Question

why it is said that the reactant is more stable in exothermic reaction ? if it is then it should be have high enthalpy?explain why it is said that the reactant is more stable in exothermic reaction ? if it is then it should be have high enthalpy?explain @Chemistry

Answer 1

Enthalpies of Reactions

Enthalpy of a reaction or energy change of a reaction DH, is the amount of energy or heat absorbed in a reaction. If the energy is required, DH is positive, and if energy is released, the DH, is negative.

_________ Products
­
|
| DH, positive for
| endothermic reaction
|
__|______ Reactants
The enthalpy can be determined by experiment, but estimates can easily be made if bond energies or standard enthalpies of formation for the reactants and products are available.
Using Formulas to Calculate DH

Due the the definitions of various types of energy related terms, formulas for evaluating enthalpies can be very confusing. For example, the formulas to calculate the enthalpy of a reaction depends on whether bond energies or enthalpies of formation are available.
When standard enthalpies of formation, Hfo, for all products and reactants are available, we have

Hreaction = SUM (Hproducts) - SUM (Hreactants)
or if you prefer using symbols
Hreaction = SHproducts - SHreactants
For simplicity in formulation we use H to represent Hfo in the above formulas.
Because bond energies are defined as the energies required to break the bonds, positive values are usually listed whereas in reality, they are energies released when chemical bonds are formed from respective atoms. Thus, using the bond energies (BE) as they are given or defined, the following formula apply:

Hreaction = SUM (BEreactants) - SUM (BEproducts)
or if you prefer using symbols
Hreaction = SBEreactants - SBEproducts
These formulas to calculate the enthalpy (heat) of a reaction can be a very confusing and you may easily get an incorrect sign for the value. Thus, remembering formulas is discouraged!
Well, if you use a diagram to help visualize the calculation or write the chemical reaction equations accompanying the thermodynamic values, you will be able to avoid the confusion. In both cases, you are applying the principle of conservation of energy in solving the problems.

Calculate Enthalpy of Reaction from Bond Energies

Due to the principle of conservation of energy the total energy before and after the reaction must not change. Thus, the energy of a reaction released or absorbed in the reaction must come from the difference in bond energies of the products and the reactants.
Example 1

The bond energy (kJ) for H2, F2, and HF are 436, 158 and 568 kJ
respectively, calculate the enthalpy (energy) of the reaction,
H2(g) + F2(g) = 2 HF
Solution
Based on the bond energies given, we have

H2 ® 2H D = 436 kJ/mol
F2 ® 2F D = 158 kJ/mol
2H + 2F ® 2HF H = -568*2 kJ/mol
Adding all three equations and energies leads to the following

H2(g) + F2(g) = 2 HF DH = -542 kJ/equation
Note that D represent bond dissociation energy, and H the enthalpy of the reaction as written. We use DH in the last equation to denote enthalpy of change of the overall reaction.
Discussion
Since bond energies are given, we use the monoatomic gases are the reference level in this calculation. The energy level diagram shown below illustrates the principle of conservatin of energy, and you are expected to have the skill to draw such a diagram.

------2 H(g) + 2 F(g)-------
­ ­­
| ||436+158 kJ
| ||
|+2*568 kJ ---H2 + F2---
| |
| | DH = -542 kJ/equation
| ¯
--------2 HF(g)------------
This diagram is very similar to those of Bohn-Haber cycle used to evaluate lattice energy.
Calculate Enthalpy of Reaction from Enthalpy of Formation

A similar cycle can be devised to calculate energy of a reaction when the standard enthalpies of formation are given. We illustrate this cycle by an an example.
Example 2

Standard enthalpies of formation are: C2H5OH(l) -228, CO2 -394, and H2O(l) -286 kJ/mol. Calculate the enthalpy of the reaction,
C2H5OH + 3 O2 ® 2 CO2 + 3 H2O
Solution
From the definition of the enthalpy of formation, we have the following equations and the energy changes of reactions.

C2H5OH(l) ® 2 C(graphite) + 3 H2(l) + 0.5 O2(g) H = 228 kJ/mol
2 C(graphite) + 2 O2(g) ® CO2(g) H = -394*2 kJ/mol
3 H2(g) + 1.5 O2(g) ® 3 H2O(l) H = -286*3 kJ/mol
Adding all three equations and energies leads to the following

C2H5OH(l) + 3 O2(g) ® 2 CO2(g) + 3 H2O (l) DH = -1418 kJ/mol
Discussion
Since the standard enthalpy of formation uses the elements as the standard, we put the elements on the top as a common level of 0 energy. The enthalpies of formation are negative, and we have the following diagram.

Thus, the enthalpy of reaction is the difference between the level of
---C2H5OH +3 O2---
and
-2 CO2 + 3 H2O-.

---2 C(graphite) + 3 H2 + 3.5 O2(g)---
| |
| | -228 kJ
| ¯
| ---C2H5OH + 3 O2---
|-394*2 -286*3 |
| |
| | DH = -394*2 - 286*3 - (-228)
| | = -1418 kJ
¯ ¯
-------------2 CO2 + 3 H2O------------

Answer 2

its too long please sshort summary

Answer 3

ok

Answer 4

A exothermic reaction world look like this A <-> B + Q (Q < 0)
Q here is energy in the form of heat.
When the reaction from A to B is caryed out the potentialenergi fall (-∆G)
Becuase of the potential energy fall the kinetic energy increase and is seen as heat. (the full energi is constant)
Becuase of the drop in potential energi there come a energybarrier, and very very few molecules have the potential energi to pass this energybarrier, and therefore becomes stable.
The higher the energy drop is, the lower is the chance for the "backwards" reaction to be done.