Data:

2Ba(s) + O2(g) ¡æ 2BaO(s) ∆H¡Æ = -1107.0 KJ
Ba(s) + CO2(g) + ¨ö O2(g) ¡æ BaCO3(s) ∆H¡Æ = -822.5 KJ
Given the data above, calculate the ∆H¡Æ for the reaction below.
BaCO3(s) ¡æ BaO(s) + CO2(g)

Your post is garbled. I think you means this

2BaO + O2 ==> 2BaO
Ba + CO2 + 1/2 O2 ==> BaCO3
============================
Take 1/2 equation 2 and add to the reverse of equation 2 and see if that isn't the equation you want.
For delta H, take 1/2 DH(eqn 1) and add to the negative of DH(eqn 2).

thank you

Your post is garbled. I think you means this

2BaO + O2 ==> 2BaO
Ba + CO2 + 1/2 O2 ==> BaCO3
============================
Take 1/2 equation 2 and add to the reverse of equation 2 and see if that isn't the equation you want.
For delta H, take 1/2 DH(eqn 1) and add to the negative of DH(eqn 2).
You can see I made a typo. That sentence should read "Take 1/2 equation 1 and add to the reverse of equation 2...." Sorry about the typo but I think you got the correct message .

To calculate the ΔH° for the reaction BaCO3(s) → BaO(s) + CO2(g), we can use the Hess's Law, which states that the enthalpy change for a reaction is the same whether it occurs in one step or several steps.

We can use the given data for the two reactions to determine the ΔH° for the desired reaction.

First, let's reverse the second reaction:

BaCO3(s) → BaO(s) + CO2(g) (reversed)

Now, let's multiply the first reaction by two:

2Ba(s) + O2(g) → 2BaO(s)

The enthalpy change for the reversed second reaction would be the negative of its original ΔH value:

ΔH° (reversed second reaction) = -(-822.5 KJ) = +822.5 KJ

Now, we add these two reactions together:

2Ba(s) + O2(g) + BaCO3(s) → 2BaO(s) + CO2(g) + BaCO3(s)

Notice that BaCO3(s) appears on both sides of the equation, so it can be canceled out:

2Ba(s) + O2(g) → 2BaO(s) + CO2(g)

The enthalpy change for the overall reaction is the sum of the ΔH° values for the individual reactions:

ΔH° (overall reaction) = ΔH° (first reaction) + ΔH° (reversed second reaction)
= -1107.0 KJ + 822.5 KJ
= -284.5 KJ

Therefore, the ΔH° for the reaction BaCO3(s) → BaO(s) + CO2(g) is -284.5 KJ.