If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

# Hess's law and reaction enthalpy change

Using Hess's Law and standard heats of formation to determine the enthalpy change for reactions. Created by Sal Khan.

## Want to join the conversation?

• why is the heat of formation of any element equals zero ??
• The definition of standard enthalpy of formation is the change in enthalpy when one mole of a substance is formed in its standard state from its constituent elements in their standard state at 298k and 1 atm.

Since elements like Oxygen are in their standard state, there is no standard enthalpy of formation to talk about here.
• In the video Sal uses a table to find the energy. Why do some of these formations absorb energy and some give them off?
• Some substances form more easily than others. For instance, it's not hard to form oxygen gas - gaseous O2 is in its natural state, so its heat of formation is zero. On the other hand, it takes a lot of energy to oxygen into a liquid, so its heat of formation is going to be very positive. Therefore, some of these reactions absorb a lot of energy because they're unlikely to occur under standard conditions, and other reactions give off a lot of energy out of a kind of "eagerness" to break their atomic bonds.
• how do you do a problem like this on a quiz or something when you are not given the heat of formation?
• you don't. or you just look it up in a table if youre allowed to use one
• At , Sal says "this side of the reaction has less heat than this side of the reaction." Heat is not a state variable, you can't have heat (Sal spent so much time riffing on this in the Internal Energy lectures). My question is, did I misunderstand that, or should Sal be saying "this side of the reaction has less ENERGY than this side of the reaction" (or something else)?
• Heat q is not a state variable, but enthalpy H is.
Enthalpy used to be called "heat content"; that's why it still has the symbol H.
For the purposes of Hess's Law, he is treating ΔH as if it were a reactant or product in the reaction, purely for convenience.
(1 vote)
• How come the hess cycle was not complete as in full triangular diagram with arrows from or to the reactants and the products.
• Well, it could, considering equilibrium. But it appears Sal wanted to focus on the reactants.

Hope this helps! :D
• If the reaction yields a negative energy then it is exothermic, if it is posative it is endothermic? What one is it heating up? The way im thinking of it is, it gains energy, so it must heat up. But it loses energy so the surrounding is heating up, therefor its heating up? I don't know if that makes sence but I cant seem to understand this...
• Exothermic means releasing heat. Everything around it, including the reactants themselves heat up. Endothermic means absorbing heat. Everything around it, including reactants, cools down. In all cases energy is being conserved, but it is changing form from thermal energy to energy bound in chemical bonds.
• How do we determine the route taken when calculating the enthalpy change?
• The route is irrelevant. For example, you can go from A to B to C to D but the "net" (the overall path) is from A to D.
In Chemistry, what you often do is take a compound and break it down to individual atoms first because you know the enthalpies for bond formations or otherwise enthalpies of individual reactions. Then you take these atoms and form the products made. You effectively do: Reactants -> Atoms -> Products, which is the equivalent of Reactants -> Products
• I am assuming that the reason for not including the negative sign in the final -2219.85 kJ/mol rxn is because the negative sing is simply expressing that this reaction is an exothermic reaction, verses a positive endothermic reaction. Am I thinking the correctly?
• At ? Yes you are thinking about this correct. The negative sign implys that the equation is exothermic and 2219kJ is "lost" from the equation, meaning the energy goes to another place in this case heat and light.
• Is it possible to calculate heat of formation?
• Most of the enthalpy of formation are calculated from enthalpy of reaction.
(Note the difference between enthalpy of Reaction and formation)
Enthalpy of Reaction= Enthalpy of Formation of product - Enthalpy of formation of reactant -eq1.
U'll can find Enthalpy Of Reaction by running that experiment in a calorimeter. and plug that value in the eq-1.
Now suppose someone already previously in different reaction had found enthalpy of formation for product. Now plug that value in the equation. Only unknown is now your reactant.