Find the lattice energy of MgH2(s), show with a series of steps when the Enthalpies of
formation for calcium hydride is given as; (ΔHf = −75.3 kJ/mol for MgH2)
Step 1: Calculate the lattice energy of MgH2(s) using the Enthalpy of formation.
Lattice energy = -75.3 kJ/mol
Step 2: Convert the lattice energy from kJ/mol to kJ/g.
Lattice energy = -75.3 kJ/g
Really?
Oh, lattice energy! That's a fun one. Okay, let's break it down step by step, but remember, I'm just a Clown Bot, so don't take my humor too seriously!
Step 1: Write the equation for the formation of MgH2(s):
Mg(s) + H2(g) → MgH2(s)
Step 2: Next, let's calculate the enthalpy change for this reaction using the enthalpies of formation:
ΔH = Σn(ΔHf(products)) - Σn(ΔHf(reactants))
Since we only have one product (MgH2), the equation becomes:
ΔH = ΔHf(MgH2)
Replace ΔHf(MgH2) with the given value of -75.3 kJ/mol:
ΔH = -75.3 kJ/mol
And there you have it! The lattice energy of MgH2(s) is -75.3 kJ/mol. Just be careful not to let it get too negative; it might start attracting bad jokes from neighboring elements!
To find the lattice energy of MgH2(s), we can use the Born-Haber cycle. The lattice energy (ΔHlattice) is defined as the energy change that occurs when one mole of an ionic compound is formed from its gaseous ions.
The Born-Haber cycle consists of several steps, so let's break it down:
Step 1: Formation of gaseous Mg atoms:
Mg(s) → Mg(g)
This step requires the sublimation energy of Mg, also known as ΔHsub.
Step 2: Formation of gaseous H atoms:
½H2(g) → H(g)
This step requires the bond dissociation energy of H2, also known as ΔHdiss.
Step 3: Formation of gaseous Mg+ ions:
Mg(g) → Mg+(g) + e-
This step requires the ionization energy of Mg, also known as ΔHion.
Step 4: Formation of gaseous H- ions:
H(g) + e- → H-(g)
This step requires the electron affinity of H, also known as ΔHea.
Step 5: Formation of solid MgH2:
Mg+(g) + 2H-(g) → MgH2(s)
This step represents the formation of the solid compound and releases the lattice energy, ΔHlattice.
According to Hess's law, the sum of the energy changes for these steps must be equal to the overall enthalpy change of formation (ΔHf) for MgH2(s):
ΔHf = ΔHsub + ΔHdiss + ΔHion + ΔHea + ΔHlattice
We are given the enthalpy of formation (ΔHf) for MgH2 as -75.3 kJ/mol. Let's assume that we are given the values for ΔHsub, ΔHdiss, ΔHion, and ΔHea. In this case, we can rearrange the equation and solve for ΔHlattice:
ΔHlattice = ΔHf - ΔHsub - ΔHdiss - ΔHion - ΔHea
By substituting the given values, you can calculate the lattice energy (ΔHlattice) for MgH2.
Note: The specific values for ΔHsub, ΔHdiss, ΔHion, and ΔHea are not provided in the question, so you would need to refer to a reliable source or specified data to obtain these values.
To calculate the lattice energy of MgH2(s), we will need to use the Born-Haber cycle. The lattice energy can be determined by calculating the various steps involved in the formation of the compound. Here are the steps to find the lattice energy of MgH2(s) using the given enthalpy of formation for calcium hydride, ΔHf = -75.3 kJ/mol for MgH2:
Step 1: Write the balanced equation for the formation of MgH2(s):
Mg(s) + H2(g) -> MgH2(s)
Step 2: Identify and write the enthalpy change for each step involved in the formation of MgH2(s). The steps are:
a) Sublimation of Mg(s):
Mg(s) -> Mg(g) ΔHsub
b) Dissociation of H2(g):
H2(g) -> 2H(g) ΔHdissoc
c) Formation of MgH2(s):
Mg(g) + 2H(g) -> MgH2(s) ΔHform
Step 3: Use the given enthalpy changes for each step:
ΔHsub is the enthalpy change for sublimation of Mg(s) which can be found in reference sources. Let's assume it to be -148.6 kJ/mol.
ΔHdissoc is the enthalpy change for the dissociation of H2(g) which can also be found in reference sources. Let's assume it to be +436.0 kJ/mol.
ΔHform is the enthalpy of formation of MgH2(s), given as -75.3 kJ/mol.
Step 4: Apply the Hess's Law, which states that the sum of the enthalpy changes for individual steps is equal to the enthalpy change for the overall process.
Therefore, the lattice energy (ΔHlattice) can be calculated using the formula:
ΔHlattice = ΔHsub + ΔHdissoc + ΔHform
Step 5: Substitute the values we assumed in Step 3 into the formula:
ΔHlattice = -148.6 kJ/mol + 436.0 kJ/mol + (-75.3 kJ/mol)
Step 6: Calculate the lattice energy:
ΔHlattice = 212.1 kJ/mol
So, the lattice energy of MgH2(s) is 212.1 kJ/mol.