Use an enthalpy diagram to calculate the lattice energy of CaCl2 from the following information. Energy needed to vaporize one mole of Ca(s) is 192 kJ. For calcium, the first IE = 589.5 kJ mol-1, the second IE = 1146 kJ mol-1. The electron affinity of Cl is -348 kJ mol-1. The bond energy of Cl2 is 242.6 kJ per mole of Cl—Cl bonds. The standard heat of formation of CaCl2 is -795 kJ mol
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Constructing the born-haber cycle and getting the relevant data like formation enthalpy of cacl2 , atomisation of ca, first and second ionisation energies of ca , atomisation of cl and then first electron affinity of cl. First equal the formation enthalpy to the sum of all the other enthalpy's and use letter X to represent the lattice enthalpy .
Well, calculating the lattice energy of CaCl2 can be quite a laborious process! In fact, the process is more complicated than changing your Netflix password after a messy breakup. But hey, don't worry, I'm here to guide you through this enthalpy diagram extravaganza!
First things first, let's take a look at the key elements in this calculation. We have Ca(s), which needs 192 kJ/mol to vaporize, Cl2(g), which has a bond energy of 242.6 kJ/mol, and Cl(g), which has an electron affinity of -348 kJ/mol.
To calculate the lattice energy, we need to break down the formation of CaCl2 into three steps. Think of it like a recipe for a disaster... I mean, a chemistry reaction!
Step 1: We need to convert Ca(s) to gaseous Ca atoms. This step requires the energy of vaporization for Ca(s), which is given as 192 kJ/mol. It's like putting Ca(s) in a sauna and watching it turn into steam - very relaxing!
Step 2: We need to dissociate Cl2(g) into individual Cl atoms. This process requires the bond energy of Cl2, which is 242.6 kJ/mol. Imagine Cl2 as a clingy couple that needs to break up - it's going to take some serious energy!
Step 3: We bring together the gaseous Ca atoms and Cl atoms to form CaCl2(s). This is where the lattice energy comes into play. The lattice energy can be calculated using the equation:
ΔHlattice = ΔHformation - ΔH1 - ΔH2
Here, ΔHformation is the standard heat of formation of CaCl2, which is given as -795 kJ/mol. ΔH1 and ΔH2 are the ionization energies of calcium, which are 589.5 kJ/mol and 1146 kJ/mol, respectively. These ionization energies represent the amount of energy needed to remove the first and second electrons from a calcium atom, like trying to pull a stubborn clown nose off your face.
So, put all these values into the equation, do some math magic, and you'll find the lattice energy of CaCl2. Just remember, it's a twisted path filled with calculations and numbers. But hey, if you can solve this problem, you deserve a standing ovation! Good luck, my scientific friend!
To calculate the lattice energy of CaCl2 using an enthalpy diagram, we need to understand the steps involved in the formation of the compound and then use the given information to calculate the lattice energy.
1. Write the formation reaction for CaCl2.
The formation of CaCl2 can be represented by the following equation:
Ca(s) + Cl2(g) → CaCl2(s) ΔHf = -795 kJ/mol
2. Break down the enthalpy change into individual steps.
The formation of CaCl2 involves several steps:
- Vaporizing one mole of Ca(s) (ΔH1 = 192 kJ/mol)
- Ionizing one mole of Ca atoms to form Ca+(g) (first ionization energy, ΔH2 = 589.5 kJ/mol)
- Ionizing one mole of Ca+ to form Ca2+(g) (second ionization energy, ΔH3 = 1146 kJ/mol)
- Dissociating one mole of Cl2(g) molecules to form 2 moles of Cl atoms (bond dissociation energy of Cl2, ΔH4 = 2 * 242.6 kJ/mol)
- Combining one mole of Ca2+(g) and 2 moles of Cl-(g) to form one mole of CaCl2(s) (lattice energy, ΔH5 = ?)
3. Construct the enthalpy diagram.
Construct an enthalpy diagram with the different steps of the formation reaction. Start with Ca(s) at the bottom and draw vertical arrows upward for each step, representing the enthalpy change for that step. The enthalpies given (ΔH1, ΔH2, ΔH3, ΔH4) can be labeled on the diagram accordingly.
ΔH1 = 192 kJ/mol
Ca(s) Ca(g)
↓ ↓
ΔH2 = 589.5 kJ/mol
Ca(g) Ca+(g)
↓ ↓
ΔH3 = 1146 kJ/mol
Ca+(g) Ca2+(g)
↓ ↓
ΔH4 = 2 * 242.6 kJ/mol
Cl2(g) → 2 Cl(g)
Now, we need to find the lattice energy, ΔH5, which is the energy released when one mole of CaCl2(s) is formed from gaseous ions.
4. Calculate the lattice energy.
The lattice energy, ΔH5, can be calculated by subtracting the sum of the enthalpies from the formation enthalpy:
ΔH5 = ΔHf - (ΔH1 + ΔH2 + ΔH3 + ΔH4)
Plugging in the given values:
ΔH5 = -795 kJ/mol - (192 kJ/mol + 589.5 kJ/mol + 1146 kJ/mol + 2 * 242.6 kJ/mol)
Calculating the expression results in:
ΔH5 = -795 kJ/mol - 2168.7 kJ/mol
Therefore, the lattice energy of CaCl2 is approximately -2963.7 kJ/mol.
Note: The negative sign indicates that the lattice energy is released as heat during the formation of CaCl2.
Explain your trouble. You set up
Ca(s) --> Ca(g) dH = +192 kJ
Ca(g) --> Ca^+ + e dH = + 589.5 kJ
Ca^+(g) --> Ca^2+ + e dH = + 1146 kJ
Cl2(g)--> 2Cl(g) dH = 242.6 kJ
2*Cl(g) + 2e -->2*Cl^-(g) dH =2*-348 kJ
Ca^2+(g) + 2Cl^-(g) --> CaCl2 dH = lattice
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Ca(s) + Cl2(g) ==> CaCl2(s) dHformation = -795 kJ