In Step 4, why is it required to add the C-C bond enthalpy? That bond was formed prior to the double bond, it doesn't disappear during the reaction. Or does the double-bond enthalpy account for both bonds instead of just the pi bond that is destroyed?

The double bond enthalpy counts for both bonds. Also remember that single, double, and triple bonds are not additive! E.g. the bond enthalpy of a single bond + single bond =/= double bond.

Why do fuels with more Carbon bonds, when combusted have a larger delta H value?

My answer is pretty long and based on my observation of calculating what happens when you burn Methane, Ethane and Propane. There is a tl;dr at the end of the answer. In Alkane series (saturated) hydrocarbons (CnH2n+2), every time you add a carbon to the chain you essentially add another C-C bond and 2 C-H bonds but you will also have to create another CO2 and H2O molecule (C02 comes from the C added and 02 used for combustion, and H2O comes from the 2 H added and 02 used for combustion). Now minding that the bond strength of C-H is 337kj/mol and C-C is 607 kj/mol you will have to add an additional 944kj/mol every time you add a Carbon to your chain to break the "fuel" compound. Now don't forget that everytime we add a Carbon we will need more oxygen for the combustion reaction, from what I saw trying to balance some combustions of Alkane fuels we will need an extra 1.5 O2 for every 1 Carbon we add to the chain. So every time we add a Carbon we need to break an extra 1.5 0=0 bond which is 1.5x498kj/mol=747kj. Now we know that for every C we add we need to add 944+747kj (1691kj) to break the bonds on the reactants side. Now remember that ever additional C yields us additional 1 CO2 and 1 H2O. Thats 2 C=0 bonds and 2 H-O bonds we need to create. The required energy to break a C=0 bond is 749kj/mol and the energy to break an H-O bond is 428kj/mol, so in order to form those bonds we have to add a - for each of those values. That means we will have to release 749x2 + 428x2 (2354kj/mol) of energy for every C we add to the chain. tl;dr: So for every carbon we add we get 1691kj to break bonds and then 2354kj is released to form bonds, so net Enthalpy change is -663kj every time you add an additional Carbon (and 2 Hydrogens). Hope this helps :)

The definition for bond enthalpy is the energy required to break 1 mole of bonds in gaseous covalent molecules under standard conditions? What are the standard conditions? High temperatures and low pressure?

standard temperature and pressure (STP) as a temperature of 273.15 K (0 °C, 32 °F) and an absolute pressure of exactly 100 000 Pa (1 bar, 14.504 psi, 0.98692 atm).

change in enthalpy during a reaction is mainly due to change in internal energy of the system which includes potential,rotational,translational KE as well. But in this case we are considering only the potential energy.So does that mean rotational,translational KE are not affected?

We are explicitly calculating *bond* enthalpy - to me this suggests that we are focusing on the potential energy. However, when we calculate an enthalpy of a reaction, we are often§ talking about how much 'heat' has been released (or absorbed) and this must be due to changes in the motion of the individual molecules. So, I think we could veiw these calculations as describing the interconversion between potential and kinetic energies of the molecules. §note: Clearly there are many exceptions such as: Chemiluminescence, where the energy is given off as light; and ATP powering molecular motors such as kinesins. However, in these cases we just replacing heat with other forms of light and (non-heat) kinetic energy, respectively (or more likely a mix of heat and these other forms of energy).

In Step 1 of the example, why would the Carbon = Carbon bond need to be broken if the resultant has a Carbon - Carbon bond? Wouldn't just one of those bonds need to break? Is there a chemical reason they both need to break & then later one re-forms? It just seems inefficient.

It is done this way as an accounting exercise. Three C-C bonds are broken and one C-C bond is formed - in others words, to do the calculation you assume that all relevant bonds are completely broken and then reformed. But this is _not_ to suggest that this is the mechanism by which the chemical reaction occurs - it certainly wouldn't go by this route. Neither, for that matter, would the H-H bond split to give two hydrogen radicals. Think of this as just a way of doing the calculation.

Is it mandatory to learn the bond enthalpy values for all chemical bonds?

It will depend on your school/teacher but usually, they won't expect you to memorize all of the bond enthalpy values. My teacher when I did AP Chemistry gave us a bond enthalpy value table whenever we had a test or quiz.

I had a few questions: - Can there be a reaction, which has ∆H = 0, so that the energy of reactants (e.g 500Kj) and products (e.g 500Kj) is equal, so that the enthalpy of reaction is zero (∆E = 500 - 500 = 0)? - I read that the enthalpy also depends on the state of the reactants, e.g when oxygen and hydrogen gases react to produce water vapor, the enthalpy change is -483.7 kJ. But, when the same reactants react to produce liquid water, the enthalpy change is -571.5 kJ. Why the difference?

Yes it is possible to have a reaction with an enthalpy change of 0. Which I suppose would make it neither exothermic or endothermic technically. Matter has different amounts of energy at different physical states. Think back to energy calculations for phase changes using Q = MCΔT. In general the gaseous version or a chemical has more energy than the liquid version of it. So producing liquid water versus gaseous water will release more energy because the liquid is at a lower energy level than the gas. Essentially that 87.8 kJ difference is how much more energy gaseous water has compared to liquid water. Hope that helps.

how to identify which carbon bond will break on reactant side and which carbon will form bond .Is it just the carbon with more bonds will break but how could we identify with which carbon bond will form on product?

This will become more obvious as you study organic chemistry. But a triple C-C bond is far more reactive than a single C-C bond and it loves to combine with hydrogen.

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Chemistry library

Course: Chemistry library > Unit 15

Lesson 2: Enthalpy

Bond enthalpies

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Learn about bond enthalpies and how they can be used to calculate the enthalpy change for a reaction.

Energy in chemical bonds

Chemical bonds represent potential energy. Quantifying the energy represented by the bonds in different molecules is an important part of understanding the overall energy implications of a reaction. In this article, we'll explore two different concepts that help describe that energy: enthalpy of reaction and bond enthalpy.

Enthalpy of reaction

During chemical reactions, the bonds between atoms may break, reform or both to either absorb or release energy. The result is a change to the potential energy of the system. The heat absorbed or released from a system under constant pressure is known as enthalpy, and the change in enthalpy that results from a chemical reaction is the enthalpy of reaction. The enthalpy of reaction is often written as delta, start text, H, end text, start subscript, start text, r, x, n, end text, end subscript.

To better understand enthalpy of reaction, let's consider the hydrogenation of propene, start text, C, end text, start subscript, 3, end subscript, start text, H, end text, start subscript, 6, end subscript, to form propane, start text, C, end text, start subscript, 3, end subscript, start text, H, end text, start subscript, 8, end subscript. In this reaction, propene gas reacts with hydrogen gas, start text, H, end text, start subscript, 2, end subscript, left parenthesis, g, right parenthesis, to form propane gas:

$~~~~~~~\text{C}_3 \text{H}_6(g)~~~~~~~~~~~~~~~~~~ ~~~~~~~~~~~~~~~~ \text H_2(g) ~~~~~~~~~~~~~~~~~~~~~~~~~~ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\text{C}_3 \text H_8 (g)$ space, space, space, space, space, space, space, start text, C, end text, start subscript, 3, end subscript, start text, H, end text, start subscript, 6, end subscript, left parenthesis, g, right parenthesis, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, start text, H, end text, start subscript, 2, end subscript, left parenthesis, g, right parenthesis, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, start text, C, end text, start subscript, 3, end subscript, start text, H, end text, start subscript, 8, end subscript, left parenthesis, g, right parenthesis

What is happening in this reaction? First we have to break the carbon start color #1fab54, start text, C, end text, equals, start text, C, end text, end color #1fab54 bond and the hydrogen start color #1fab54, start text, H, end text, minus, start text, H, end text, end color #1fab54 bond of the reactants. As a rule, breaking bonds between atoms requires adding energy. The stronger the bond, the more energy it takes to break the bond. To make the product propane, a new start color #aa87ff, start text, C, negative, C, end text, end color #aa87ff bond and two new start color #aa87ff, start text, C, negative, H, end text, end color #aa87ff bonds are then formed. Since breaking bonds requires adding energy, the opposite process of forming new bonds always releases energy. The stronger the bond formed, the more energy is released during the bond formation process. In this particular reaction, because the newly formed bonds release more energy than was needed to break the original bonds, the resulting system has a lower potential energy than the reactants. This means the enthalpy of reaction is negative.

Mathematically, we can think of the enthalpy of reaction as the difference between the potential energy from the product bonds and the potential energy of the reactant bonds:

$\begin{aligned}\Delta\text H_{\text{rxn}}&=\text{potential energy of product bonds}-\text{potential energy of reactant bonds}\\ \\ &=\text{energy added to break reactant bonds}+\text{energy released when making product bonds}\end{aligned}$

Reactions where the products have a lower potential energy than the reactants, such as the hydrogenation of propene described above, are exothermic. Reactions where the products have a higher potential energy than the reactants are endothermic.

In an exothermic reaction, the released energy doesn't simply disappear. Instead it is converted to kinetic energy, which produces heat. This is observed as an increase in temperature as the reaction progresses. On the other hand, endothermic reactions often require the addition of energy to favor the formation of products. In practice, this often means running a reaction at a higher temperature with a heat source.

In order to quantify the enthalpy of reaction for a given reaction, one approach is to use the standard enthalpies of formation for all of the molecules involved. These values describe the change in enthalpy to form a compound from its constituent elements. Subtracting the standard enthalpies of formation for the reactants from the standard enthalpies of the products approximates the enthalpy of reaction for the system. To learn more about enthalpies of formation (which are also called heats of formation) and how to use them to calculate the enthalpy of reaction, you can check out our video on standard heat of formation and the video on using heats of formation to calculate reaction enthalpies.

An alternative approach is to estimate the enthalpy of reaction by looking at the individual bonds involved. If we know how much energy we need to make and break each of the bonds, then we can add those values to find the enthalpy of reaction. We will discuss this method in more detail in the remainder of this article.

Bond enthalpy

Bond enthalpy (which is also known as bond-dissociation enthalpy, average bond energy, or bond strength) describes the amount of energy stored in a bond between atoms in a molecule. Specifically, it's the energy that needs to be added for the homolytic or symmetrical cleavage of a bond in the gas phase. A homolytic or symmetrical bond breaking event means that when the bond is broken, each atom that originally participated in the bond gets one electron and becomes a radical, as opposed to forming an ion.

Chemical bonds form because they're thermodynamically favorable, and breaking them inevitably requires adding energy. For this reason, bond enthalpy values are always positive, and they usually have units of start text, k, J, slash, m, o, l, end text or start text, k, c, a, l, slash, m, o, l, end text. The higher the bond enthalpy, the more energy is needed to break the bond and the stronger the bond. To determine how much energy will be released when we form a new bond rather than break it, we simply make the bond enthalpy value negative.

Because bond enthalpy values are so useful, average bond enthalpies for common bond types are readily available in reference tables. While in reality the actual energy change when forming and breaking bonds depends on neighboring atoms in a specific molecule, the average values available in the tables can still be used as an approximation.

Tip: The bond values listed in tables are for a mole of reaction for a single bond. This means that if there are multiples of the same bond breaking or forming in a reaction, you will need to multiply the bond enthalpy in your calculation by how many of that type of bond you have in the reaction. This also means it's important to make sure the equation is balanced and that the coefficients are written as the smallest possible integer values so the correct number of each bond is used.

Using bond enthalpies to estimate enthalpy of reaction

Once we understand bond enthalpies, we use them to estimate the enthalpy of reaction. To do this, we can use the following procedure:

Step 1. Identify which bonds in the reactants will break and find their bond enthalpies.

Step 2. Add up the bond enthalpy values for the broken bonds.

Step 3. Identify which new bonds form in the products and list their negative bond enthalpies. Remember we have to switch the sign for the bond enthalpy values to find the energy released when the bond forms.

Step 4. Add up the bond enthalpy values for the formed product bonds.

Step 5. Combine the total values for breaking bonds (from Step 2) and forming bonds (from Step 4) to get the enthalpy of reaction.

Example: Hydrogenation of propene

Let's find the enthalpy of reaction for the hydrogenation of propene, our example from the beginning of the article.

Step 1: Identify bonds broken

This reaction breaks one start text, C, end text, equals, start text, C, end text bond and one start text, H, end text, minus, start text, H, end text bond.

Using a reference table, we find that the bond enthalpy of a start text, C, end text, equals, start text, C, end text bond is 610, start text, k, J, slash, m, o, l, end text, while the bond enthalpy of a start text, H, end text, minus, start text, H, end text bond is 436, start text, k, J, slash, m, o, l, end text.

Step 2: Find total energy to break bonds

Combining the values from Step 1 gives us:

start color #1fab54, start text, E, n, e, r, g, y, space, a, d, d, e, d, space, t, o, space, b, r, e, a, k, space, b, o, n, d, s, end text, end color #1fab54, equals, 610, start text, k, J, slash, m, o, l, end text, plus, 436, start text, k, J, slash, m, o, l, end text, equals, 1046, start text, k, J, slash, m, o, l, end text

as the total energy required to break the necessary bonds in propene and hydrogen gas.

Step 3: Identify bonds formed

This reaction forms one new start text, C, end text, minus, start text, C, end text bond and two new start text, C, end text, minus, start text, H, end text bonds.

Using a reference table, we find that the bond enthalpy of a start text, C, end text, minus, start text, C, end text bond is 346, start text, k, J, slash, m, o, l, end text, while the bond enthalpy of a start text, C, end text, minus, start text, H, end text bond is 413, start text, k, J, slash, m, o, l, end text. To find how much energy is released when these bonds are formed, we'll need to multiply each bond enthalpy by minus, 1. Also, since two new start text, C, end text, minus, start text, H, end text bonds are formed, we'll need to multiply the start text, C, end text, minus, start text, H, end text bond enthalpy by 2.

Step 4: Find total energy released to form new bonds

Combining the values from Step 3 gives us:

start color #aa87ff, start text, E, n, e, r, g, y, space, r, e, l, e, a, s, e, d, space, t, o, space, m, a, k, e, space, p, r, o, d, u, c, t, space, b, o, n, d, s, end text, end color #aa87ff, equals, minus, 346, start text, k, J, slash, m, o, l, end text, plus, left parenthesis, 2, times, minus, start text, 413, end text, start text, k, J, slash, m, o, l, end text, right parenthesis, equals, minus, 1172, start text, k, J, slash, m, o, l, end text

for the total energy that will be released by forming the new bonds.

Step 5: Add up energy for bonds broken and formed

From Step 2 and Step 4, we have 1046, start text, k, J, end text of energy required to break bonds and minus, 1172, start text, k, J, end text of energy released from forming bonds. Combining these values, we get for the enthalpy of reaction:

$\begin{aligned}\Delta\text H_{\text{rxn}}&=\text{energy added to break reactant bonds}+\text{energy released when making product bonds}\\ \\ &=1046\,\text{kJ/mol}+(-1172\,\text{kJ/mol})\\ \\ &=-126\,\text{kJ/mol}\end{aligned}$

Since the enthalpy of reaction for the hydrogenation of propene is negative, we know that the reaction is exothermic.

Summary

Bond enthalpy and enthalpy of reaction help us understand how a chemical system uses energy during reactions. The bond enthalpy describes how much energy is needed to break or form a bond, and it is also a measure of bond strength. By combining the bond enthalpy values for all of the bonds broken and formed during a reaction, it's possible to estimate the total change in potential energy of the system, which is delta, start text, H, end text, start subscript, start text, r, x, n, end text, end subscript for a reaction at constant pressure. Depending on whether the enthalpy of reaction is positive or negative, we can determine whether a reaction will be endothermic or exothermic.

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josericardoodias
Posted 7 years ago. Direct link to josericardoodias's post “In Step 4, why is it requ...”
In Step 4, why is it required to add the C-C bond enthalpy? That bond was formed prior to the double bond, it doesn't disappear during the reaction. Or does the double-bond enthalpy account for both bonds instead of just the pi bond that is destroyed?
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- Amanda O'Hara
  Posted 7 years ago. Direct link to Amanda O'Hara's post “The double bond enthalpy ...”
  The double bond enthalpy counts for both bonds. Also remember that single, double, and triple bonds are not additive! E.g. the bond enthalpy of a single bond + single bond =/= double bond.
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kgrot127
Posted 5 years ago. Direct link to kgrot127's post “Why do fuels with more Ca...”
Why do fuels with more Carbon bonds, when combusted have a larger delta H value?
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- contactmike110
  Posted 5 years ago. Direct link to contactmike110's post “My answer is pretty long ...”
  My answer is pretty long and based on my observation of calculating what happens when you burn Methane, Ethane and Propane. There is a tl;dr at the end of the answer.
  
  In Alkane series (saturated) hydrocarbons (CnH2n+2), every time you add a carbon to the chain you essentially add another C-C bond and 2 C-H bonds but you will also have to create another CO2 and H2O molecule (C02 comes from the C added and 02 used for combustion, and H2O comes from the 2 H added and 02 used for combustion).
  Now minding that the bond strength of C-H is 337kj/mol and C-C is 607 kj/mol you will have to add an additional 944kj/mol every time you add a Carbon to your chain to break the "fuel" compound.
  Now don't forget that everytime we add a Carbon we will need more oxygen for the combustion reaction, from what I saw trying to balance some combustions of Alkane fuels we will need an extra 1.5 O2 for every 1 Carbon we add to the chain.
  So every time we add a Carbon we need to break an extra 1.5 0=0 bond which is 1.5x498kj/mol=747kj.
  Now we know that for every C we add we need to add 944+747kj (1691kj) to break the bonds on the reactants side.
  Now remember that ever additional C yields us additional 1 CO2 and 1 H2O.
  Thats 2 C=0 bonds and 2 H-O bonds we need to create.
  The required energy to break a C=0 bond is 749kj/mol and the energy to break an H-O bond is 428kj/mol, so in order to form those bonds we have to add a - for each of those values.
  That means we will have to release 749x2 + 428x2 (2354kj/mol) of energy for every C we add to the chain.
  
  tl;dr: So for every carbon we add we get 1691kj to break bonds and then 2354kj is released to form bonds, so net Enthalpy change is -663kj every time you add an additional Carbon (and 2 Hydrogens).
  
  Hope this helps :)
  Comment on contactmike110's post “My answer is pretty long ...”
  (11 votes)
Arushi
Posted 7 years ago. Direct link to Arushi's post “The definition for bond e...”
The definition for bond enthalpy is the energy required to break 1 mole of bonds in gaseous covalent molecules under standard conditions? What are the standard conditions? High temperatures and low pressure?
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- Andres Romero
  Posted 7 years ago. Direct link to Andres Romero's post “standard temperature and ...”
  standard temperature and pressure (STP) as a temperature of 273.15 K (0 °C, 32 °F) and an absolute pressure of exactly 100 000 Pa (1 bar, 14.504 psi, 0.98692 atm).
  Comment on Andres Romero's post “standard temperature and ...”
  (3 votes)
Radsiacobazzi
Posted 6 years ago. Direct link to Radsiacobazzi's post “In Step 1 of the example,...”
In Step 1 of the example, why would the Carbon = Carbon bond need to be broken if the resultant has a Carbon - Carbon bond? Wouldn't just one of those bonds need to break? Is there a chemical reason they both need to break & then later one re-forms? It just seems inefficient.
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- RogerP
  Posted 6 years ago. Direct link to RogerP's post “It is done this way as an...”
  It is done this way as an accounting exercise. Three C-C bonds are broken and one C-C bond is formed - in others words, to do the calculation you assume that all relevant bonds are completely broken and then reformed. But this is not to suggest that this is the mechanism by which the chemical reaction occurs - it certainly wouldn't go by this route. Neither, for that matter, would the H-H bond split to give two hydrogen radicals.
  
  Think of this as just a way of doing the calculation.
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nidhi bharadwaj
Posted 7 years ago. Direct link to nidhi bharadwaj's post “change in enthalpy during...”
change in enthalpy during a reaction is mainly due to change in internal energy of the system which includes potential,rotational,translational KE as well. But in this case we are considering only the potential energy.So does that mean rotational,translational KE are not affected?
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(3 votes)
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- tyersome
  Posted 6 years ago. Direct link to tyersome's post “We are explicitly calcula...”
  We are explicitly calculating bond enthalpy - to me this suggests that we are focusing on the potential energy.
  
  However, when we calculate an enthalpy of a reaction, we are often§ talking about how much 'heat' has been released (or absorbed) and this must be due to changes in the motion of the individual molecules.
  
  So, I think we could veiw these calculations as describing the interconversion between potential and kinetic energies of the molecules.
  
  §note: Clearly there are many exceptions such as: Chemiluminescence, where the energy is given off as light; and ATP powering molecular motors such as kinesins. However, in these cases we just replacing heat with other forms of light and (non-heat) kinetic energy, respectively (or more likely a mix of heat and these other forms of energy).
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Pankaj Yargal
Posted 3 years ago. Direct link to Pankaj Yargal's post “Is it mandatory to learn ...”
Is it mandatory to learn the bond enthalpy values for all chemical bonds?
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- Seongjoo
  Posted 3 years ago. Direct link to Seongjoo's post “It will depend on your sc...”
  It will depend on your school/teacher but usually, they won't expect you to memorize all of the bond enthalpy values. My teacher when I did AP Chemistry gave us a bond enthalpy value table whenever we had a test or quiz.
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William Shiuk
Posted 7 months ago. Direct link to William Shiuk's post “I had a few questions: - ...”
I had a few questions:
- Can there be a reaction, which has ∆H = 0, so that the energy of reactants (e.g 500Kj) and products (e.g 500Kj) is equal, so that the enthalpy of reaction is zero (∆E = 500 - 500 = 0)?
- I read that the enthalpy also depends on the state of the reactants, e.g when oxygen and hydrogen gases react to produce water vapor, the enthalpy change is -483.7 kJ. But, when the same reactants react to produce liquid water, the enthalpy change is -571.5 kJ. Why the difference?
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- Richard
  Posted 6 months ago. Direct link to Richard's post “Yes it is possible to hav...”
  Yes it is possible to have a reaction with an enthalpy change of 0. Which I suppose would make it neither exothermic or endothermic technically.
  
  Matter has different amounts of energy at different physical states. Think back to energy calculations for phase changes using Q = MCΔT. In general the gaseous version or a chemical has more energy than the liquid version of it. So producing liquid water versus gaseous water will release more energy because the liquid is at a lower energy level than the gas. Essentially that 87.8 kJ difference is how much more energy gaseous water has compared to liquid water.
  
  Hope that helps.
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  (2 votes)
Ritik
Posted 7 years ago. Direct link to Ritik's post “how to identify which car...”
how to identify which carbon bond will break on reactant side and which carbon will form bond .Is it just the carbon with more bonds will break but how could we identify with which carbon bond will form on product?
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- RogerP
  Posted 6 years ago. Direct link to RogerP's post “This will become more obv...”
  This will become more obvious as you study organic chemistry. But a triple C-C bond is far more reactive than a single C-C bond and it loves to combine with hydrogen.
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kooryan03
Posted 5 years ago. Direct link to kooryan03's post “At Step 3, when we break ...”
At Step 3, when we break the carbon double bond, are we then sharing the electrons from one of the bonds from the double bond broken, with the new hydrogens to form new C—H bonds?
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eli
Posted 4 years ago. Direct link to eli's post “what are the conditions t...”
what are the conditions that are required in order to obtain an accurate value of Hrxn using the bond energy method of calculations?

Is it the multiplying of the number of bonds and the addition of product and reactant.
Or the use of the right equation
delta Hrxn = sum of BE (reactant) - sum of BE (product)

can the wrong state of the element also affect the answer
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