Did they provide you the enthalpies for those equations you listed above?
I found these from internet so I will be using these in the answer:
1) SrO + CO2 -> SrCO3 ∆H = +234 kJ
2) 2SrO -> 2Sr + O2 ∆H = +1184 kJ
3) 2SrCO3 -> 2Sr + 2C + 3O2 ∆H = +2440 kJ
Generally, speaking the definition of enthalpy of formation is the energy required to make some compound from the elements that make it up i.e.
for carbon dioxide that would be C (carbon) and O2 (oxygen)
This means you need to make the equation C(s) + O2(g) --> CO2(g), from your equations
This question if they are giving you enthalpies which I assume they did, it's a Hess' Law Question.
So if you look at equation 1 it has co2 as a reactant, but in our equation we want we need it as a product. This means you need to reverse the equation and its enthalpy sign will also be reversed
SrCO3 --> SrO + CO2 ∆H = +234 kJ ----> label this as 1a
Now if you look at equation 3. We have 2 moles of C but we want 1 mole in our final equation. So divide the whole equation by 2. This means the enthalpy is halved
SrCO3 --> Sr + C + 3/2 O2 ∆H = +1220 kJ
But we need to reverse this equation because we want C as a reactant in our final equation. Similarly, to above this reverse the enthalpy sign
Sr + C + 3/2 O2 --> SrCO3 ∆H = -1220 kJ ---> label this as 3a
Now add together 1a and 3a. This means the enthalpies will also add.
SrCO3 + Sr + C + 3/2 O2 --> SrO + CO2 + SrCO3 ∆H = -1220 + 234
Components on both sides of the equation cancel as it would similarly in a normal maths equation (i.e. if you had x+z + y = x + w + a,
the x bit would cancel, giving z + y = w + a)
SrCO3 + Sr + C + 3/2 O2 --> SrO + CO2 + SrCO3 ∆H = -986 kJ
Sr + C + 3/2 O2 --> SrO + CO2 ∆H = -986 kJ ---> label this as 4
Now If you look at equation 2 we need to use this equation to eliminate S and SrO from equation 4. To do this we need to first halve equation 2.
Which halves the enthalpy value
SrO -> Sr + 1/2 O2 ∆H = +592 kJ --> label this as 2a
Add equations 2a and 4, enthalpies also add
Sr + C + 3/2 O2 + SrO --> SrO + CO2 + Sr + 1/2 O2 ∆H = -986 + 592 kJ
Same as before things on left and right will cancel.
This means Sr and SrO cancel completely
This gives us C + 3/2 O2 --> CO2 + 1/2 O2 ∆H = -394 kJ
But note that you have oxygen on left and right. the equation above is similar to a maths equation of a + 3/2 b = c + 1/2 b
This simplified gives a + b = c
Where a represents C, b represents O2, c represents CO2
Therefore as we were aiming for we get
C + O2 --> CO2 ∆H = -394 kJ
