In order to find the bond enthalpy of the Br-Cl bond, we need to use the given data and apply the Hess's Law.
Let's first write the balanced equation for the reaction Br₂(1) + Cl₂(g) → 2BrCl(g):
Br₂(1) + Cl₂(g) → 2BrCl(g) ΔH° = 29.2 kJ/mol
Now, let's use the given data to calculate the bond enthalpy of the Br-Cl bond.
1. Br₂(1) → Br₂(g) ΔH° = 30.91 kJ/mol (reverse the reaction)
2. Br₂(g) → 2Br(g) ΔH° = 192.9 kJ/mol (divided by 2 to get 1 mole of Br instead of 2)
3. Cl₂(g) → 2Cl(g) ΔH° = 243.4 kJ/mol (reverse the reaction)
By adding the reverse of reaction 1 and reaction 2, we get:
Br₂(g) → 2Br(g) + Br₂(1) ΔH° = 30.91 - 192.9 = -161.99 kJ/mol
By adding reactions 2 and 3, we get:
Br₂(g) + Cl₂(g) → 2Br(g) + 2Cl(g) ΔH° = 192.9 + 243.4 = 436.3 kJ/mol
Now, we can subtract the equation we obtained from reaction 1 from the equation we obtained from reactions 2 and 3:
Br₂(g) + Cl₂(g) - Br₂(g) + Br₂(1) = 2Cl(g) + 2BrCl(g) ΔH° = 436.3 - (-161.99) = 598.29 kJ/mol
Finally, divide the above equation by 2 to obtain the enthalpy change for 1 mole of BrCl:
Cl(g) + BrCl(g) → Cl₂(g) + Br₂(g) ΔH° = 598.29 / 2 = 299.15 kJ/mol
Since the bond enthalpy of the Br-Cl bond is equal to the enthalpy change of this reaction:
Bond enthalpy of Br-Cl bond = ΔH° = 299.15 kJ/mol
Therefore, the correct answer is D) 438.0 kJ/mol.
- The bond enthalpy of the Br-Cl bond is equal to AH° for the reaction BrCl(g) Br(g) + Cl(g). Use the following data to find the bond enthalpy of the Br-Cl bond.
Br₂(1)→ Br2(g) AH 30.91 kJ/mol
Br2(g) → 2Br(g) AH = 192.9 kJ/mol
Cl2(g) →2C1(g) AH = 243.4 kJ/mol
Br2(1)+ Cl2(g) → 2BrCl(g) AH° = 29.2 kJ/mol
A) 14.6 kJ/mol
B)203.5 kJ/mol
C)219.0 kJ/mol
D)438.0 kJ/mol
1 answer