

for calculating this ΔH

subtract the equation 1 from 2

ΔH = -241.84 - (- 285.77)

ΔH = +43.93KJ/mol