Asked by harsha
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
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
Answers
Answered by
mandeep
i got 269 kJ
but im not sure...it would be better if a professor confirned it
but im not sure...it would be better if a professor confirned it
Answered by
mandeep
what do you think DrBob?
Answered by
DrBob222
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.
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.
Answered by
DrBob222
269 it is.
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