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### Topic: Valence bond theory  (Read 871 times)

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#### Win,odd Dhamnekar

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##### Valence bond theory
« on: May 12, 2023, 11:02:59 AM »
Question:
Draw a curve that describes the energy of a system with H and Cl atoms at varying distances. Then, find the minimum energy of this curve two ways.
(a) Use the bond energy found in Table of representative bond energies and bond length  to calculate the energy for one single HCl bond (Hint: How many bonds are in a mole?)
(b) Use the enthalpy of reaction and the bond energies for H2 and Cl2 to solve for the energy of one mole of HCl bonds $$H_2(g) + Cl_2 (g) \rightleftharpoons 2HCl(g) \Delta H°_{rxn} =-184.7 kJ/mol$$

The energy of a system with H and Cl atoms at varying distances can be represented by a curve that shows the potential energy of the system as a function of the distance between the two atoms. At very large distances, the potential energy is zero because there is no interaction between the atoms. As the atoms get closer together, they start to attract each other and the potential energy decreases. At some point, the potential energy reaches a minimum value, which corresponds to the most stable configuration of the system. This is the bond length of the H-Cl molecule. If the atoms get even closer together, they start to repel each other and the potential energy increases again.

(a) According to my sources, the bond energy for H-Cl is 431 kJ/mol. This means that it takes 431 kJ of energy to break one mole of H-Cl bonds. Since there are Avogadro's number (6.022 x 1023) of molecules in one mole, this means that it takes 431 kJ / (6.022 x 1023 )= 7.16 x 10-19 kJ to break a single H-Cl bond.

(b) The enthalpy of reaction for H2(g) + Cl2(g)  2HCl (g) is given as -184.7 kJ/mol. This means that when one mole of H2 reacts with one mole of Cl2 to form two moles of HCl, 184.7 kJ of heat is released. We can use Hess's law and the bond energies for H2 and Cl2 to solve for the bond energy of HCl.

Let's say that the bond energy for H2 is D{H-H} and for Cl2 is D{Cl-Cl}. The bond energy for HCl can be represented as D{H-Cl}. The enthalpy change for breaking one mole of H2 bonds and one mole of Cl2 bonds can be written as D{H-H} + D{Cl- Cl}. The enthalpy change for forming two moles of HCl bonds can be written as -2D{H-Cl}. According to Hess's law, we can write an equation for the enthalpy change of reaction as:

$$\Delta H^°_{rxn} = (D_{H-H} + D_{Cl-Cl}) + (-2D_{H-Cl})$$

Substituting the known values into this equation, we get:

$$-184.7 = (D_{H-H} + D_{Cl-Cl}) + (-2D_{H-Cl})$$

Rearranging this equation and solving for D{H-Cl}, we get:

$$D_{H-Cl} = \frac{1}{2}(D_{H-H} + D_{Cl-Cl} - (-184.7))$$

According to my sources,  the bond energy for H2 is 432 kJ/mol and for Cl2 is 243 kJ/mol. Substituting these values into our equation, we get:

$$D_{H-Cl} = \frac{1}{2}(432 + 243 - (-184.7)) = \frac{1}{2}(859.7) = 429.85 \text{ kJ/mol}$$

So, using this method we find that the bond energy for HCl is approximately 429.85 kJ/mol.

How to draw a curve?

« Last Edit: May 12, 2023, 12:17:23 PM by Win,odd Dhamnekar »
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