The value of ΔS° for reaction is - 22.2 J/K.mol
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Calculation,
Given value of S°(J/K.mol) for
= 248.5
= 240.5
= 210.6
= 256.2
Formula used:
ΔS° (Reaction) = ∑S°(Product) - ∑S°(Reactant)
ΔS° = (256.2 + 210.6 ) - ( 248.5 + 240.5) = 466.8 - 489 = - 22.2 J/K.mol
The change in stander entropy of reaction is - 22.2 J/K.mol. The negative sign indicates the that entropy of reaction is decreases when reactant converted into product.
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This problem is providing a comparison on the atoms forming formaldehyde, CH₂O, in terms of the groups they are in, C in group 4A, O in group 6A and H in group 1A, and it is asking for a description about the bonds once they are bonded in a formal Lewis dot structure.
In such a way, the attached file shows the Lewis dot structure, in which we can find two bonds between the two hydrogen atoms and the carbon one, and also, a double bond between carbon and oxygen.
The aforementioned is explained by bearing to mind the concept of octet, which states that elements in groups 4A, 5A, 6A and 7A are able to complete 8 valence electrons after bonding.
Thus, since carbon is in group 4A we infer it needs 4 more electrons to complete the octet, and that is achieved due to the 4 bonds it is having with the two hydrogen atoms and double bond with the oxygen atom.
In addition, since oxygen is in group 6A, we infer it needs two more electrons to complete its octet, which is attained via the double bond it has with the carbon atom. For hydrogen, however, it can complete one bond at maximum, because it is in group 1A, which means it can complete two valence electrons after bonding.
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