Given the standard heats of reaction
Reaction ∆H0
M(s) + 2 X2(g) → MX4(g) −123.7 kJ/mol
X2(g) → 2 X(g) +297.3 kJ/mol
M(g) → M(s) −25.1 kJ/mol
calculate the average bond energy for a single
M X bond.
Answer in units of kJ/mol
To calculate the average bond energy for a single MX bond, we can use Hess's law and the given standard heats of reaction.
Let's denote the average bond energy of a single MX bond as "E(MX)", in kJ/mol.
We can start by writing the overall reaction as the sum of the individual reactions:
M(s) + 2 X2(g) → MX4(g) ∆H1 = -123.7 kJ/mol (1st given reaction)
X2(g) → 2 X(g) ∆H2 = +297.3 kJ/mol (2nd given reaction)
M(g) → M(s) ∆H3 = -25.1 kJ/mol (3rd given reaction)
Now, we can manipulate these reactions to isolate the formation of a single MX bond:
2 X(g) → X2(g) (reverse reaction of the 2nd reaction)
M(s) + X2(g) → MX2(g) (multiply the 1st reaction by 2 and reverse)
MX2(g) → MX4(g) (multiply the 1st reaction by 2)
Let's assign the average bond energy of a single X-X bond as "E(X-X)", in kJ/mol.
The overall reaction becomes:
M(s) + 2 X(g) → MX4(g) ∆H1 = -123.7 kJ/mol (1st given reaction)
M(s) + X2(g) → MX2(g) ∆H4 = -2 * ∆H1 = +247.4 kJ/mol (manipulated reaction)
MX2(g) → MX4(g) ∆H5 = - ∆H1 = +123.7 kJ/mol (manipulated reaction)
X2(g) → 2 X(g) ∆H2 = +297.3 kJ/mol (2nd given reaction)
M(g) → M(s) ∆H3 = -25.1 kJ/mol (3rd given reaction)
To simplify the calculation, we'll sum ∆H4, ∆H5, and ∆H2 together, and cancel out M(s) in the process:
2 X(g) → X2(g) + MX2(g) ∆H24 = ∆H4 + ∆H5 = 247.4 + 123.7 = 371.1 kJ/mol
X2(g) + MX2(g) → MX4(g) ∆H245 = ∆H24 + ∆H2 = 371.1 + 297.3 = 668.4 kJ/mol
Finally, we can relate these enthalpy changes to the bond dissociation energies:
∆H245 = 2 * E(X-X) + E(MX)
668.4 = 2 * E(X-X) + E(MX)
Solving for E(MX), the average bond energy for a single MX bond:
E(MX) = (668.4 - 2 * E(X-X))
Therefore, the average bond energy for a single MX bond is (668.4 - 2 * E(X-X)) kJ/mol.
To calculate the average bond energy for a single M-X bond, we need to use the concept of Hess's law and the given standard heats of reaction.
Hess's law states that the change in enthalpy for a reaction is independent of the pathway taken, as long as the initial and final conditions are the same. In other words, we can add or subtract chemical equations to obtain a desired overall reaction.
In this case, we will use Hess's law to manipulate the given reactions to obtain the desired M-X bond formation.
1. We start with the given reaction:
M(s) + 2X2(g) → MX4(g) (∆H1 = -123.7 kJ/mol)
2. We need to cancel out any unwanted species to isolate the desired M-X bond. Looking at the reaction, we have 2X2 on both sides, so we can cancel it out:
M(s) → MX4(g) (∆H2 = -123.7 kJ/mol)
3. To further isolate the M-X bond, we need to eliminate M(s). For that, we must use the third given reaction:
M(g) → M(s) (∆H3 = -(-25.1) kJ/mol)
Since we want M(s) on the product side, we must reverse the third reaction:
M(s) → M(g) (∆H3' = +25.1 kJ/mol)
4. Now we can combine reactions 2 and 3' to obtain the desired M-X bond reaction:
M(s) + MX4(g) → M(g) + MX4(g) (∆H4 = ∆H2 + ∆H3' = -123.7 + 25.1 kJ/mol)
5. Finally, we can convert the above equation to represent the formation of a single M-X bond:
M(s) + X(g) → M(g) + X(g) (∆H5 = (∆H4) / 4)
The reaction now represents the formation of a single M-X bond. The enthalpy change for this reaction will give us the average bond energy for a single M-X bond.
Calculating ∆H5:
∆H5 = (∆H4) / 4
∆H5 = (-123.7 + 25.1 kJ/mol) / 4
∆H5 = -98.6 kJ/mol / 4
∆H5 = -24.65 kJ/mol
Therefore, the average bond energy for a single M-X bond is approximately -24.65 kJ/mol.