Given that H 2 ( g ) + F 2 ( g ) ⟶ 2 HF ( g ) Δ H ∘ rxn = − 546.6 kJ H2(g)+F2(g)⟶2HF(g)ΔHrxn°=−546.6 kJ 2 H 2 ( g ) + O 2 ( g )

Question

3 years ago

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Given that H 2 ( g ) + F 2 ( g ) ⟶ 2 HF ( g ) Δ H ∘ rxn = − 546.6 kJ H2(g)+F2(g)⟶2HF(g)ΔHrxn°=−546.6 kJ 2 H 2 ( g ) + O 2 ( g )

⟶ 2 H 2 O ( l ) Δ H ∘ rxn = − 571.6 kJ 2H2(g)+O2(g)⟶2H2O(l)ΔHrxn°=−571.6 kJ calculate the value of Δ H ∘ rxn ΔHrxn∘ for 2 F 2 ( g ) + 2 H 2 O ( l ) ⟶ 4 HF ( g ) + O 2 ( g )

Chemistry

2 answers:

ValentinkaMS [17] · Answer 1 · 2022-08-16T10:36:25+03:00

Answer:

ΔH° = -521,6kJ

Explanation:

It is possible to obtain the ΔH° of a reaction using Hess's law that consist in the algebraic sum of the ΔH° of semireactions.

For the semireactions:

<em>(1) </em>H₂(g) + F₂(g) ⟶ 2HF(g) ΔHrxn° = −546.6kJ

<em>(2) </em>2H₂(g) + O₂(g) ⟶ 2H₂O(l) Δ Hrxn° = −571.6kJ

The sum of 2×(1) - (2) gives:

The ΔH° for this reaction is:

ΔH° = -546,6kJ×2 - (-571,6kJ)

<em>ΔH° = -521,6kJ</em>

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I hope it helps!

Fudgin [204] · Answer 2 · 2022-08-16T12:48:28+03:00

ΔHrxn ° for 2 F₂ (g) + 2 H₂O (l) ⟶ 4HF (g) + O₂ (g): <u>-521.6 kJ </u>

<h3>Further explanation </h3>

The change in enthalpy in the formation of 1 mole of the elements is called enthalpy of formation

The enthalpy of formation measured in standard conditions (25 ° C, 1 atm) is called the standard enthalpy of formation (ΔHf °)

Determination of the enthalpy of formation of a reaction can be through a calorimetric experiment, based on the principle of Hess's Law, enthalpy of formation table, or from bond energy data

Delta H reaction (ΔH) is the amount of heat / heat change between the system and its environment

(ΔH) can be positive (endothermic = requires heat) or negative (exothermic = releasing heat)

The value of ° H ° can be calculated from the change in enthalpy of standard formation:

<h3>∆H ° rxn = ∑n ∆Hf ° (product) - ∑n ∆Hf ° (reactants) </h3>

ΔH∘rxn = ΔH∘f of the product (s) if ∆Hf ° (reactants) = 0

Based on the principle of Hess's Law, <em>the change in enthalpy of a reaction will be the same even though it is through several stages or ways </em>

Known ΔHrxn from reaction:

H₂ (g) + F₂ (g) ⟶2HF (g) ΔHrxn ° = −546.6 kJ reaction 1 (R1)

2H₂ (g) + O₂ (g) ⟶2H₂O (l) ΔHrxn ° = −571.6 kJ reaction 2 (R2)

ΔHrxn from reaction:

2 F₂ (g) + 2 H₂O (l) ⟶ 4HF (g) + O₂ (g) reaction 3 (R3)

Can be searched from ΔHrxn ° R1 and R2

From R3 it is known that the reaction coefficient of F₂ is 2, so we can multiply R1 by 2 (include ΔHrxn °)

2H₂ (g) + 2F₂ (g) ⟶4HF (g) ΔHrxn ° = −1093.2 kJ reaction 4 (R4)

R3 H₂O lies in the reactants, so that we can reverse R2,and so ΔHrxn ° is marked +

2H₂O (l) ⟶2H₂(g) + O₂ (g) ΔHrxn ° = + 571.6 kJ reaction 5 (R5)

We add R4 to R5 to get R3, by removing 2H₂ (g) because it is located in the reactants and products

2H₂ (g) + 2F₂ (g) ⟶4HF (g) ΔHrxn ° = −1093.2 kJ