For the decomposition of NaHCO3, estimate the temperature at which the total pressure of the gases above the solids is 1 atm.
2NaHCO3 (s) <--> Na2CO3(s) + H2O(g) + CO2(g)
Additional Details
5 hours ago
Answer is 393 K.
If possible, please show work.
35 minutes ago
We'd need enthalpies of formation and standard entropies for those substances. And are you sure it's the total pressure is 1 atm or the partial pressure of each gas?
The question is the exact wording of the book.
NaHCO3 (s)
dHf* = -950.8 kJ mol^-1
dGf* = -851 kJ mol^-1
S* = 102 J mol^-1 K^-1
Na2CO3 (s)
dHf* = -1131 kJ mol^-1
dGf* = -1044 kJ mol^-1
S* = 135.0 J^-1 mol^-1 K^-1
H20 (g)
dHf* = -241.8 kJ mol^-1
dGf* = -228.6 kJ mol^-1
S* = 188.7 J mol^-1 K^-1
CO2(g)
dHf* = -393.5 kJ mol^-1
dGf* = -394.4 kJ mol^-1
S* = 213.6 J mol^-1 K^-1
2007-07-16
15:56:39
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Science & Mathematics
➔ Chemistry