The bond enthalpy of the Br−Cl bond is equal to ƒ¢H�‹ for the reaction
BrCl(g) �¨ Br(g) + Cl(g).
Use the following data to find the bond enthalpy of the Br−Cl bond.
Br2(l)--->Br2(g) ĢH=30.91 KJ/mol
Br2(g)--->2Br2(g) ĢH=192.9 KJ/mol
Cl2(g)---->2Cl(g) ĢH=243.4 KJ/mol
Br2(l) + Cl2(g)---->2BrCl(g) ĢH=29.2 KJ/mol
A. 219.0 kJ/mol
B. 203.5 kJ/mol
C. 14.6 kJ/mol
D. 438.0 kJ/mol
E. 407.0 kJ/mol
To find the bond enthalpy of the Br−Cl bond, we can use the concept of Hess's Law and the given data.
Hess's Law states that the total enthalpy change of a reaction is independent of the route taken, as long as the initial and final conditions are the same.
In this case, we can see that the desired reaction BrCl(g) ⟶ Br(g) + Cl(g) is the reverse of the reaction 2Br(g) + Cl(g) ⟶ 2BrCl(g).
To calculate the bond enthalpy of the Br−Cl bond, we need to find the sum of the bond energies of Br2(g) and Cl2(g) and subtract the bond energy of the Br−Br bond.
Using the given data:
1. Br2(l) ⟶ Br2(g) ƒ¢H = 30.91 KJ/mol
2. Br2(g) ⟶ 2 Br(g) ƒ¢H = 192.9 KJ/mol
3. Cl2(g) ⟶ 2 Cl(g) ƒ¢H = 243.4 KJ/mol
4. Br2(l) + Cl2(g) ⟶ 2 BrCl(g) ƒ¢H = 29.2 KJ/mol
Step 1: Calculate the enthalpy change for the reaction Br2(g) ⟶ 2 Br(g) and reverse the sign since we want the reverse reaction:
-192.9 KJ/mol
Step 2: Calculate the enthalpy change for the reaction Cl2(g) ⟶ 2 Cl(g) and reverse the sign since we want the reverse reaction:
-243.4 KJ/mol
Step 3: Add the reversed enthalpy changes from Step 1 and Step 2:
-192.9 KJ/mol + (-243.4 KJ/mol) = -436.3 KJ/mol
Step 4: Subtract the enthalpy change of the reaction Br2(l) ⟶ Br2(g):
-436.3 KJ/mol - 30.91 KJ/mol = -467.21 KJ/mol
Step 5: Divide the result by 2 since there are 2 mole ratios in the given reaction:
-467.21 KJ/mol ÷ 2 = -233.605 KJ/mol
Step 6: Reverse the sign to make the enthalpy positive:
233.605 KJ/mol
Therefore, the bond enthalpy of the Br−Cl bond is approximately 233.605 KJ/mol.
The closest option to this value is A. 219.0 kJ/mol.