One of the examples of homologous series is given below:
1 answer:
<h3>Answer & Explanation:</h3>
<em><u>Homologous series</u></em><em>:</em><em> </em>Homologous series is the series in which the functional group of any compound and their chemical properties are similar, the difference between any two successive members is same.
<em>Above homologous series is of </em><em>Alkanol</em><em> </em>
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Above homologous series is differed by methylene (-CH2-) group. hence the mass diffrence between them would be the mass of -CH2- group.
The compound X is next to CH3CH2OH hence, X will have molecular formula CH3CH2CH2OH which is obtained by addition of a methylene group.
The reason behind the highest boiling point of butanol is his molecular mass, the main reason is the surface area, as the molecular mass increases surface area also increases which lead to increase in the vanderwall forces, due to which boiling point of compound increases. the same compound having least solubility is due to lyophilic/Hydrophobic ( tendency to get dissolved in non aqueous solution) increases & solubility of compound in polar solvents decreases.
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Answer:
The correct answer is -1085 KJ/mol
Explanation:
To calculate the formation enthalphy of a compound by knowing its lattice energy, you have to draw the Born-Haber cycle step by step until you obtain each element in its gaseous ions. Find attached the correspondent Born-Haber cycle.
In the cycle, Mg(s) is sublimated (ΔHsub= 150 KJ/mol) to Mg(g) and then atoms are ionizated twice (first ionization: ΔH1PI= 735 KJ/mol, second ionization= 1445 KJ/mol) to give the magnesium ions in gaseous state.
By other hand, the covalent bonds in F₂(g) are broken into 2 F(g) (Edis= 154 KJ/mol) and then they are ionizated to give the fluor ions in gaseous state 2 F⁻(g) (2 x ΔHafinity=-328 KJ/mol). The ions together form the solid by lattice energy (ΔElat=-2913 KJ/mol).
The formation enthalphy of MgF₂ is:
ΔHºf= ΔHsub + Edis + ΔH1PI + ΔH2PI + (2 x ΔHaffinity) + ΔElat
ΔHºf= 150 KJ/mol + 154 KJ/mol + 735 KJ/mol + 1445 KJ/mol + (2 x (-328 KJ/mol) + (-2913 KJ/mol).
ΔHºf= -1085 KJ/mol
