Question

Given that H 2 ( g ) + F 2 ( g ) ⟶ 2 HF ( g ) Δ H ∘ rxn = − 546.6 kJ 2 H 2 ( g ) + O 2 ( g ) ⟶ 2 H 2 O ( l ) Δ H ∘ rxn = − 571.6 kJ calculate the value of Δ H ∘ rxn for 2 F 2 ( g ) + 2 H 2 O ( l ) ⟶ 4 HF ( g ) + O 2 ( g )

Answer 1

Answer: - 521.6kJ

Step-by-step explanation:

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

`H_2(g)+F_2(g)\rightarrow 2HF(g)`
`\Delta H^0_1=-546.6kJ` (1)

`2H_2(g)+O_2(g)\rightarrow 2H_2O(l)`
`\Delta H^0_2=-571.6kJ` (2)

The final reaction is:

`2F_2(g)+2H_2O(l)\rightarrow 4HF(g)+O_2(g)`
`\Delta H^0_3=?` (3)

By multipling (1) by 2

`2H_2(g)+2F_2(g)\rightarrow 4HF(g)`
`\Delta H^0_1'=-2* 546.6kJ=-1093.2kJ` (1')

Subtracting (2) from (1')

`2F_2(g)+2H_2O(l)\rightarrow 4HF(g)+O_2(g)`
`\Delta H^0_3=?` (3)

Hence
`\Delta H^0_3=\Delta H^0_1'-\Delta H^0_2=-1093.2-(-571.6)kJ=-521.6kJ`.

The enthalpy for the reaction is -521.6kJ