Question

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)

Answer 1

Your post is garbled. I think you means this
2BaO + O2 ==> 2BaO
Ba + CO2 + 1/2 O2 ==> BaCO3
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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).

Answer 2

thank you

Answer 3

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).
<b>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 .</b>