Energetics Alevel chem multiple choice

Q5.
Which reaction has an enthalpy change equal to the standard enthalpy of formation of lithium fluoride?
A
Li(g) + F2(g) → LiF(s)
B Li+(g) + F–(g) → LiF(s)
C Li+(aq) + F–(aq) → LiF(s)
D
Q6.
Two reactions of iron with oxygen are shown.
(1) (Total 13 marks)
Li(s) + F2(g) → LiF(s)
(Total 1 mark)
Fe(s) + O2(g) → FeO(s)
ΔH = – 272 kJ mol–1 ΔH = – 822 kJ mol–1
2 Fe(s) + O2(g) → Fe2O3(s)
What is the enthalpy change, in kJ mol–1, for this reaction?
2 FeO(s) + O2(g) → Fe2O3(s)
A +550 B –278 C –1094 D –1372

The first question requires you to know the definition of formation enthalpy. If you don't know it, look it up.

The second question contains a typo: Fe(s) + O₂(g) → FeO(s) is not balanced

Enthalpy of formation is the enthalpy change when 1 mole of a substance is formed from its constituent elements but I’m not sure how that applies to the question

Also for the second one it’s :
Fe(s) + 1/2O2 (g) -> FeO(s)

Two reactions of iron with oxygen are shown.
(Total 1 mark)
equation 1: Fe(s) + 1/2 O2(g) → FeO(s) ΔH = – 272 kJ mol–1
equation 2: 2 Fe(s) + 3/2 O2(g) → Fe2O3(s) ΔH = – 822 kJ mol–1
What is the enthalpy change, in kJ mol–1, for this reaction?
2 FeO(s) + 1/2 O2(g) → Fe2O3(s)
A +550 B –278 C –1094 D –1372
that is the actual question without typos, can someone explain why the answer is B

In order to generate the required equation, you must take equation 1:
Fe(s) + 1/2 O₂(g) → FeO(s)
reverse it and multiply by 2 to give:
2FeO(s) → 2Fe(s) + O₂(g) ΔH = +544 kJ mol^–1

Then add this equation to equation 2:
2FeO(s) → 2Fe(s) + O₂(g) ΔH = +544 kJ mol^–1t
2 Fe(s) + 3/2 O₂(g) → Fe₂O₃(s) ΔH = – 822 kJ mol^–1
---------------------------------------------- add
2FeO(s) + 1/2 O₂(g) → Fe₂O₃(s) ΔH = +544 - 822 = -278 kJ

In order to generate the required equation, you must take equation 1:
Fe(s) + 1/2 O₂(g) → FeO(s)
reverse it and multiply by 2 to give:
2FeO(s) → 2Fe(s) + O₂(g) ΔH = +544 kJ mol^–1
Then add this equation to equation 2:
2FeO(s) → 2Fe(s) + O₂(g) ΔH = +544 kJ mol^–1
2 Fe(s) + 3/2 O₂(g) → Fe₂O₃(s) ΔH = – 822 kJ mol^–1
---------------------------------------------- add
2FeO(s) + 1/2 O₂(g) → Fe₂O₃(s) ΔH = +544 - 822 = -278 kJ

In order to generate the required equation, you must take equation 1:
Fe(s) + 1/2 O₂(g) → FeO(s)
reverse it and multiply by 2 to give:
2FeO(s) → 2Fe(s) + O₂(g) ΔH = +544 kJ mol^–1
Then add this equation to equation 2:
2FeO(s) → 2Fe(s) + O₂(g) ΔH = +544 kJ mol^–1
2 Fe(s) + 3/2 O₂(g) → Fe₂O₃(s) ΔH = – 822 kJ mol^–1
---------------------------------------------- add
2FeO(s) + 1/2 O₂(g) → Fe₂O₃(s) ΔH = +544 - 822 = -278 kJ

hi, why would you reverse equation 1's sign?

The law of conservation of energy. If you reverse an equation you change the sign.
The same energy is released going one way as is absorbed going the other way.

thank you! would this apply to any of this type of question? e.g. dissociation, enthalpy changes

In order to generate the required equation, you must take equation 1:
Fe(s) + 1/2 O₂(g) → FeO(s)
reverse it and multiply by 2 to give:
2FeO(s) → 2Fe(s) + O₂(g) ΔH = +544 kJ mol^–1
Then add this equation to equation 2:
2FeO(s) → 2Fe(s) + O₂(g) ΔH = +544 kJ mol^–1
2 Fe(s) + 3/2 O₂(g) → Fe₂O₃(s) ΔH = – 822 kJ mol^–1
---------------------------------------------- add
2FeO(s) + 1/2 O₂(g) → Fe₂O₃(s) ΔH = +544 - 822 = -278 kJ

Why do you reverse the first equation and then add it to the 2nd