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
Well, if we want to calculate the average bond energy for a single M X bond, we can use the concept of Hess's Law and the given standard heats of reaction.
Let's represent the formation of MX4(g) as:
M(s) + 2 X2(g) → MX4(g)
And the decomposition of X2(g) as:
X2(g) → 2 X(g)
Now, we can rearrange these two reactions and cancel out the common species (X(g)) to get the formation of MX4(g) directly from M(s) and X2(g):
M(s) + X2(g) → MX4(g)
The enthalpy change (∆H) for this new reaction would be the sum of the enthalpy changes of the original reactions:
∆H = -123.7 kJ/mol + 2(297.3 kJ/mol) + (-25.1 kJ/mol)
Simplifying this, we get:
∆H = 446.8 kJ/mol
Now, we need to divide this ∆H value by the number of M X bonds formed in the reaction, which is one:
Average bond energy = ∆H / Number of M X bonds
Average bond energy = 446.8 kJ/mol / 1 bond
Therefore, the average bond energy for a single M X bond is 446.8 kJ/mol.
Now, that wasn't too "bonding," was it?
To calculate the average bond energy for a single MX bond, we need to consider the enthalpy changes in the reactions and the stoichiometric coefficients.
From the given reactions, we can see that:
1. For the formation of MX4, we have:
M(s) + 2 X2(g) → MX4(g) ∆H0 = -123.7 kJ/mol
2. For the dissociation of X2 into X:
X2(g) → 2 X(g) ∆H0 = +297.3 kJ/mol
3. For the deposition of M:
M(g) → M(s) ∆H0 = -25.1 kJ/mol
We can use these reactions to determine the bond energy for a single MX bond.
First, let's reverse reaction 2 to obtain the formation reaction for X2:
2 X(g) → X2(g) ∆H1 = -297.3 kJ/mol
Next, let's multiply reaction 1 by 2 to balance the X atoms:
2 M(s) + 4 X2(g) → 2 MX4(g) ∆H2 = -247.4 kJ/mol
Now, let's combine reactions 2 and 3 to determine the bond energy for a single MX bond:
2 X(g) + M(s) → MX4(g) + M(g) ∆H3 = ∆H2 + ∆H1 - ∆H0
∆H3 = -247.4 kJ/mol - (-297.3 kJ/mol) - (-123.7 kJ/mol)
∆H3 = -247.4 kJ/mol + 297.3 kJ/mol - 123.7 kJ/mol
∆H3 = -73.8 kJ/mol
Finally, since we have 2 X-M bonds in MX4, we can calculate the average bond energy for a single MX bond:
Average bond energy = ∆H3 / 2
Average bond energy = -73.8 kJ/mol / 2
Average bond energy = -36.9 kJ/mol
Therefore, the average bond energy for a single MX bond is -36.9 kJ/mol.
To calculate the average bond energy for a single M-X bond, we need to consider the given standard heats of reaction.
First, let's write out the balanced equations for the reactions:
1. M(s) + 2X2(g) → MX4(g)
2. X2(g) → 2 X(g)
3. M(g) → M(s)
The standard heat of reaction (∆H0) for each reaction is also given:
1. ∆H01 = -123.7 kJ/mol
2. ∆H02 = +297.3 kJ/mol
3. ∆H03 = -25.1 kJ/mol
Now, let's consider the reactions. In the first reaction, one M-X bond is broken since 2 X2 molecules are involved. The energy required to break these bonds is given by the equation:
∆H01 = + (Energy required to break the M-X bond) - 2*(Energy required to break the X-X bond)
In the second reaction, two X-X bonds are broken to form two X atoms:
∆H02 = 2*(Energy required to break the X-X bond) - (Energy required to form two X atoms)
In the third reaction, one M-X bond is formed:
∆H03 = - (Energy released when the M-X bond is formed)
From these equations, we can find the energy required to break one M-X bond:
(Energy required to break the M-X bond) = ∆H01 + 2*(Energy required to break the X-X bond)
And the energy released when one M-X bond is formed:
(Energy released when the M-X bond is formed) = -∆H03
To calculate the average bond energy for a single M-X bond, we can use the average values of the bond energies over a large number of bond dissociation events:
Average bond energy = [(Energy required to break the M-X bond) + (Energy released when the M-X bond is formed)] / 2
By substituting the values from the given reactions, we can calculate the average bond energy for a single M-X bond.