Find DHo for BaCO3(s) --> BaO(s) + CO2(g) given

2 Ba(s) + O2(g) --> 2 BaO(s) DHo = -1107.0 kJ
Ba(s) + CO2(g) + 1/2 O2(g) --> BaCO3(s) DHo = -822.5 kJ

1. -1929.5 kJ
2. 537.0 kJ
3. 269.0 kJ
4. -284.5 kJ
5. -1376.0 kJ

i got 537.0 kJ....is that right :S
HELP PLZZ FAST

I was working on it when the phone range about an hour ago.

Use equation 1 as is. Reverse equation 2, multiply it by 2 (change the sign of delta H for equation 2 also and multiply it by 2) then add the two equtions. That will give you twice the equation you want, so divide all of the coefficients by 2 and divide delta H sum by 2. That should be the correct answer.

i got 269 kJ

but im not sure...it would be better if a professor confirned it

what do you think DrBob?

269 it is.

To find the enthalpy change (ΔH) for the reaction BaCO3(s) --> BaO(s) + CO2(g), you can use Hess's Law. According to Hess's Law, if a reaction can be expressed as a sum of two or more other reactions, then the enthalpy change of the overall reaction is equal to the sum of the enthalpy changes of the individual reactions.

Considering the given reactions:
1. 2 Ba(s) + O2(g) --> 2 BaO(s) ΔH₁ = -1107.0 kJ
2. Ba(s) + CO2(g) + 1/2 O2(g) --> BaCO3(s) ΔH₂ = -822.5 kJ

You need to reverse the first reaction to make it align with the desired reaction. Multiply the first reaction by -1/2 to cancel out the appearance of BaO(s), as you have it in the desired reaction. This gives us:

-1/2 (2 Ba(s) + O2(g) --> 2 BaO(s)) ΔH₃ = 1/2(1107.0 kJ) = -553.5 kJ

By adding reactions 2 and 3, you get the desired reaction:

BaCO3(s) --> BaO(s) + CO2(g) ΔH = ΔH₂ + ΔH₃

Plugging in the values, we have:

ΔH = -822.5 kJ + (-553.5 kJ) = -1376.0 kJ

Therefore, the correct answer is 5. -1376.0 kJ, and not 537.0 kJ.