✔ 最佳答案
Actually i just memorized it as one of the typical HCO3- compound properties.
Just as the unstable H2CO3, once be heated, it will release CO2 and the same as HCO3- compound.
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Besides, i found some data from others, hoping that can help you
(he said Na2CO3 can also release CO2 under a greater energy given)
NaHCO3 is less thermally stable than Na2CO3.
That does NOT mean Na2CO3 will not decompose.
The decomposition of NaHCO3 is shown below:
2NaHCO3(s) -Δ->Na2CO3(s)+H2O(l)+CO2(g)
The mechanism is elimination/dehydration.
You can imagine that the -OH in HCO3- is attacked by the lone pair electrons of O in the -OH of another HCO3- . Hence, O=C(OH2)+O- and CO32- are formed.
The formed CO32- becomes Na2CO3. As for the former molecule, The group -OH2+ detaches from the carbon, leaving a O=C-O-.
The -OH2+ group becomes water and the O=C-O- becomes CO2 by resonance.
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The decomposition of Na2CO3 is shown below:
Na2CO3(s) -Δ->Na2O(s)+CO2(g)
This involves the electrostatic attraction of the cation and anion.
Whether a carbonate decompose depends on high degree of distortion of the electron cloud between anion and cation.
Sodium has a small charge density (hence weak polarising power) that electron cloud of carbonate is not quite distorted.
This carbonate is stable at lower temperature. Though, it is not the case for high temperature, at which the compound is more unstable with more energy.
One C-O bond may become C=O and the other C-O bond in carbonate would be broken, forming the products listed above.
2010-03-20 10:26:54 補充 decomposition of NaHCO3 嗰度,
the O in one -OH attackes the H in another -OH, electrons of the -OH are given back to the O in the bond-breaking process.
2012-01-01 12:56:19 補充:
more resonance extent somehow means more stable