- 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
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.
To find the bond enthalpy of the Br-Cl bond, we can use the given reaction and the enthalpies of the individual reactions.
The given reaction is:
Br₂(1) + Cl₂(g) → 2BrCl(g) AH° = 29.2 kJ/mol
Now, let's break down the given reaction into the individual reactions:
1. Br₂(1) → Br₂(g) AH = 30.91 kJ/mol
2. Br₂(g) → 2Br(g) AH = 192.9 kJ/mol
3. Cl₂(g) → 2Cl(g) AH = 243.4 kJ/mol
Now, let's add these reactions to obtain the overall reaction:
Br₂(1) + Cl₂(g) → 2Br(g) + 2Cl(g)
To obtain the given reaction, we need to multiply the third reaction by half to balance the number of atoms:
0.5 Cl₂(g) → Cl(g) AH = 0.5 * 243.4 kJ/mol = 121.7 kJ/mol
Now, let's add the individual reactions:
Br₂(1) + 0.5 Cl₂(g) → Br₂(g) + Cl(g) Total AH = 30.91 + 121.7 kJ/mol = 152.61 kJ/mol
Finally, let's compare the obtained reaction with the given reaction:
Br₂(1) + Cl₂(g) → 2BrCl(g) AH° = 29.2 kJ/mol
As the reaction we obtained has a total enthalpy of 152.61 kJ/mol, which is greater than the enthalpy of the given reaction (29.2 kJ/mol), it means that energy is absorbed in the formation of Br-Cl bond.
Therefore, the bond enthalpy of the Br-Cl bond is:
152.61 kJ/mol - 29.2 kJ/mol = 123.4 kJ/mol
Among the given options, the closest value to this bond enthalpy is 203.5 kJ/mol (Option B).
To find the bond enthalpy of the Br-Cl bond, we need to break down the reaction Br₂(1)+ Cl₂(g) → 2BrCl(g) into a series of reactions for which we have the bond enthalpy data.
Let's start by breaking down the reaction as follows:
1) Br₂(1) → Br₂(g) (Given: AH = 30.91 kJ/mol)
2) Br₂(g) → 2Br(g) (Given: AH = 192.9 kJ/mol)
3) Cl₂(g) → 2Cl(g) (Given: AH = 243.4 kJ/mol)
4) Br₂(1) + Cl₂(g) → 2BrCl(g) (Given: AH° = 29.2 kJ/mol)
Now, let's rearrange these reactions to match the overall reaction:
1) Br₂(1) → Br₂(g) (multiplied by 2)
2) 2Br₂(g) → 4Br(g) (multiplied by 2)
3) Cl₂(g) → 2Cl(g)
4) 2Br₂(g) + Cl₂(g) → 4BrCl(g) (multiplied by 2)
Now, let's sum up these reactions to cancel out the intermediates:
2Br₂(1) + Cl₂(g) → 4BrCl(g) + 4Br(g) + 2Cl(g)
We can cancel out the intermediates to get:
2Br₂(1) + Cl₂(g) → 4BrCl(g)
Now, we compare this final reaction with the given reaction:
Br₂(1) + Cl₂(g) → 2BrCl(g)
By comparing the two reactions, we can see that they have the same stoichiometry. Therefore, the enthalpy change for the final reaction is simply double the given enthalpy change:
2 x (29.2 kJ/mol) = 58.4 kJ/mol
So, the bond enthalpy of the Br-Cl bond is 58.4 kJ/mol.
None of the provided answer choices matches with 58.4 kJ/mol. This suggests that there might be an error in the given data or calculations.