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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
Original post by zara ijaz
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
Reply 2
Original post by charco
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)
Original post by zara ijaz
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)


So, using the definition of formation enthalpy, you can now write down the equation that represents the formation of lithium fluoride.
However, there is also a typo in both the question and the answer ...
Two reactions of iron with oxygen are shown.
(Total 1 mark)
Fe(s) + 1/2 O2(g) FeO(s) ΔH = 272 kJ mol–1
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
(edited 10 months ago)
Original post by student72876
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–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
Original post by charco
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


thankyouuu :smile:

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