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Offline fatimasaboor123

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Enthalpy of Reaction
« on: March 21, 2018, 12:29:15 AM »
I have some confusion regarding the exact formula of the enthalpy of reaction.

In a lecture by Khan Academy, ∆H=∑∆H(bonds broken in reactants)−∑∆H(bonds made in products) is the formula but in my textbook and other web pages ∆H=∑∆H(Products)−∑∆H(Reactants).

The latter is used everywhere but the former seems to make more sense.

This is the example given there:
ΔH value negative --> energy released --> exothermic reaction ΔH value positive --> energy absorbed --> endothermic reaction
∆H=∑∆H(bonds broken in reactants)−∑∆H(bonds made in products)
Let's understand this through an example. We can calculate the enthalpy change (ΔHΔH) for the following reaction:

H2 + F2------.> 2HF
To break one mole of H2  energy absorbed is 436436436 kJ.
To break one mole of F2 energy absorbed is 158 kJ.
To form two moles of 2HF, energy released is 2 X (568) kJ.
So applying the equation, ∆H=∑∆H(bondsbrokeninreactants)−∑∆H(bondsmadeinproducts)
ΔHreaction=(436+158)–(2X568)=−542kJ
The overall enthalpy of the reaction is negative, i.e., it’s an exothermic reaction where energy is released in the form of heat.
The example is precise and all the data is correct but if we apply the other formula (products-reactants) then the sign comes positive and the reaction will be endothermic but the formation of HF is exothermic.

My current understanding of heat of reaction is:
That in a chemical reaction, first energy is absorbed to break the bonds in reactants and then energy is released to make the bonds in products.
Exothermic reactions is when a chemical reaction releases more energy than it absorbs.

Endothermic reactions is when a chemical reaction absorbs more energy than it releases.

∆H=∑∆H(bondsbrokeninreactants)−∑∆H(bondsmadeinproducts)

It can also be written as :

∆H=∑∆H(energy absorbed)−∑∆H(energy released)

So, this formula seems to be perfectly correct but my textbook and web pages are kind of not following this equation.
Please *delete me*

Offline Enthalpy

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Re: Enthalpy of Reaction
« Reply #1 on: March 22, 2018, 09:46:17 PM »
Hi Fatimasaboor123!

Did you notice that in the formula that doesn't mention bonds, ΔH is a standard heat of formation, not a bond energy?

Understanding the heat of reaction based on bonds is more direct, clearer, OK, fine for a first attempt. But it is inconvenient for practical uses, because
  • Forming a molecule from individual atoms releases a huge energy, both at the reactants and products, but the difference of both, which is the interesting heat of reaction, is much smaller. Computing from the bond energies would then be inaccurate.
  • It would need to know the energy necessary to atomize all elements, but this is poorly known. Even for carbon: vaporizing it is difficult, and what proportion of C, C2, C4 and others the gas contains is still debated. How to make a measure then?
  • The bond energy depends not only on the two atoms, but on their neighbours, the strain, and so on. Pretty unmanageable.

This has lend to measure and tabulate a different energy, more convenient and accurate: the standard enthalpy of formation. It tells the heat of reaction to form a compound from the elements in their standard state, that is, the state they take commonly at 1atm and 298K.

The energy to form a molecule from its atoms would be the sum of all bond energies (with signs possibly), but here "standard state" implies to form it from "molecules" of elements most often. Nitrogen is the gaseous molecule N2, carbon is solid graphite written only C but where the atoms belong to huge molecules, iron written just Fe involves the atoms in metallic bonds, and so on. This makes a big numerical change, as these molecular elements have released much energy when formed from atoms.

The standard heat of formation answers the drawbacks. Heats of reaction are measured, tabulated and computed for reactions from molecules to molecules, which avoids the differences of big and poorly known values.

A resulting subtlety is that heats of formation have a sign, as opposed to bond energies. Depending of whether the molecule has stronger or weaker bonds than its molecular elements, the heat of formation is negative or positive. In the example you chose:
H2 ΔH=0 because it's the element in its normal state
F2 ΔH=0 same reason
HF ΔH=-568kJ/mol (minus because the formation releases heat)
so the enthalpy for the reaction is computed as
H2+F2 :rarrow: 2HF
-0 -0 +2*(-568k) = 2*(-568k)
and it is negative.

My suggestion about the signs: it's necessary to understand them once, but only computers can manage them without errors. Humans think more reliably by wondering for each compound of a reaction whether to add or subtract, rather than applying formulas.

Offline dheepankarthik

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Re: Enthalpy of Reaction
« Reply #2 on: November 19, 2021, 04:46:34 AM »
Hello, fatimasaboor123.

I believe that :
  • An endothermic reaction is a reaction where heat is taken in.
  • An exothermic reaction is a reaction where heat is given out.
Whether heat is taken in or given out, can be easily identified.
This is how:
  • If the product side has a lower bond energy than the reactant side, it can be cognized that heat has been given out.
  • If it's the other way round, i.e. If the product side has a higher bond energy than the reactant side, it can be cognized that heat has been taken in.
This can be generalized as the formula: Enthalpy Change = Heat/Bond energy on product side - Heat/Bond energy on reactant side.

Note: The reactant side is the LHS of the reaction equation. The product side is the RHS of the equation.



In respect to our formula:
  • The answer will be positive if the Heat/Bond energy on the product side is higher, meaning "heat is taken in", meaning a endothermic reaction.
  • The answer will be negative if the Heat/Bond energy on the product side is lower, meaning "heat is given out", meaning a exothermic reaction.




Let us use this knowledge by solving a problem:


The equation is

H2 + F2  :rarrow: HF
The bond energies for this equations are:
  • F - F bond: 155 kilojoules per mole
  • H - H bond: 432 kilojoules per mole
  • F - H bond: 565 kilojoules per mole

Step 1: Let's balance the equation:

H2 + F2  :rarrow: HF
First, let's write down the no. of atoms of each element in both the LHS and RHS:

  • LHS:
    • H : 2
    • F: 2
  • RHS:
    • H : 1
    • F: 1
Let's add a coefficient of 2 to the HF molecule.

H2 + F2  :rarrow: 2HF

Again, let's write down the no. of atoms of each element in both the LHS and RHS:

  • LHS:
    • H : 2
    • F: 2
  • RHS:
    • H : 2
    • F: 2
The equation has been balanced.


Step 2: Let's replace the formulas of atoms with the bond structures.

H - H + F - F  :rarrow: H - F  H - F


Step 3: Let's calculate the bond energies for the LHS and RHS:

H - H + F - F  :rarrow: H - F  H - F
There is one H - H bond and one F - F bond on the LHS. There are two H - F bonds on RHS.

  • F - F bond: 155 kilojoules per mole
  • H - H bond: 432 kilojoules per mole
  • F - H bond: 565 kilojoules per mole

So, the bond energy on the LHS will be 155 + 432 kilojoules per mole (kJ/mol), which will be 587 kJ/mol.

And, the bond energy on the RHS will be 565 * 2 kJ/mol, which will be 1130 kJ/mol.

So we have the bond energies:
  • LHS: 587 kJ/mol
  • RHS: 1130 kJ/mol


Step 4: Use the formula.

Enthalpy Change = Heat/Bond energy on product side - Heat/Bond energy on reactant side.

When we use the equation:

Enthalpy change = 1130 kJ/mol - 587 kJ/mol = 543 kJ/mol.
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As said earlier, the answer will be positive if the Heat/Bond energy on the product side is higher, meaning "heat is taken in", meaning a endothermic reaction.

So, this reaction is a endothermic reaction i.e. it has taken in heat.



Thank you for reading this article cum blog. I hope this is informative. Here are the sources I used for gathering information.

https://www.bbc.co.uk/bitesize/guides/zwfr2nb/revision/5

https://www.wiredchemist.com/chemistry/data/bond_energies_lengths.html

Thank you. :)



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