Given the standard heats of reaction
M(s) + 2 X2(g) → MX4(g) −113.7 kJ/mol
X2(g) → 2 X(g) +317.3 kJ/mol
M(g) → M(s) −23.1 kJ/mol
calculate the average bond energy for a single M-X bond.
Answer in units of kJ/mol.
i have an idea of how but im not fully sure
Add eqn 1 to 2x the reverse of eqn 2 and add to eqn 3 to get this, I think, but check it to make since I'm doing this in my head.
M(s) + 2 X2(g) → MX4(g) −113.7 kJ/mol
4X(g) → 2X2(g) -634.6 kJ/mol
M(g) → M(s) −23.1 kJ/mol
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M(g) + 4X(g) ==> MX4(g) ==> -114.7 + (-634.6) + (-23.1) = ? which is for 4 MX bonds so divide that number by 4 to get one bond. Check me out on all of this.
Well, it seems like you're in a bit of a bond-ing situation here! Don't worry, I'm here to help you through it with a sprinkle of humor.
To find the average bond energy for a single M-X bond, we need to use the concept of Hess's Law. It's like the law of thermodynamics, but with a lot fewer lawyers involved.
The first step is to calculate the overall energy change for the reaction M(g) + 2X2(g) → MX4(g). We can do this by adding up the individual energy changes:
Energy change of M(g) → M(s): -23.1 kJ/mol
Energy change of X2(g) → 2X(g): +317.3 kJ/mol
Adding these energy changes gives us the overall energy change for the reaction:
-23.1 kJ/mol + 317.3 kJ/mol = 294.2 kJ/mol
Now that we have the overall energy change, we divide it by the number of M-X bonds formed in the reaction, which is 4 in this case (based on the MX4 product):
294.2 kJ/mol ÷ 4 = 73.55 kJ/mol
So, the average bond energy for a single M-X bond is approximately 73.55 kJ/mol. Keep in mind that this is just an average, so individual M-X bonds may vary slightly in energy. Now, isn't that bon(d)-afide chemistry fun?
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.
The enthalpy change of a reaction is equal to the sum of the enthalpies of formation of the products minus the sum of the enthalpies of formation of the reactants.
Let's consider the given reactions:
1. M(s) + 2 X2(g) → MX4(g) ΔH1 = -113.7 kJ/mol
2. X2(g) → 2 X(g) ΔH2 = +317.3 kJ/mol
3. M(g) → M(s) ΔH3 = -23.1 kJ/mol
We first need to manipulate these equations to determine the enthalpy change for one M-X bond.
Adding equation 3 to equation 1 and equation 2:
M(g) + 2 X2(g) → MX4(g) ΔH1 + ΔH3 = -136.8 kJ/mol
X2(g) → 2 X(g) ΔH2 = +317.3 kJ/mol
Now, let's divide the overall reaction ΔH1 + ΔH3 by the stoichiometric coefficient of X2(g) to isolate the desired bond:
(M(g) + 2 X2(g))/2 → MX4(g) (-136.8 kJ/mol) / 2 = -68.4 kJ/mol
This equation represents the enthalpy change for one M-X bond formation.
Therefore, the average bond energy for a single M-X bond is -68.4 kJ/mol.
To calculate the average bond energy for a single M-X bond, you can use Hess's law.
Hess's law states that the change in enthalpy of a reaction is the same whether the reaction takes place in one step or a series of steps.
Let's assume the average bond energy for a single M-X bond is denoted as E(M-X) kJ/mol.
From the given reactions, we can see that the formation of MX4 from M and X2 involves breaking two M-X bonds and forming four X-X bonds.
Using the given standard heats of reaction, we can set up the equation:
(-113.7 kJ/mol) = 2E(M-X) + 4×(0 kJ/mol) [since X-X bond formation does not involve any energy change]
Simplifying the equation:
-113.7 kJ/mol = 2E(M-X)
Dividing both sides of the equation by 2:
E(M-X) = -113.7 kJ/mol / 2
E(M-X) = -56.85 kJ/mol
Therefore, the average bond energy for a single M-X bond is approximately -56.85 kJ/mol.
Note: The negative sign indicates that the bond formation is exothermic, meaning energy is released during the formation of the bond.