Step-by-step explanation:
Given the following reaction:
H₂ (g) + Cl₂ (g) --> 2 HCl (g)
We have to find the ΔH for that reaction. We can estimate the change in enthalpy using the bond enthalpies.
We need to add energy to break a bond and energy will be released when a bond is broken.
ΔH = energy added to break bonds + energy released when making bonds
Now we have to find the bonds that are broken.
The reactants are H₂ and Cl₂. Hydrogen gas is a diatomic gas that has a single bond between both atoms of H (H-H). Chlorine gas also is a diatomic gas that has a single bond between both atoms of Cl (Cl-Cl).
Bonds broken = 1 (H-H) + 1 (Cl-Cl)
The product of our reaction is HCl. In this case it is a gas that has a single bond between the atoms of Cl and H. Since the coefficient of the reaction is two, we will have to form two bonds.
Bonds formed = 2 (H-Cl)
If we search for the values of the different bond enthalpies we can find that they are:
Bond enthalpy H-Cl = 431 kJ/mol
Bond enthalpy H-H = 436 kJ/mol
Bond enthalpy Cl-Cl = 242 kJ/mol
Finally we can replace these values in the formula and we can estimate the ΔH of the reaction. We have to pay attention that for breaking the bonds we have to add energy, so the bonds enthalpies are positive. When forming the bonds the energy is released so the bond enthalpies should be negative (we have to change the sign).
ΔH = energy added to break bonds + energy released when making bonds
ΔH = 1 (H-H) + 1 (Cl-Cl) - 2 (H-Cl)
ΔH = 436 kJ/mol + 242 kJ/mol - 2 * 431 kJ/mol
ΔH = -184 kJ/mol
Answer: the change in enthalpy for the reaction is -184 kJ/mol.