Data given are related to the Born-Haber cycle for HCl
Calculate the amount of energy lost when the ionic species return to their molecular form.
Atomization of ½H2(g)=217.6KJ/mol
1st Ionisation of ½H2(g)=1312KJ/mol
Atomization of ½Cl2(g)=121 KJ/mol
1st Ionisation of ½Cl2(g)=-364 KJ/mol
ÄHfè(298K) of HCl(g)=-92.3
To calculate the amount of energy lost when the ionic species return to their molecular form in the Born-Haber cycle for HCl, we need to consider the following steps:
1. Atomization of H2(g) into H atoms:
- The enthalpy change for the atomization of 1 mole of H2(g) is given as 217.6 KJ/mol. However, we have ½H2, so the enthalpy change for ½H2(g) is half of the given value.
- Therefore, the enthalpy change for the atomization of ½H2(g) is 217.6 KJ/mol / 2 = 108.8 KJ/mol.
2. First ionization of H atoms:
- The enthalpy change for the first ionization of 1 mole of H2(g) is given as 1312 KJ/mol. However, we have ½H2, so the enthalpy change for ½H2(g) is half of the given value.
- Therefore, the enthalpy change for the first ionization of ½H2(g) is 1312 KJ/mol / 2 = 656 KJ/mol.
3. Atomization of Cl2(g) into Cl atoms:
- The enthalpy change for the atomization of 1 mole of Cl2(g) is given as 121 KJ/mol. However, we have ½Cl2, so the enthalpy change for ½Cl2(g) is half of the given value.
- Therefore, the enthalpy change for the atomization of ½Cl2(g) is 121 KJ/mol / 2 = 60.5 KJ/mol.
4. First ionization of Cl atoms:
- The enthalpy change for the first ionization of 1 mole of Cl2(g) is given as -364 KJ/mol. However, we have ½Cl2, so the enthalpy change for ½Cl2(g) is half of the given value.
- Therefore, the enthalpy change for the first ionization of ½Cl2(g) is -364 KJ/mol / 2 = -182 KJ/mol.
5. Formation of HCl(g):
- The enthalpy change for the formation of 1 mole of HCl(g) is given as -92.3 KJ/mol.
Now, to calculate the amount of energy lost when the ionic species return to their molecular form, we need to sum up the enthalpy changes of the individual steps:
ΔH = (Enthalpy change for atomization of ½H2(g)) + (Enthalpy change for first ionization of ½H2(g)) + (Enthalpy change for atomization of ½Cl2(g)) + (Enthalpy change for first ionization of ½Cl2(g)) + (Enthalpy change for formation of HCl(g))
ΔH = 108.8 KJ/mol + 656 KJ/mol + 60.5 KJ/mol + (-182 KJ/mol) + (-92.3 KJ/mol)
By plugging these values into the equation and performing the arithmetic, you should be able to calculate the total amount of energy lost when the ionic species return to their molecular form in the Born-Haber cycle for HCl.