Calculate the lattice energy for CaCl2 from the following information: Energy needed to vaporize one mole of Ca(s) is 192kJ. For
calcium the first ionization energy is 589.5kJ/mol and the second ionization energy is 1146kJ/mol. The electron affinity of Cl is-348kJ/mol. The bond energy of Cl2 is 242.6kJ/mol of CI-Cl bonds. The standard heat of formation of CaCl2 is -795kJ/mol. (include the sign and your numerical answer, do not include units, do not use scientific notation, and round your answer to 1 decimal point)
1 answer:
Answer:
The correct answer is -2268.5 KJ/mol
Explanation:
Find attached the diagram for Born-Haber cycle. The solid calcium (Ca (s)) can be first sublimated into its atoms in gaseous phase and then the atoms are ionizated two times to give the cations in gaseous phase. The gaseous chlorine (Cl₂(g)) is first dissociated into its gaseous ions and then the atoms are ionizated to give anions in gaseous state. The lattice energy (LE) is the energy involved in the formation of the solid CaCl₂ from the elementary ions in gaseous state.
From Born-Haber cycle we have:
ΔHºf= ΔHvap + IE₁ + IE₂ + BE + 2 x EA + LE
⇒ LE= ΔHºf - ΔHvap - IE₁ - IE₂ - BE - (2 x EA)
LE= -795 KJ/mol - 192 KJ/mol - 589 KJ/mol - 1146 KJ/mol - 242 KJ/mol - (2 x (-348 KJ/mol))
LE= -2268.5 KJ/mol
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Answer: A volume of 455 mL from 0.550 M KBr solution can be made from 100.0 mL of 2.50 M KBr.
Explanation:
Given: = ?,
= 0.55 M
= 100.0 mL,
= 2.50 M
Formula used to calculate the volume of KBr is as follows.
Substitute the values into above formula as follows.
Thus, we can conclude that a volume of 455 mL from 0.550 M KBr solution can be made from 100.0 mL of 2.50 M KBr.
Explanation:
Strength of intermolecular forces depends on the number of carbon atoms present in a compound. More is the number of carbon atoms attached linearly to each other more will be the surface area occupied by it. Hence, more is the strength of the compound.
This means that more is the branching present in a compound or lesser is the number of carbon atoms present in it then less will be the strength of intermolecular forces in the compound.
Thus, we can conclude that given compounds are placed in order of decreasing strength of intermolecular forces as follows.
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