Learn about bond enthalpies and how they can be used to calculate the enthalpy change for a reaction.
Log in josericardoodias 8 years agoPosted 8 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? • (9 votes) Amanda O'Hara 8 years agoPosted 8 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. (23 votes) kgrot127 6 years agoPosted 6 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? • (6 votes) contactmike110 6 years agoPosted 6 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). 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 :) Arushi 8 years agoPosted 8 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? • (5 votes) Andres Romero 8 years agoPosted 8 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). (3 votes) Radsiacobazzi 7 years agoPosted 7 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. • (2 votes) RogerP 7 years agoPosted 7 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. (5 votes) nidhi bharadwaj 8 years agoPosted 8 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? • tyersome 7 years agoPosted 7 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). (3 votes) Pankaj Yargal 4 years agoPosted 4 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? • (1 vote) Seongjoo 4 years agoPosted 4 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. (2 votes) William Shiuk a year agoPosted a year ago. Direct link to William Shiuk's post “I had a few questions:- ...” I had a few questions: • (1 vote) Richard a year agoPosted a year 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. (2 votes) Ritik 7 years agoPosted 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? • (1 vote) RogerP 7 years agoPosted 7 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. (2 votes) kooryan03 6 years agoPosted 6 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? • (1 vote) JANAK 9 months agoPosted 9 months ago. Direct link to JANAK's post “Many times enthalpy of th...” Many times enthalpy of the reaction is taken as the : (Enthalpy of formation of reactants)- (Enthalpy of formation of the products) For standard reactions it is (Enthalpy of Products -Enthalpy of reactants) • (1 vote) Jiya Gupta 7 months agoPosted 7 months ago. Direct link to Jiya Gupta's post “To calculate bond energy ...” To calculate bond energy of compound, the formula (Bond energy of Reactants) - (bond energy of products) is used, while to calculate enthalpy of formation = (enthalpy of formation of products - enthalpy of formation of reacts) is used (1 vote)Want to join the conversation?
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.
- 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?
Why it is take reverse ?
