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 )

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

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:

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

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

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

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

 +

2F₂ (g) + 2H₂O (l) ⟶ 4HF (g) + O₂ (g) ΔHrxn ° = -521.6 kJ

Delta H solution

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an exothermic reaction

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as endothermic or exothermic

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an exothermic dissolving process

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the answer is:  

line graphs show patterns in data with variables that change over time; whereas, circle graphs show data as parts out of a whole (such as out of 100%).

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