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) 16.2 dm^3/mol*h
b) 6.1 × 10^3 s, 2.5 × 10^3 s (it is different to the hint)
Explanation:
We can use the integrated rate equation in order to obtain k.
For the reaction A+ B --> P the reaction rate is written as
Rate =
If and
are the inital concentrations and x the concentration reacted at time t, so
and
and the rate at time t is written as:
Separating variables and integrating
The integral in left side is solved by partial fractions, it can be used integral tables
Using logarithm properties (ln x - ln y = ln(x/y))
Using the given values k can be calculated. But the data seems inconsistent since if the concentration of A changes from 0.075 to 0.02 mol dm^-3 it implies that 0.055 mol dm^-3 of A have reacted after 1 h, so according to the reaction given the same quantity of B should react, and we only have a of 0.05 mol dm^-3.
Assuming that the concentration of B fall to 0.02 mol dm^-3 (and not the concentration the A). So we arrive to the answer given in the Hint.
So, the values given are t= 1, ,
,
, it implies that the quantity reacted, x, is 0.03 and
. Then, the value of k would be
b) the question b requires calculate the time when the concentration of the specie is half of the initial concentration.
For reactant A, It is solved with the same equation
but suppossing that so
, k=16.2 and the same initial concentrations. Replacing in the equation
For reactant B, so
, k=16.2 and the same initial concentrations. Replacing in the equation
Note: The procedure presented is correct, despite of the answer be something different to the given in the hint, I obtain that result if the k is 19.2... (maybe an error in calculation of given numbers)